Question 12 Marks
If $\theta$ is the angle between two vectors $\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\ \text{and}\ 3\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}},$ find $\sin\theta.$
Answer$\vec{\text{a}}\cdot\vec{\text{b}}=|\vec{\text{a}}||\vec{\text{b}}|\cos\theta$
$\vec{\text{a}}\cdot\vec{\text{b}}=(\hat{\text{i}}-2\hat{\text{j}}+3\vec{\text{k}})\cdot(3\hat{\text{i}}-2\hat{\text{i}}+\hat{\text{k}})$
$=3+4+3$
$=10$
$10=\big(\sqrt{1+4+9}\big)\big(\sqrt{9+4+1}\big)\cos\theta$
$\cos\theta=\frac{10}{\sqrt{14}\sqrt{14}}$
$\cos\theta=\frac{10}{14}$
$\sin\theta=\sqrt{1-\cos^2\theta}$
$=\sqrt{{1-\frac{25}{49}}}=\sqrt{\frac{49-25}{49}}$
$=\frac{\sqrt{24}}{7}=\frac{2\sqrt{6}}{7}$
View full question & answer→Question 22 Marks
If $\theta$ is the angle between two vectors $\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\ \text{and}\ 3\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}},$ find $\sin\theta.$
Answer$\vec{\text{a}}\cdot\vec{\text{b}}=|\vec{\text{a}}||\vec{\text{b}}|\cos\theta$
$\vec{\text{a}}\cdot\vec{\text{b}}=(\hat{\text{i}}-2\hat{\text{j}}+3\vec{\text{k}})\cdot(3\hat{\text{i}}-2\hat{\text{i}}+\hat{\text{k}})$
$=3+4+3$
$=10$
$10=\big(\sqrt{1+4+9}\big)\big(\sqrt{9+4+1}\big)\cos\theta$
$\cos\theta=\frac{10}{\sqrt{14}\sqrt{14}}$
$\cos\theta=\frac{10}{14}$
$\sin\theta=\sqrt{1-\cos^2\theta}$
$=\sqrt{{1-\frac{25}{49}}}=\sqrt{\frac{49-25}{49}}$
$=\frac{\sqrt{24}}{7}=\frac{2\sqrt{6}}{7}$
View full question & answer→Question 32 Marks
If $\theta$ is the angle between two vectors $\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\ \text{and}\ 3\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}},$ find $\sin\theta.$
Answer$\vec{\text{a}}\cdot\vec{\text{b}}=|\vec{\text{a}}||\vec{\text{b}}|\cos\theta$
$\vec{\text{a}}\cdot\vec{\text{b}}=(\hat{\text{i}}-2\hat{\text{j}}+3\vec{\text{k}})\cdot(3\hat{\text{i}}-2\hat{\text{i}}+\hat{\text{k}})$
$=3+4+3$
$=10$
$10=\big(\sqrt{1+4+9}\big)\big(\sqrt{9+4+1}\big)\cos\theta$
$\cos\theta=\frac{10}{\sqrt{14}\sqrt{14}}$
$\cos\theta=\frac{10}{14}$
$\sin\theta=\sqrt{1-\cos^2\theta}$
$=\sqrt{{1-\frac{25}{49}}}=\sqrt{\frac{49-25}{49}}$
$=\frac{\sqrt{24}}{7}=\frac{2\sqrt{6}}{7}$
View full question & answer→Question 42 Marks
If $|\vec{\text{a}}|=2,\big|\vec{\text{b}}\big|=3$ and $\vec{\text{a}}.\vec{\text{b}}=3,$ find the projection of $\vec{\text{b}}$ on $\vec{\text{a}}.$
AnswerWe have
$|\vec{\text{a}}|=2$ and $\vec{\text{a}}.\vec{\text{b}}=3$
So, the projection of $\vec{\text{b}}$ on $\vec{\text{a}}$ is
$\Big(\frac{\vec{\text{a}}.\vec{\text{b}}}{|\vec{\text{a}}|}\Big)$
$=\frac{3}{2}$
View full question & answer→Question 52 Marks
Represent the following graphically:
- A displacement of 40km, 30º east of north.
- A displacement of 50km south-east.
- A displacement of 70km, 40º north of west.
Answer
- The vector $\overrightarrow{\text{OP}}$ represents the required displacement vector.
- The vector $\overrightarrow{\text{OQ}}$ represents the required vector.
- The vector $\overrightarrow{\text{OR}}$ represents the required vector.
View full question & answer→Question 62 Marks
Find the angle at which the following vectors are inclined to each of the coordinate axes:
$4\hat{\text{i}}+8\hat{\text{j}}+\hat{\text{k}}$
AnswerLet $\vec{\text{r}}$ be the given vector, and let it make an angle $\alpha,\beta,\gamma$ with OX, OY, OZ respectively. Then, its direction cosines are $\cos\alpha,\cos\beta,\cos\gamma$.So direction ratios of $\vec{\text{r}}=4\hat{\text{i}}+8\hat{\text{j}}+\hat{\text{k}}$ are proportional to 4, 8, 1. Therefore,
Direction cosine of $\vec{\text{r}}$ are $\frac{4}{\sqrt{4^2+8^2+1^2}},\frac{8}{\sqrt{4^2+8^2+1^2}},\frac{1}{\sqrt{4^2+8^2+1^2}}$ or $\frac{4}{9},\frac{8}{9},\frac{1}{9}.$
$\therefore\alpha=\cos^{-1}=\Big(\frac{4}{9}\Big),\beta=\cos^{-1}=\Big(\frac{8}{9}\Big),\gamma=\cos^{-1}=\Big(\frac{1}{9}\Big)$.
View full question & answer→Question 72 Marks
Find $\vec{\text{a}}.\vec{\text{b}}$ when
$\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=4\hat{\text{i}}-4\hat{\text{j}}+7\hat{\text{k}}$
Answer$\vec{\text{a}}.\vec{\text{b}}$
$=(\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}).(4\hat{\text{i}}-4\hat{\text{j}}+7\hat{\text{k}})$
$=(1)(4)+(-2).(-4)+(1)(7)$
$=4+8+7$
$=19$
$\vec{\text{a}}.\vec{\text{b}}=19$
View full question & answer→Question 82 Marks
For any two vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ write the value of $\big(\vec{\text{a}}.\vec{\text{b}}\big)^2+\big|\vec{\text{a}}\times\vec{\text{b}}\big|^2$ in terms of their magnitudes.
Answer$\big(\vec{\text{a}}.\vec{\text{b}}\big)^2+\big|\vec{\text{a}}\times\vec{\text{b}}\big|^2$
$=\big(|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta\big)^2+\big(|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta\big)^2$
$=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2$ $\big(\cos^2\theta+\sin^2\theta\big)$
$=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2$ (1)
$=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2$
View full question & answer→Question 92 Marks
Write the projection of the vector $7\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}}$ on the vector $2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}.$
AnswerLet $\vec{\text{a}}=7\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}};\vec{\text{b}}=2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}$
The projection of $\vec{\text{a}}$ on $\vec{\text{b}}$ is
$\Bigg(\frac{\vec{\text{a}}\vec{\text{b}}}{\big|\vec{\text{b}}\big|}\Bigg)$
$=\frac{\big(7\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}}\big).\big(2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}\big)}{\big|2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}\big|}$
$=\frac{14+6-12}{\sqrt{4+36+9}}$
$=\frac{8}{7}$
View full question & answer→Question 102 Marks
Find $\big|\vec{\text{a}}-\vec{\text{b}}\big|$ if
$|\vec{\text{a}}|=3,\big|\vec{\text{b}}\big|=4$ and $\vec{\text{a}}.\vec{\text{b}}=1$
AnswerGiven that$|\vec{\text{a}}|=3,\big|\vec{\text{b}}\big|=4$ and $\vec{\text{a}}.\vec{\text{b}}=1\dots(1)$
We know that
$\big|\vec{\text{a}}-\vec{\text{b}}\big|^2=|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-2\vec{\text{a}}.\vec{\text{b}}$
$=3^2+4^2-2(1)$ [using (1)]
$=9+16-2$
$=23$
$\therefore \big|\vec{\text{a}}-\vec{\text{b}}\big|=\sqrt{23}$
View full question & answer→Question 112 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are two vectors of the same magnitude inclined at an angle of 60° such that $\vec{\text{a}}.\vec{\text{b}}=8,$
write the value of their magnitude.
AnswerGiven that
$|\vec{\text{a}}|=\big|\vec{\text{b}}\big|$
and $\vec{\text{a}}$ and $\vec{\text{b}}$ are inclined at an angle of 60°
Also, given that
$\vec{\text{a}}.\vec{\text{b}}=8$
$\Rightarrow|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos60^\circ=8$
$\Rightarrow|\vec{\text{a}}||\vec{\text{a}}|\big(\frac{1}{2}\big)=8$
$\Rightarrow|\vec{\text{a}}|^2=16$
$\Rightarrow|\vec{\text{a}}|=4$
View full question & answer→Question 122 Marks
Find the sum of the vectors $\vec{a}=\hat{i}-2\hat{j}+\hat{k,}\ \ \vec{b}=-2\hat{i}+4\hat{j}+5\hat{k}\ \text{and}\ \vec{c}= \hat{i}-6\hat{j}-7\hat{k}.$
AnswerThe given vectors are $\vec{a}=\hat{i}-2\hat{j}+\hat{k,}\ \ \vec{b}=-2\hat{i}+4\hat{j}+5\hat{k}\ \text{and}\ \vec{c}= \hat{i}-6\hat{j}-7\hat{k}$
$\therefore\vec{a}+\vec{b}+\vec{c}=(1-2+1)\hat{i}+(-2+4-6)\hat{j}+(1+5-7)\hat{k}$
$=0\cdot\hat{i}-4\hat{j}-1\cdot\hat{k}$
$=-4\hat{j}-\hat{k}$
View full question & answer→Question 132 Marks
Find the sum of the following vectors: $\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}},\vec{\text{b}}=2\hat{\text{i}}-3\hat{\text{j}},\vec{\text{c}}=2\hat{\text{i}}-3\hat{\text{k}}$.
AnswerGiven: $\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}},\vec{\text{b}}=2\hat{\text{i}}-3\hat{\text{j}},\vec{\text{c}}=2\hat{\text{i}}-3\hat{\text{k}}$So, Sum of the three vectors$=\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{i}}+3\hat{\text{k}}$
$=5\hat{\text{i}}-5\hat{\text{j}}+3\hat{\text{k}}$
View full question & answer→Question 142 Marks
$\text{If}\ \vec{a}\cdot\vec{a}=0\ \text{and}\ \vec{a}\cdot\vec{b}=0,$ then what can be concluded about the vector $\vec{b}?$
Answer$\text{Given:}\ \ \vec{a}.\vec{a}=0\Rightarrow\ \ \big|\vec{a}\big|^2=0$ $ \ \Rightarrow\ \ \ \ \big|\vec{a}\big|=0$ $\text{Again}\ \ \vec{a}.\vec{b}=0\ \Rightarrow\ \ \big|\vec{a}\big|.\Big|\vec{b}\Big|\text{cos}\ \theta=0$ $\ \Rightarrow\ \ 0.\Big|\vec{b}\Big|\text{cos}\theta=0\ \Big[\because\big|\vec{a}\big|=0\Big]$ $\Rightarrow\ $ 0 = 0 for all (any vector $\vec{b}.$)Therefore, $\vec{b}$ can be any vector.
View full question & answer→Question 152 Marks
Find the position vector of the mid-point of the vector joining the points P(2, 3, 4) and Q(4, 1, -2).
AnswerThe position vector of mid-point R of the vector joining points P(2, 3, 4) and Q(4, 1, -2) is given by,
$\overrightarrow{\text{OR}}=\frac{\big(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}\big)+\big(4\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}\big)}{2}$
$=\frac{(2+4)\hat{\text{i}}+(3+1)\hat{\text{j}}+(4-2)\hat{\text{k}}}{2}$
$=\frac{6\hat{\text{i}}+4\hat{\text{j}}+2\hat{\text{k}}}{2}$
$=3\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$
View full question & answer→Question 162 Marks
Find the value of $\theta\in(0,\frac{\pi}{2})$ for which vectors $\vec{\text{a}}=(\sin\theta)\hat{\text{i}}+(\cos\theta)\hat{\text{j}}$ and $\vec{\text{b}}=\hat{\text{i}}-\sqrt{3}\hat{\text{j}}+2\hat{\text{k}}$ are perpendicular.
AnswerWe have
$\vec{\text{a}}=(\sin\theta)\hat{\text{i}}+(\cos\theta)\hat{\text{j}}$
and
$\vec{\text{b}}=\hat{\text{i}}-\sqrt{3}\hat{\text{j}}+2\hat{\text{k}}$
It is given that the vectors are perpendicular.
$\Rightarrow\vec{\text{a}}.\vec{\text{b}}=0$
$\Rightarrow\sin\theta-\sqrt{3}\cos\theta=0$
$\Rightarrow\sin\theta=\sqrt{3}\cos\theta$
$\Rightarrow\tan\theta=\sqrt{3}$
$\Rightarrow\theta=\frac{\pi}{3}$
View full question & answer→Question 172 Marks
For what value of 'a' the vectors $2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$ and $\text{a}\hat{\text{i}}+6\hat{\text{j}}-8\hat{\text{k}}$ are collinear?
AnswerGiven: Two vectors, let $\vec{\text{p}}=2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}$ and $\vec{\text{q}}=\text{a}\hat{\text{i}}+6\hat{\text{j}}-8\hat{\text{k}}$
Since the given vectors are collinear, we have,
$\vec{\text{p}}=\lambda\vec{\text{q}}$
$\Rightarrow\ 2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}=\lambda\big(\text{a}\hat{\text{i}}+6\hat{\text{j}}-8\hat{\text{k}}\big)$
$\Rightarrow\ 2\hat{\text{i}}-3\hat{\text{j}}+4\hat{\text{k}}=\text{a}\lambda\hat{\text{i}}+6\lambda\hat{\text{j}}-8\lambda\hat{\text{k}}$
$\Rightarrow\ \lambda\text{a}=2,6\lambda=-3$ and $-8\lambda=4$
$\Rightarrow\ \lambda=-\frac{1}2$ and $\text{a}= -4$
View full question & answer→Question 182 Marks
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are the position vectors of the vertices of an equilateral triangle whose orthocentre is at the origin, then write the values of $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}$.
AnswerLet, ABC be a given equilateral ttriangle and its vertices are $\text{A}(\vec{\text{a}}),\text{B}(\vec{\text{b}})$ and $\text{C}(\vec{\text{c}})$.
Also, $\text{O}(\vec{0})$ be the orthocentre of trianglre ABC.
We know that centroid and orthocentre of equilateral triangle coincide at one point.
Orthocentre of $\triangle\text{ABC}=\vec0$
$\Rightarrow$ Centroid $\triangle\text{ABC}=\vec0$
$\Rightarrow\ \frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3=\vec0$
$\therefore\ \vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=\vec0$
View full question & answer→Question 192 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are vectors of equal magnitude, write the value of $\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big).$
AnswerWE have
$|\vec{\text{a}}|=\big|\vec{\text{b}}\big|\dots(1)$
Now,
$\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)$
$=|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2$
$=|\vec{\text{a}}|^2-|\vec{\text{a}}|^2$ [using (1)]
$=0$
View full question & answer→Question 202 Marks
Find the value of $\lambda$ is the vectors $2\hat{\text{i}}+\lambda\hat{\text{j}}+3\hat{\text{k}}$ and $3\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}$ are perpendicular to each other.
AnswerGiven: $2\hat{\text{i}}+\lambda\hat{\text{j}}+3\hat{\text{k}}$ and $3\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}$ are perpendicular to each other.
So, their dot product is zero.
$\big(2\hat{\text{i}}+\lambda\hat{\text{j}}+3\hat{\text{k}}\big).(3\text{i}+2\text{j}-4\text{k})$
$\Rightarrow6+2\lambda-12=0$
$\Rightarrow2\lambda-6=0$
$\Rightarrow\lambda=3$
View full question & answer→Question 212 Marks
Write the direction cosines of the vector $\vec{\text{r}}=6\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$.
AnswerGiven: $\vec{\text{r}}=6\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$
Then, direction cosines of $\hat{\text{r}}$ are $\frac{6}{\sqrt{6^2+(-2)^2+3^2}},\frac{-2}{\sqrt{6^2+(-2)^2+3^2}},\frac{3}{\sqrt{6^2+(-2)^2+3^2}}$ or, $\frac{6}7,\frac{-2}7,\frac{3}7$
View full question & answer→Question 222 Marks
If $\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}$ and $\vec{\text{b}}=2\hat{\text{i}}+4\hat{\text{j}}+9\hat{\text{k}}$, find a unit vector parallel to $\vec{\text{a}}+\vec{\text{b}}$.
AnswerGiven:$\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}},\ \vec{\text{b}}=2\hat{\text{i}}+4\hat{\text{j}}+9\hat{\text{k}}$
Now, $\vec{\text{a}}+\vec{\text{b}}=3\hat{\text{i}}+6\hat{\text{j}}+6\hat{\text{k}}$
$\big|\vec{\text{a}}+\vec{\text{b}}\big|=\sqrt{3^2+6^2+6^2}$
$=\sqrt{9+36+36}$ $=\sqrt{81}$ $=9$Unit vector parallel to $\vec{\text{a}}+\vec{\text{b}}=\frac{\vec{\text{a}}+\vec{\text{b}}}{\big|\vec{\text{a}}+\vec{\text{b}}\big|}=\frac{3\hat{\text{i}}+6\hat{\text{j}}+6\hat{\text{k}}}{9}$
$=\frac{1}9\times3\big(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big)=\frac{1}3\big(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big)$
View full question & answer→Question 232 Marks
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are non-coplanar vectors, then find the value of $\frac{\vec{\text{a}}.\big(\vec{\text{b}}\times\vec{\text{c}}\big)}{\big(\vec{\text{c}}\times\vec{\text{a}}\big).\vec{\text{b}}}+\frac{\vec{\text{b}}.\big(\vec{\text{a}}\times\vec{\text{c}}\big)}{\vec{\text{c}}.\big(\vec{\text{a}}\times\vec{\text{b}}\big)}$
View full question & answer→Question 242 Marks
If $\hat{\text{a}},\hat{\text{b}}$ are unit vectors such that $\hat{\text{a}}+\hat{\text{b}}$ is a unit vector, write the value of $\big|\hat{\text{a}}-\hat{\text{b}}\big|.$
AnswerGiven that $\hat{\text{a}}$ and $\hat{\text{b}}$ are unit vectors such that $\hat{\text{a}}+\hat{\text{b}}$ is a unit vector.
$\Rightarrow|\hat{\text{a}}|=\big|\hat{\text{b}}\big|=\big|\hat{\text{a}}+\hat{\text{b}}\big|=1\dots(1)$
Now,
$\big|\hat{\text{a}}+\vec{\text{b}}\big|=1$
Squaring both sides, we get
$|\vec{\text{a}}|^2+\big|\hat{\text{b}}\big|^2+2\hat{\text{a}}.\hat{\text{b}}=1$
$\Rightarrow1+1+2\hat{\text{a}}.\hat{\text{b}}=1$ [Form (1)]
$\Rightarrow\hat{\text{a}}.\hat{\text{b}}=\frac{-1}{2}\dots(2)$
Now,
$\big|\hat{\text{a}}-\hat{\text{b}}\big|^2=|\text{a}|^2+\big|\hat{\text{b}}\big|^2-2\hat{\text{a}}.\hat{\text{b}}$
$=1+1-2\big(\frac{-1}{2}\big)=3$ [From (1) and (2)]
$\therefore\big|\hat{\text{a}}-\hat{\text{b}}\big|=\sqrt{3}$
View full question & answer→Question 252 Marks
Find the angle between the vectors $\vec{\text{a}} $ and $\vec{\text{b}},$ where$\vec{\text{a}}=2\hat {\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$ and $\vec{\text{b}} =4\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}$
AnswerLet $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$.
$\big|\vec{\text{a}}\big|=\sqrt{(2)^2+(-1)^2+(2)^{2}}=\sqrt{9}=3$
$\big|\vec{\text{b}}\big|=\sqrt{(4)^2+(4)^2+(-2)^{2}}=\sqrt{36}=6$
$\vec{\text{a}}.\vec{\text{b}}=8-4-4=0$
$\cos\theta=\frac{\vec{\text{a }}.\vec{\text{ b}}}{\big|\vec{a}\big|\big|\vec{b}\big|}=\frac{0}{(3)(6)}=0$
$\Rightarrow\theta=\cos^{-1}(0)=\frac{\pi}{2}$
View full question & answer→Question 262 Marks
If $\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}$ and $\vec{\text{b}}=-\hat{\text{j}}+\hat{\text{k}},$ find the projection of $\vec{\text{a}}$ on $\vec{\text{b}}.$
AnswerWe have
$\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}$ and $\vec{\text{b}}=-\hat{\text{j}}+\hat{\text{k}}$
The projection of $\vec{\text{a}}$ on $\vec{\text{b}}$ is
$\frac{\vec{\text{a}}.\vec{\text{b}}}{\big|\vec{\text{b}}\big|}$
$=\frac{(\hat{\text{i}}-\hat{\text{j}}).(-\hat{\text{j}}+\hat{\text{k}})}{\big|-\hat{\text{j}}+\hat{\text{k}}\big|}$
$=\frac{0+1+0}{\sqrt{1+1}}$
$=\frac{1}{\sqrt{2}}$
View full question & answer→Question 272 Marks
For any two vectors $\vec{\text{a}}$ and $\vec{\text{b}},$ find $\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{b}}.$
AnswerLet:$\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$
$\vec{\text{b}}=\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{a}_1&\text{a}_2&\text{a}_3\\\text{b}_1&\text{b}_2&\text{b}_3\end{vmatrix}$
$=\hat{\text{i}}(\text{a}_2\text{b}_3-\text{a}_3\text{b}_2)-\hat{\text{j}}(\text{a}_1\text{b}_3-\text{a}_3\text{b}_1)+\hat{\text{k}}(\text{a}_1\text{b}_2-\text{a}_2\text{b}_1)$
$\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{b}}$
$=\big[\hat{\text{i}}(\text{a}_2\text{b}_3-\text{a}_3\text{b}_2)-\hat{\text{j}}(\text{a}_1\text{b}_3-\text{a}_3\text{b}_1)+\hat{\text{k}}(\text{a}_1\text{b}_2-\text{a}_2\text{b}_1)\big]\\.\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$
$=\text{a}_2\text{b}_1\text{b}_3-\text{a}_3\text{b}_1\text{b}_2-\text{a}_1\text{b}_2\text{b}_3+\text{a}_3\text{b}_1\text{b}_2+\text{a}_1\text{b}_2\text{b}_3-\text{a}_2\text{b}_1\text{b}_3$
$=\text{b}_1(\text{a}_2\text{b}_3-\text{a}_3\text{b}_2)-\text{b}_2(\text{a}_1\text{b}_3-\text{a}_3\text{b}_1)+\text{b}_3(\text{a}_1\text{b}_2-\text{a}_2\text{b}_1)$
$=\text{a}_2\text{b}_1\text{b}_3-\text{a}_3\text{b}_1\text{b}_2-\text{a}_1\text{b}_2\text{b}_3+\text{a}_3\text{b}_1\text{b}_2+\text{a}_1\text{b}_2\text{b}_3-\text{a}_2\text{b}_1\text{b}_3$
$=0$
View full question & answer→Question 282 Marks
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ represent the sides of a triangle taken in order, then write the value of $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}$.
AnswerLet ABC be a triangle such that $\overrightarrow{\text{BC}}=\vec{\text{a}},\ \overrightarrow{\text{CA}}=\vec{\text{b}},\ \overrightarrow{\text{AB}}=\vec{\text{c}}$. Then,$\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=\overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}+\overrightarrow{\text{AB}}$
$=\overrightarrow{\text{BA}}+\overrightarrow{\text{AB}}$ $\Big[\because \overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}=\overrightarrow{\text{BA}}\Big]$
$=\vec0$
View full question & answer→Question 292 Marks
Find a vector of magnitude 4 units which is parallel to the vector $\sqrt3\hat{\text{i}}+\hat{\text{j}}$.
AnswerLet $\vec{\text{a}}=\sqrt3\hat{\text{i}}+\hat{\text{j}}$
Then, $\big|\vec{\text{a}}\big|=\sqrt{\big(\sqrt3\big)^2+1}=\sqrt{3+1}=\sqrt4=2$
A unit vector parallel to $\vec{\text{a}}=\hat{\text{a}}=\frac{\vec{\text{a}}}{|\vec{\text{a}}|}=\frac{1}2\big(\sqrt3\hat{\text{i}}+\hat{\text{j}}\big)$
Hence, Required vector $=4\hat{\text{a}}=4\times\frac{1}2\big(\sqrt3\hat{\text{i}}+\hat{\text{j}}\big)=2\sqrt3\hat{\text{i}}+2\hat{\text{j}}$
View full question & answer→Question 302 Marks
Find the components along the coordinate axis of the position vector of the following point:
P(3, 2)
AnswerHere, P = (3, 2)
Position vector of $\text{P}=3\hat{\text{i}}+2\hat{\text{j}}$
Component of P along x-axis $=3\hat{\text{i}}$
Component of P along x-axis $=2\hat{\text{j}}$
View full question & answer→Question 312 Marks
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are the position vectors of the vertices of a triangle, then write the position vector of its centroid.
AnswerLet ABC be a triangle and D, E and F are the midpoints of the sides BC, CA and AB respectively.
Also, Let $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are the position vectors of A, B, C respectively. Then the position vectors of D, E, F are $\Big(\frac{\vec{\text{b}}+\vec{\text{c}}}2\Big),\Big(\frac{\vec{\text{c}}+\vec{\text{a}}}2\Big),\Big(\frac{\vec{\text{a}}+\vec{\text{b}}}2\Big)$ respectively.
The position vector of a point divides AD in the ratio of 2; is $\frac{1.\vec{\text{a}}+2\frac{\vec{\text{b}}+\vec{\text{c}}}{2}}{2}=\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3$
Similarly, Position vectors of the points divides BE, CF in the ratio of 2 : 1 are equal to $\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3$.
Thus, the point dividing AD in the ratio 2 : 1 also divides BE, CF in the same ratio.
Hence, the medians of a triangle are concurrent and the position vector of the centroid is $\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3$.
View full question & answer→Question 322 Marks
Write the projection of $\vec{\text{r}}=3\hat{\text{i}}-4\hat{\text{j}}+12\hat{\text{k}}$ on the coordinate axes.
AnswerWe have
$\vec{\text{r}}=3\hat{\text{i}}-4\hat{\text{j}}+12\hat{\text{k}}$
Projection of $\vec{\text{r}}$ on x-axis $=\frac{\vec{\text{r}}.\hat{\text{i}}}{|\hat{\text{i}}|}=\frac{3}{1}=3$
Projection of $\vec{\text{r}}$ on y-axis $=\frac{\vec{\text{r}}.\hat{\text{j}}}{|\hat{\text{j}}|}=\frac{-4}{1}=-4$
Projection of $\vec{\text{r}}$ on z-axis $=\frac{\vec{\text{r}}.\hat{\text{k}}}{|\hat{\text{k}}|}=\frac{12}{1}=12$
View full question & answer→Question 332 Marks
For what value of $\lambda$ are the vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ perpendicular to each other if
$\vec{\text{a}}=\lambda\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=4\hat{\text{i}}-9\hat{\text{j}}+2\hat{\text{k}}$
AnswerIf the vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ are perpendicular to each other, then
$\vec{\text{a}}.\vec{\text{b}}=0$
$\Rightarrow\Big(\lambda\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\Big).\Big(4\hat{\text{i}}-9\hat{\text{j}}+2\hat{\text{k}}\Big)=0$
$\Rightarrow4\lambda-18+2=0$
$\Rightarrow4\lambda-16=0$
$\Rightarrow4\lambda=16$
$\Rightarrow\lambda=4$
View full question & answer→Question 342 Marks
If $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ are non-coplanar vectors, prove that the given vectors are non-coplanar:
$\vec{\text{a}}+2\vec{\text{b}}+3\vec{\text{c}},\ 2\vec{\text{a}}+\vec{\text{b}}+3\vec{\text{c}}$ and $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}$
AnswerLet if possible the given vectors are coplanar. Then one of the vector is expressible in the terms of the other two.
We have,
$\vec{\text{a}}+2\vec{\text{b}}+3\vec{\text{c}}=\text{x}\big(2\vec{\text{a}}+\vec{\text{b}}+3\vec{\text{c}}\big)+\text{y}\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big)$
$=\vec{\text{a}}(\text{2x + y})+\vec{\text{b}}(\text{x + y})+\vec{\text{c}}(3\text{x}+\text{y})$
$\Rightarrow\text{2x + y}=1,\ \text{x + y}=2,\ 3\text{x}+\text{y}=3$
On solving the first two equations we get x = -1, y = 3. Clearly the values of x, y does not satisfy the third equation.
Hence, the given vectors are non-coplanar.
View full question & answer→Question 352 Marks
If $\vec{\text{a}}=3\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$ and $\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}},$ then find $\big(\vec{\text{a}}\times\vec{\text{b}}\big)\vec{\text{a}}.$
AnswerSince $\vec{\text{a}}\times\vec{\text{b}}$ is a vector, $\big(\vec{\text{a}}\times\vec{\text{b}}\big)\vec{\text{a}}$ without any dot or cross product in between is meaningless.
View full question & answer→Question 362 Marks
Write the projection of $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ along the vector $\hat{\text{j}}.$
AnswerProjection of $\vec{\text{a}}$ on $\vec{\text{b}}=\frac{\vec{\text{a}}.\vec{\text{b}}}{\big|\vec{\text{b}}\big|}$
Projection of $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ along $\hat{\text{j}}$
$=\frac{\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big).\vec{\text{j}}}{|\vec{\text{j}}|}$
$=\frac{1}{1}$
$=1$
View full question & answer→Question 372 Marks
Write a unit vector in the direction of the sum of the vectors $\vec{\text{a}}=2\hat{\text{i}}+2\hat{\text{j}}-5\hat{\text{k}}$ and $\vec{\text{b}}=2\hat{\text{i}}+\text{y}\hat{\text{j}}-7\hat{\text{k}}$.
AnswerWe have, $\vec{\text{a}}=2\hat{\text{i}}+2\hat{\text{j}}-5\hat{\text{k}}$ and $\vec{\text{b}}=2\hat{\text{i}}+\text{y}\hat{\text{j}}-7\hat{\text{k}}$
$\therefore\ \vec{\text{a}}+\vec{\text{b}}=\big(2\hat{\text{i}}+2\hat{\text{j}}-5\hat{\text{k}}\big)+\big(2\hat{\text{i}}+\hat{\text{j}}-7\hat{\text{k}}\big)$
$=4\hat{\text{i}}+3\hat{\text{j}}-12\hat{\text{k}}$
$\Rightarrow\ \big|\vec{\text{a}}+\vec{\text{b}}\big|=\sqrt{4^2+3^2+(-12)^2}$
$=\sqrt{16+9+144}$
$=\sqrt{169}$
$=13$
$\therefore$ Required unit vector $=\frac{\vec{\text{a}}+\vec{\text{b}}}{\big|\vec{\text{a}}+\vec{\text{b}}\big|}=\frac{4\hat{\text{i}}+3\hat{\text{j}}-12\hat{\text{k}}}{13}$
$=\frac{4}{13}\hat{\text{i}}+\frac{3}{13}\hat{\text{j}}-\frac{12}{13}\hat{\text{k}}$
View full question & answer→Question 382 Marks
Show that the vector $\hat{i}+\hat{j}+\hat{k}$ is equally inclined to the axes OX, OY and OZ.
Answer$\text{Let}\ \vec{a}=\hat{i}+\hat{j}+\hat{k}.$ Then,$|\vec{a}|=\sqrt{1^2+1^2+1^2}=\sqrt{3}$
Therefore, the direction cosines of $\vec{a}=\text{are}\ \bigg(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\bigg).$
Now, let $\alpha,\ \beta,$ and $\gamma$ be the angles formed by $\vec{a}$ with the positive directions of x, y, and z axes.
Then, we have $\text{cos}\ \alpha=\frac{1}{\sqrt{3}},\text{cos}\beta=\frac{1}{\sqrt{3}},\text{cos}\ \gamma=\frac{1}{\sqrt{3}}.$ Hence, the given vector is equally is equally inclined to axes OX, OY, and OZ.
View full question & answer→Question 392 Marks
Find the components along the coordinate axis of the position vector of the following point:S(4,-3)
AnswerHere, S = (4, -3)
Position vector of $\text{S}=4\hat{\text{i}}-3\hat{\text{j}}$
Component of S along x-axis $=4\hat{\text{i}}$
Component of S along x-axis $=-3\hat{\text{j}}$
View full question & answer→Question 402 Marks
If $\big|\vec{\text{a}}\times\vec{\text{b}}\big|^2+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2=144$ and $|\vec{\text{a}}|=4,$ find $\big|\vec{\text{b}}\big|.$
AnswerWe know
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|^2+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2$
$\Rightarrow144=4^2\big|\vec{\text{b}}\big|^2$
$\Rightarrow144=16\big|\vec{\text{b}}\big|^2$
$\Rightarrow\big|\vec{\text{b}}\big|^2=9$
$\Rightarrow\big|\vec{\text{b}}\big|=3$
View full question & answer→Question 412 Marks
If the position vector of a point (-4, -3) be $\vec{\text{a}}$, find $\big|\vec{\text{a}}\big|$.
AnswerGiven a point (-4, -3) such that its position vector $\vec{\text{a}}$ is given by
$\vec{\text{a}}=-4\hat{\text{i}}-3\hat{\text{j}}$
Then,
$\big|\vec{\text{a}}\big|=\sqrt{(-4)^2+(-3)^2}$
$=\sqrt{16+9}$
$=\sqrt{25}$
$=5$
View full question & answer→Question 422 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are two vectors such that $\big|\vec{\text{a}}\big|=4,\big|\vec{\text{b}}\big|=3$ and $\vec{\text{a}}.\vec{\text{b}}=6,$ find the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$.
AnswerLet $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$.Given that
$\vec{\text{a}}.\vec{\text{b}}=6$
$\Rightarrow\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\cos\theta=6$
$\Rightarrow(4)(3)\cos\theta=6$
$\Rightarrow12\cos\theta=6$
$\Rightarrow\cos\theta=\frac{6}{12}=\frac{1}{2}$
$\Rightarrow\theta=\cos^{-1}\big(\frac{1}{2}\big)=\frac{\pi}{3}$
View full question & answer→Question 432 Marks
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are three mutually perpendicular unit vectors, then prove that $\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|=\sqrt{3}$
AnswerGiven that $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ are unit vectors.
So, $\big|\vec{\text{a}}\big|=1,\Big|\vec{\text{b}}\big|=1$ and $\big|\vec{\text{c}}\big|=1$
Since they are mutually perpendicular,
$\vec{\text{a}}.\vec{\text{b}}=\vec{\text{b}}.\vec{\text{c}}=\vec{\text{c}}.\vec{\text{a}}=0$
Now,
$\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|^2=|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+|\vec{\text{c}}|+2\vec{\text{a}}.\vec{\text{b}}+2\vec{{\text{b}}}.\vec{\text{c}}+2\vec{\text{c}}.\vec{\text{a}}$
$=1+1+1+0+0+0$
$=3$
$\therefore\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|=\sqrt{3}$
View full question & answer→Question 442 Marks
If $\vec{\text{a}}=3\hat{\text{i}}-\hat{\text{j}}-4\hat{\text{k}},\ \vec{\text{b}}=-2\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}}$ and $\vec{\text{c}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$, find $\big|3\vec{\text{a}}-2\vec{\text{b}}+4\vec{\text{c}}\big|$.
AnswerGiven: $\vec{\text{a}}=3\hat{\text{i}}-\hat{\text{j}}-4\hat{\text{k}},\ \vec{\text{b}}=-2\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}} $ and $\vec{\text{c}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$
Now, $3\vec{\text{a}}-2\vec{\text{b}}+4\vec{\text{c}}=3\big(\vec{\text{a}}=3\hat{\text{i}}-\hat{\text{j}}-4\hat{\text{k}}\big)\\-2\big(\vec{\text{b}}=-2\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}}\big)+4\big(\vec{\text{c}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}\big)$
$=9\hat{\text{i}}-3\hat{\text{j}}-12\hat{\text{k}}+4\hat{\text{i}}-8\hat{\text{j}}+6\hat{\text{k}}+4\hat{\text{i}}+8\hat{\text{j}}-4\hat{\text{k}}$
$=17\hat{\text{i}}-3\hat{\text{j}}-10\hat{\text{k}}$
Hence, $\big|3\vec{\text{a}}-2\vec{\text{b}}+4\vec{\text{c}}\big|$
$=\sqrt{17^2+(-3)^2+(-10)^2}$
$=\sqrt{289+9+100}$
$=\sqrt{398}$
View full question & answer→Question 452 Marks
Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).
AnswerVertices of $\triangle\text{ABC}$ are A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).
$\therefore\ \ \ \ \ \ \text{Position vector of point A}={(1, 1, 2) }=\hat{i}+\hat{j}+2\hat{k}$
$\text{Position vector of point B}={(2, 3, 5) }=2\hat{i}+3\hat{j}+5\hat{k}$
$\text{Position vector of point C}={(1, 5, 5) }=\hat{i}+5\hat{j}+5\hat{k}$
$\text{Now}\ \ \overrightarrow{\text{AB}}\ \ $ = Position vector of point B - Position vector of point A
$= 2\hat{i}+3\hat{j}+5\hat{k}-\big(\hat{i}+\hat{j}+2\hat{k}\big)$ $=2\hat{i}+3\hat{j}+5\hat{k}-\hat{i}-\hat{j}-2\hat{k}$
$=\hat{i}+2\hat{j}+3\hat{k}$
$\text{And}\ \ \overrightarrow{\text{AC}}\ \ $ = Position vector of point C - Position vector of point A
$=\hat{i}+5\hat{j}+5\hat{k}\ -(\hat{i}+\hat{j}+2\hat{k})$ $=\hat{i}+5\hat{j}+5\hat{k}\ -\hat{i}-\hat{j}-2\hat{k}$
$=0\hat{i}+4\hat{j}+3\hat{k}$
$\therefore\ \ \ \ \ \ \overrightarrow{\text{AB}}\ \text{x}\ \overrightarrow{\text{AC}}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&2&3\\0&4&3\end{vmatrix}=\hat{i}(6-12)-\hat{j}(3-0)+\hat{k}(4-0)=-6\hat{i}-3\hat{j}+4\hat{k}$
$\text{Now}\ \ \ \ \text{Area of triangle ABC}=\frac{1}{2}\bigg|\overrightarrow{\text{AB}}\times\overrightarrow{\text{AC}}\bigg|$ $=\frac{1}{2}\sqrt{36+9+16}$
$=\frac{1}{2}\sqrt{61}\ \text{sq.units}$
View full question & answer→Question 462 Marks
If $\vec{\text{a}}$ is a unit vector, then find $|\vec{\text{x}}|$ in each of the following.$\big(\vec{\text{x}}-\vec{\text{a}}\big).\big(\vec{\text{x}}+\vec{\text{a}}\big)=12$
AnswerGiven that $\vec{\text{a}}$ is a unit vector.$\big(\vec{\text{x}}-\vec{\text{a}}\big).\big(\vec{\text{x}}+\vec{\text{a}}\big)=12$
$\Rightarrow|\vec{\text{x}}|^2-|\vec{\text{a}}|^2=12$
$\Rightarrow|\vec{\text{x}}|^2-1^2=12$ [From (1)]
$\Rightarrow|\vec{\text{x}}|^2=13$
$\Rightarrow|\vec{\text{x}}|=\sqrt{13}$
View full question & answer→Question 472 Marks
For what value of $\lambda$ are the vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ perpendicular to each other if
$\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}$ and $\vec{\text{b}}=3\hat{\text{i}}+2\hat{\text{j}}+\lambda\hat{\text{k}}$
AnswerIf the vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ are perpendicular to each other, then$\vec{\text{a}}.\vec{\text{b}}=0$
$\Rightarrow\Big(2\hat{\text{i}}+3 \hat{\text{j}}+4\hat{\text{k}}\Big).\Big(3\hat{\text{i}}+2\hat{\text{j}}-\lambda\hat{\text{k}}\Big)=0$
$\Rightarrow6+6-4\lambda=0$
$\Rightarrow12-4\lambda=0$
$\Rightarrow4\lambda=12$
$\Rightarrow\lambda=3$
View full question & answer→Question 482 Marks
Find the angle between the vectors $\vec{\text{a}} $ and $\vec{\text{b}},$ where$\vec{\text{a}}=\hat {\text{i}}+2\hat{\text{j}}-\hat{\text{k}},$ and $\vec{\text{b}} =\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
AnswerLet $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$.$\big|\vec{\text{a}}\big|=\sqrt{(1)^2+(2)^2+(-1)^{2}}=\sqrt{6}$
$\big|\vec{\text{b}}\big|=\sqrt{(1)^2+(-1)^2+(1)^{2}}=\sqrt{3}$
$\vec{\text{a}}.\vec{\text{b}}=1-2-1=-2$
$\cos\theta=\frac{\vec{\text{a }}.\vec{\text{ b}}}{\big|\vec{a}\big|\big|\vec{b}\big|}=\frac{-2}{\sqrt{6}\sqrt{3}}=\frac{-2}{\sqrt{18}}=\frac{-\sqrt{2}\times\sqrt{2}}{\sqrt{2}\times\sqrt{9}}=\frac{-\sqrt{2}}{3}$
$\Rightarrow\theta=\cos^{-1}\Big(\frac{-\sqrt{2}}{{3}}\Big)$
View full question & answer→Question 492 Marks
If $\vec{\text{a}}$ is a unit vector, then find $|\vec{\text{x}}|$ in each of the following.$\big(\vec{\text{x}}-\vec{\text{a}}\big).\big(\vec{\text{x}}+\vec{\text{a}}\big)=8$
AnswerGiven that $\vec{\text{a}}$ is a unit vector.
$\Rightarrow|\vec{\text{a}}|=1\dots(1)$
$\big(\vec{\text{x}}-\vec{\text{a}}\big).\big(\vec{\text{x}}+\vec{\text{a}}\big)=8$
$\Rightarrow|\vec{\text{x}}|^2-|\vec{\text{a}}|^2=8$
$\Rightarrow|\vec{\text{x}}|^2-1^2=8$ [From (1)]
$\Rightarrow|\vec{\text{x}}|^2=9$
$\Rightarrow|\vec{\text{x}}|=3$
View full question & answer→Question 502 Marks
Write two different vectors having same magnitude.
AnswerLet $\vec{\text{a}}=2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$ and $\vec{\text{b}}=-2\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}}$
It can be observed that
$|\vec{\text{a}}|=\sqrt{2^2+(-1)^2+3^2}=\sqrt{14}$
$\big|\vec{\text{b}}\big|=\sqrt{(-2)^2+1^2+(-3)^2}=\sqrt{14}$
Hence, $\vec{\text{a}}\text{ and }\vec{\text{b}}$ are two vectors having same direction.
View full question & answer→Question 512 Marks
Write the direction cosines of the vectors $-2\hat{\text{i}}+\hat{\text{j}}-5\hat{\text{k}}$.
AnswerGiven: $-2\hat{\text{i}}+\hat{\text{j}}-5\hat{\text{k}}$
Then, its direction cosines are:
$\frac{-2}{\sqrt{(-2)^2+1^2+(-5)^2}},\frac{1}{\sqrt{(-2)^2+1^2+(-5)^2}},\frac{-5}{\sqrt{(-2)^2+1^2+(-5)^2}}$ or, $\frac{-2}{\sqrt{30}},\frac{1}{\sqrt{30}},\frac{-5}{\sqrt{30}}$
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Write the value of $\hat{\text{i}}.\big(\hat{\text{j}}\times\hat{\text{k}}\big)+\hat{\text{j}}.\big(\hat{\text{k}}\times\hat{\text{i}}\big)+\hat{\text{k}}.\big(\hat{\text{i}}\times\hat{\text{j}}\big).$
Answer$\hat{\text{i}}.\big(\hat{\text{j}}\times\hat{\text{k}}\big)+\hat{\text{j}}.\big(\hat{\text{k}}\times\hat{\text{i}}\big)+\hat{\text{k}}.\big(\hat{\text{j}}\times\hat{\text{i}}\big)$
$=\hat{\text{i}}.\hat{\text{i}}+\hat{\text{j}}.\hat{\text{j}}+\hat{\text{k}}.\hat{\text{k}}$
$=|\hat{\text{i}}|^2+|\hat{\text{j}}|^2+|\hat{\text{k}}|^2$
$=1+1+1$
$=3$
View full question & answer→Question 532 Marks
If $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},\ \vec{\text{b}}=4\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$and $\vec{\text{c}}=\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$, find a vecctor of magnitude 6 units which is parallel to the vector $2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}$.
AnswerWe have, $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},\ \vec{\text{b}}=4\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$and $\vec{\text{c}}=\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$Then,
$2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}=2\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)\\-\big(4\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\big)+3\big(\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\big)$ $=\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}$ $\therefore$ A unit vector parallel to $2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}$ is $\frac{2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}}{\big|2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}\big|}=\frac{(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}})}{\sqrt{1^2+(-2)^2+2^2}}$ $=\frac{(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}})}{\sqrt9}$ $=\frac{(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}})}3$ Hence, Required vector $=\frac{6}3\big(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}\big)$ $=2\hat{\text{i}}-4\hat{\text{j}}+4\hat{\text{k}}$
View full question & answer→Question 542 Marks
Find the magnitude of the vector $\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}}$.
AnswerGiven: $\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}-6\hat{\text{k}}$$\therefore$ Magnitude of the vector $=\big|\vec{\text{a}}\big|=\sqrt{2^2+3^2+(-6)^2}$
$=\sqrt{4+9+36}$
$=\sqrt{49}$
$=7$
View full question & answer→Question 552 Marks
For what value of $\lambda$ are the vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ perpendicular to each other if
$\vec{\text{a}}=\lambda\hat{\text{i}}+3\hat{\text{j}}+2\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$
AnswerIf the vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ are perpendicular to each other, then
$\vec{\text{a}}.\vec{\text{b}}=0$
$\Rightarrow\Big(\lambda\hat{\text{i}}+3\hat{\text{j}}+2\hat{\text{k}}\Big).\Big(\hat{\text{i}}-1\hat{\text{j}}+3\hat{\text{k}}\Big)=0$
$\Rightarrow\lambda-3+6=0$
$\Rightarrow\lambda+3=0$
$\Rightarrow\lambda=-3$
View full question & answer→Question 562 Marks
vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ are such that $|\vec{\text{a}}|=\sqrt{3},\big|\vec{\text{b}}\big|=\frac{2}{3}$ and $\big(\vec{\text{a}}\times\vec{\text{b}}\big)$ is a unit vector. write the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
AnswerGiven: $\vec{\text{a}}\times\vec{\text{b}}$ is a unit vector.
$\Rightarrow\big|\vec{\text{a}}\times\vec{\text{b}}\big|=1\dots(1)$
Let $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
We know
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta$
From (1), we get
$1=(\sqrt{3})\Big(\frac{2}{3}\Big)\sin\theta$
$\Rightarrow\sin\theta=\frac{\sqrt{3}}{2}$
$\Rightarrow\theta=\frac{\pi}{3}$
View full question & answer→Question 572 Marks
Find the angle betwwen two vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ if$|\vec{\text{a}}|=3,\big|\vec{\text{b}}\big|=3$ and $\vec{\text{a}}.\vec{\text{b}}=1$
AnswerLet the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ is $\theta,$ then
$\cos\theta=\frac{\vec{\text{a}}.\vec{\text{b}}}{|\vec{\text{a}}|\big|\vec{\text{b}}\big|}$
$=\frac{1}{3.3}$
$\cos\theta=\frac{1}{9}$
$\theta=\cos^{-1}\big(\frac{1}{9}\big)$
View full question & answer→Question 582 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are unit vectors such that $\vec{\text{a}}\times\vec{\text{b}}$ is also a unit vector, find the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
AnswerLet $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
Given:
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=1$
$|\vec{\text{a}}|=1$
$|\vec{\text{b}}|=1$
We know
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta$
$\Rightarrow1=(1)(1)\sin\theta$
$\Rightarrow\sin\theta=1$
$\Rightarrow\theta=\frac{\pi}{2}$
View full question & answer→Question 592 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are two vectors of magnitudes 3 and $\frac{\sqrt{2}}{3}$ repectively such that $\vec{\text{a}}\times\vec{\text{b}}$ is a unit vector. write the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
AnswerWrite the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
Let $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
It is given that $\vec{\text{a}}\times\vec{\text{b}}$ is a unit vector.
$\Rightarrow\big|\vec{\text{a}}\times\vec{\text{b}}\big|=1$
We know
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta$
$\Rightarrow1=(3)\Big(\frac{\sqrt{2}}{3}\big)\sin\theta$
$\Rightarrow\sin\theta=\frac{1}{\sqrt{2}}$
$\Rightarrow\theta=45^\circ,135^\circ$
View full question & answer→Question 602 Marks
Find the angle between the vectora $\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}.$
AnswerWe have
$\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$
Let $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
$|\vec{\text{a}}|=\sqrt{(1)^2+(-1)^2+(1)^2}=\sqrt{3}$
$\big|\vec{\text{b}}\big|=\sqrt{(1)^2+(1)^2+(-1)^2}=\sqrt{3}$
and
$\vec{\text{a}}.\vec{\text{b}}=1-1-1=-1$
Now,
$\cos\theta=\frac{\vec{\text{a}}.\vec{\text{b}}}{|\vec{\text{a}}|\big|\vec{\text{b}}\big|}=\frac{-1}{\sqrt{3}\sqrt{3}}=\frac{-1}{3}$
$\therefore\theta=\cos^-1\big(\frac{-1}{3}\big)$
View full question & answer→Question 612 Marks
Find the volume of the parallelopiped with its edges represented by the vectors
$\hat{\text{i}}+\hat{\text{j}},\hat{\text{i}}+2\hat{\text{j}}$ and $\hat{\text{i}}+\hat{\text{j}}+\pi{\text{k}}.$
AnswerLet:
$\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}$
$\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}$
$\vec{\text{c}}=\hat{\text{i}}+\hat{\text{j}}+\pi\hat{\text{k}}$
We know that the volume of a parallelopiped whose three adjecent edges are $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ is equal to $\big|\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]\big|.$
We have
$\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=\begin{vmatrix}1&1&0\\1&2&0\\1&1&\pi \end{vmatrix}$
$=1(2\pi-0)-1(\pi-0)+0(1-2)$
$=2\pi-\pi=\pi$
$\therefore$ volume $=\big|\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]\big|=|\pi|=\pi$ cubic units
View full question & answer→Question 622 Marks
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are position vectors of the vertices A, B and C respectively, of a triangle ABC, write the value of$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}$.
AnswerGiven: $\vec{\text{a}},\vec{\text{b}}\text{ and }\vec{\text{c}}$ are the position vectors of A, B and C respectively. Then,
$\overrightarrow{\text{AB}}=\vec{\text{b}}-\vec{\text{a}}$
$\overrightarrow{\text{BC}}=\vec{\text{c}}-\vec{\text{b}}$
$\overrightarrow{\text{CA}}=\vec{\text{a}}-\vec{\text{c}}$
Consider,
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}\\=\vec{\text{b}}-\vec{\text{a}}+\vec{\text{c}}-\vec{\text{b}}+\vec{\text{a}}-\vec{\text{c}}$
$=\vec0$
View full question & answer→Question 632 Marks
If $\vec{\text{a}}\text{ and }\vec{\text{b}}$ represent two adjacent sides of a parallelogram, then write vectors representing its diagonals.
AnswerLet $\vec{\text{a}}\text{ and }\vec{\text{b}}$ represents two adjacent sides of a parallelogram ABCD.
$\therefore$ AB = DC and AD = BC
$\Rightarrow\ \overrightarrow{\text{DC}}=\overrightarrow{\text{AB}}=\vec{\text{a}}$ and $\Rightarrow\ \overrightarrow{\text{AD}}=\overrightarrow{\text{BC}}=\vec{\text{b}}$
In $\triangle\text{ABC}$
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}=\overrightarrow{\text{AC}}$
$\Rightarrow\ \vec{\text{a}}+\vec{\text{b}}=\overrightarrow{\text{AC}}$
In $\triangle\text{ABD}$
$\ \overrightarrow{\text{AD}}+\overrightarrow{\text{DB}}=\overrightarrow{\text{AB}}$
$\Rightarrow\ \vec{\text{b}}+\overrightarrow{\text{DB}}=\vec{\text{a}}$
$\Rightarrow\ \overrightarrow{\text{DB}}=\vec{\text{a}}-\vec{\text{b}}$
View full question & answer→Question 642 Marks
Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$.
AnswerSuppose the vector makes equal angle with the coordinate axis.
Then, its direction cosines are . Therefore,
$\text{l}=\text{m}=\text{n}.$
$\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Rightarrow\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Rightarrow3\text{l}^2=1$
$\Rightarrow\text{l}^2=\frac{1}{3}$
$\Rightarrow\text{l}=\frac{1}{\sqrt{3}}$
Hence, direction cosines are $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}.$
View full question & answer→Question 652 Marks
Find the value of 'p' for which the vectors $3\hat{\text{i}}+2\hat{\text{j}}+9\hat{\text{k}}$ and $\hat{\text{i}}-2\text{p}\hat{\text{j}}+3\hat{\text{k}}$ are parallel.
AnswerLet $\vec{\text{a}}=3\hat{\text{i}}+2\hat{\text{j}}+9\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\text{p}\hat{\text{j}}+3\hat{\text{k}}$ be the two given vectors.
If $\vec{\text{a}}\text{ and }\vec{\text{b}}$ are parallel, then
$\vec{\text{b}}=\lambda\vec{\text{a}}$ for some scalar $\lambda$
$\therefore\ \hat{\text{i}}-2\text{p}\hat{\text{j}}+3\hat{\text{k}}=\lambda\big(3\hat{\text{i}}-2\hat{\text{j}}+9\hat{\text{k}}\big)$
$\Rightarrow\ \hat{\text{i}}-2\text{p}\hat{\text{j}}+3\hat{\text{k}}=3\lambda\hat{\text{i}}+2\lambda\hat{\text{j}}+9\lambda\hat{\text{k}}$
$\Rightarrow\ 1=\lambda3$ and $-2\text{p}=2\lambda$ $\big($ Equating coefficients of $\hat{\text{i}},\hat{\text{j}},\hat{\text{k}}\big)$
$\Rightarrow\ \text{p}=-\lambda=-\frac{1}3$
Thus, the value of p is $-\frac{1}3$.
View full question & answer→Question 662 Marks
If $\vec{\text{b}}$ is a unit vector such that $\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=8,$ find $|\vec{\text{a}}|.$
AnswerGiven that $\vec{\text{b}}$ is a unit vector.
$\therefore\big|\vec{\text{b}}\big|=1$
And
$\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=8$ (Given)
$\Rightarrow|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2=8$
$\Rightarrow|\vec{\text{a}}|^2-1^2=8$
$\Rightarrow|\vec{\text{a}}|^2=9$
$\therefore|\vec{\text{a}}|=3$
View full question & answer→Question 672 Marks
If $\vec{\text{a}}$ is a vector and m is a scalar such that m $\vec{\text{a}}=\vec0$, then what are the alternatives for m and $\vec{\text{a}}$?
AnswerGiven: $\vec{\text{a}}$ is a vector and m is a scalar such that, $\text{m}\vec{\text{ a}}=\vec0$
Then either $\text{m}=0\text{ or, } \vec{\text{a}}=\vec0$
View full question & answer→Question 682 Marks
Find a vector in the direction of $\vec{\text{a}}=2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$, which has magnitude of 6 units.
AnswerGiven:
$\vec{\text{a}}=2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$
$|\vec{\text{a}}|=\sqrt{2^2+(-1)^2+2^2}$
$=\sqrt{4+1+4}$
$=\sqrt{9}$
$=3$
$\therefore$ Required Vector $=6\times\frac{\vec{\text{a}}}{|\vec{\text{a}}|}=6\times\frac{\big(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\big)}3$
$=4\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}}$
View full question & answer→Question 692 Marks
If $\vec{\text{a}}=\hat{\text{i}}+3\hat{\text{j}}-2\hat{\text{k}}$ and $\vec{\text{b}}=-\hat{\text{i}}+3\hat{\text{k},}$ find $\big|\vec{\text{a}}\times\vec{\text{b}}\big|.$
AnswerIf $\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{b}_1\hat{\text{j}}+\text{c}_1\hat{\text{k}}$ and
$\vec{\text{b}}=\text{a}_2\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{c}_2\hat{\text{k}},$ then
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{a}_1&\text{b}_1&\text{c}_2\\\text{a}_2&\text{b}_2&\text{c}_2 \end{vmatrix}$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&3&-2\\-1&0&3 \end{vmatrix}$
$=\hat{\text{i}}(9-0)-\hat{\text{j}}(3-2)+\hat{\text{k}}(0+3)$
$\vec{\text{a}}\times\vec{\text{b}}=9\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\sqrt{(9)^2+(-1)^2+(3)^2}$
$=\sqrt{81+1+9}$
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\sqrt{91}$
View full question & answer→Question 702 Marks
If $\vec{\text{a}}.\vec{\text{a}}=0$ and $\vec{\text{a}}.\vec{\text{b}}=0,$ what can you conclude about the vector $\vec{\text{b}}$?
AnswerIt is given that $\vec{\text{a}}.\vec{\text{a}}=0$ and $\vec{\text{a}}.\vec{\text{b}}=0.$
Now,
$\vec{\text{a}}.\vec{\text{a}}=0\Rightarrow|\vec{\text{a}}|^2=0\Rightarrow|\vec{\text{a}}|=0$
$\therefore \vec{\text{a}}$ is a zero vector.
Hence, vector $\vec{\text{b}}$ satisfying $\vec{\text{a}}.\vec{\text{b}}=0.$ can be any vector
View full question & answer→Question 712 Marks
Write the value of $\big[2\hat{\text{i }} 3\hat{\text{j }}4\hat{\text{k}}\big].$
AnswerWe have
$\big[2\hat{\text{i }}3\hat{\text{j }}4\hat{\text{k}}\big]$
$=\big(2\hat{\text{i}}\times3\hat{\text{j}}\big).4\hat{\text{k}}$ $\big(\therefore\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}\big)$
$=6\hat{\text{k}}.4\hat{\text{k}}$
$=24$
View full question & answer→Question 722 Marks
Write the expression for the area of the parallelogram having $\vec{\text{a}}$ and $\vec{\text{b}}$ as its diagonals.
AnswerGiven: $\vec{\text{a}}$ and $\vec{\text{b}}$ are diagonals of a parallelogram.
Area of the parallelogram $=\frac{1}{2}\big|\vec{\text{a}}\times\vec{\text{b}}\big|$
View full question & answer→Question 732 Marks
Write the value of p for which $\vec{\text{a}}=3\hat{\text{i}}+2\hat{\text{j}}+9\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}+\text{p}\hat{\text{j}}+3\hat{\text{k}}$ are parallel vectors.
AnswerWe have$\vec{\text{a}}=3\hat{\text{i}}+2\hat{\text{j}}+9\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}+\text{p}\hat{\text{j}}+3\hat{\text{k}}$
Given that $\vec{\text{a}}$ and $\vec{\text{b}}$ are parallel.
$\Rightarrow\vec{\text{a}}=\text{t}\vec{\text{b}}$ for some t.
$\Rightarrow3\hat{\text{i}}+2\hat{\text{j}}+9\hat{\text{k}}=\text{t}\big(\hat{\text{i}}+\text{p}\hat{\text{j}}+3\hat{\text{k}}\big)$
$\Rightarrow3\hat{\text{i}}+2\hat{\text{j}}+9\hat{\text{k}}=\text{t}\hat{\text{i}}+\text{pt}\hat{\text{j}}+3\text{t}\hat{\text{k}}$
Comparing both sides, we get
$3=\text{t,}2=\text{pt}$ and $9=3\text{t}$
$\Rightarrow\text{t}=3$ and $\text{pt}=2$
$\Rightarrow3\text{t}=2$
$\therefore\text{t}=\frac{2}{3}$
View full question & answer→Question 742 Marks
Find the unit vector parallel to the vector $\hat{\text{i}}+\sqrt3\hat{\text{j}}$.
AnswerLet $\vec{\text{a}}=\hat{\text{i}}+\sqrt3\hat{\text{j}}$
Then, $\big|\vec{\text{a}}\big|=\sqrt{1^2+\big(\sqrt3\big)^2}$
$=\sqrt{1+3}$
$=\sqrt4$
$=2$
Unit vector parallel to $\vec{\text{a}}=\hat{\text{a}}=\frac{\vec{\text{a}}}{|\vec{\text{a}}|}=\frac{1}{2}\big(\hat{\text{i}}+\sqrt3\hat{\text{j}}\big)=\frac{1}2\hat{\text{i}}+\frac{\sqrt3}2\hat{\text{j}}$
View full question & answer→Question 752 Marks
Find the cosine of the angle between the vectors $4\hat{\text{i}}-3\hat{\text{j}}+3\hat{\text{k}}$ and $2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}.$
AnswerLet, $\vec{\text{a}}=4\hat{\text{i}}-3\hat{\text{j}}+3\hat{\text{k}}$
and $\vec{\text{b}}=2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}$
Let $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
$|\vec{\text{a}}|=\sqrt{(4)^2+(-3)^2+(3)^2}=\sqrt{34}$
$\big|\vec{\text{b}}\big|=\sqrt{(2)^2+(-1)^2+(-1)^2}=\sqrt{6}$
$\therefore\vec{\text{a}}.\vec{\text{b}}=8+3-3=8$
$\cos\theta=\frac{\vec{\text{a}}.\vec{\text{b}}}{|\vec{\text{a}}|\big|\vec{\text{b}}\big|}=\frac{8}{\sqrt{34}\sqrt{6}}=\frac{8}{2\sqrt{51}}=\frac{4}{\sqrt{51}}$
View full question & answer→Question 762 Marks
Find the value of x for which $\text{x}\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$ is a unit vector.
AnswerWe have, $\text{x}\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$ is a unit vector.
$\therefore\ \sqrt{\text{x}^2+\text{x}^2+\text{x}^2}=1$
$\Rightarrow\sqrt3|\text{x}|=1$
$\Rightarrow|\text{x}|=\frac{1}{\sqrt3}$
$\Rightarrow\text{x}=\pm\frac{1}{\sqrt3}$
View full question & answer→Question 772 Marks
For any three vectors $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ write the value of $\vec{\text{a}}\times\big(\vec{\text{b}}+\vec{\text{c}}\big)+\vec{\text{b}}\times\big(\vec{\text{c}}+\vec{\text{a}}\big)+\vec{\text{c}}\times\big(\vec{\text{a}}+\vec{\text{b}}\big).$
Answer$\vec{\text{a}}\times\big(\vec{\text{b}}+\vec{\text{c}}\big)+\vec{\text{b}}\times\big(\vec{\text{c}}+\vec{\text{a}}\big)+\vec{\text{c}}\times\big(\vec{\text{a}}+\vec{\text{b}}\big)$
$=\big(\vec{\text{a}}\times\vec{\text{b}}\big)+\big(\vec{\text{a}}\times\vec{\text{c}}\big)+\big(\vec{\text{b}}\times\vec{\text{c}}\big)+\big(\vec{\text{b}}\times\vec{\text{a}}\big)+\big(\vec{\text{c}}\times\vec{\text{a}}\big)+\big(\vec{\text{c}}\times\vec{\text{b}}\big)$
$=\big(\vec{\text{a}}\times\vec{\text{b}}\big)+\big(\vec{\text{a}}\times\vec{\text{c}}\big)+\big(\vec{\text{b}}\times\vec{\text{c}}\big)-\big(\vec{\text{a}}\times\vec{\text{b}}\big)-\big(\vec{\text{a}}\times\vec{\text{c}}\big)-\big(\vec{\text{b}}\times\vec{\text{c}}\big)$
$=\vec{0}$
View full question & answer→Question 782 Marks
Write the value of $\big[\hat{\text{i}}-\hat{\text{j }}\hat{\text{j}}-\hat{\text{k }}\hat{\text{k}}-\hat{\text{i}}\big].$
AnswerWe have
$\big[\hat{\text{i}}-\hat{\text{j }}\hat{\text{j}}-\hat{\text{k }}\hat{\text{k}}-\hat{\text{i}}\big]=\big[\big(\hat{\text{i}}-\hat{\text{j}}\big)\times\big(\hat{\text{j}}-\hat{\text{k}}\big)\big].\big(\hat{\text{k}}-\hat{\text{i}}\big)$
$\big(\therefore\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}\big)$
$\big[\big(\hat{\text{i}}\times\hat{\text{j}}\big)-\big(\hat{\text{i}}\times\hat{\text{k}}\big)-\big(\hat{\text{j}}\times{\hat{\text{j}}}\big)+\big(\hat{\text{j}}\times\hat{\text{k}}\big)\big].\big(\hat{\text{k}}-\hat{\text{i}}\big)$
$=\big[\hat{\text{k}}+\hat{\text{j}}+\hat{\text{i}}\big].\big(\hat{\text{k}}-\hat{\text{i}}\big)$
$=\big[\big(\hat{\text{k}}.\hat{\text{k}}\big)-\big(\hat{\text{k}}.\hat{\text{i}}\big)+\big(\hat{\text{j}}.\hat{\text{k}}\big)-\big(\hat{\text{j}}.\hat{\text{i}}\big)+\big(\hat{\text{i}}.\hat{\text{k}}\big)-\big(\hat{\text{i}}.\hat{\text{i}}\big)\big]$
$=1-0+0-0+0-1=0$
View full question & answer→Question 792 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are two vectors such that $\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\sqrt{3}$ and $\vec{\text{a}}.\vec{\text{b}}=1,$ find the angle between.
Answer$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\sqrt{3}$
$\Rightarrow|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta=\sqrt{3}\dots(1)$
$\vec{\text{a}}.\vec{\text{b}}=1$
$\Rightarrow|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta=1\dots(2)$
Dividing (1) by (2), we get
$\frac{|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta}{|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta}=\sqrt{3}$
$\Rightarrow\tan\theta=\sqrt{3}$
$\Rightarrow\theta=60^\circ$
View full question & answer→Question 802 Marks
Find the components along the coordinate axis of the position vector of the following point:R(-11, -9)
AnswerHere, R = (-11, -9)
Position vector of $\text{R}=-11\hat{\text{i}}-9\hat{\text{j}}$
Component of R along x-axis $=-11\hat{\text{i}}$
Component of R along x-axis $=-9\hat{\text{j}}$
View full question & answer→Question 812 Marks
If the vectors $3\hat{\text{i}}-2\hat{\text{j}}-4\hat{\text{k}}$ and $18\hat{\text{i}}-12\hat{\text{j}}-\text{m}\hat{\text{k}}$ are parallel, find the value of m.
AnswerTHe given vectors are parallel.
$\therefore3\hat{\text{i}}-2\hat{\text{j}}-4\hat{\text{k}}=\text{t}\big(18\hat{\text{i}}-12\hat{\text{j}}-\text{m}\hat{\text{k}}\big)$
$\Rightarrow3\hat{\text{i}}-2\hat{\text{j}}-4\hat{\text{k}}=18\text{t}\hat{\text{i}}-12\text{t}\hat{\text{j}}-\text{t}\text{m}\hat{\text{k}}$
Comparing both sides, we get
$18\text{t}=3,-12\text{t}=-2,-4=\text{tm}$
$\Rightarrow\text{t}=\frac{1}{6}$
Substituting the value of m in -4 = -tm, we get
$-4=-\text{m}\big(\frac{1}{6}\big)$
$\therefore\text{m}=24$
View full question & answer→Question 822 Marks
Write two different vectors having same magnitude.
AnswerConsider $\vec{a}=(\hat{i}-2\hat{j}+3\hat{k})\ \text{and}\ \vec{b}=(2\hat{i}+\hat{j}-3\hat{k}).$ It can be observed that $\Big|\vec{a}\Big|=\sqrt{1^2+(-2)^2+3^2}=\sqrt{1+ 4+9}=\sqrt{14}\ \text{and}$$\Big|\vec{b}\Big|=\sqrt{2^2+1^2+(-3)^2}=\sqrt{4+ 1+9}=\sqrt{14}.$
Hence, $\vec{a}\ \text {and}\ \vec{b}$ are two different vectors having the same magnitude. The vectors are different because they have different directions.
View full question & answer→Question 832 Marks
If $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}},\ \vec{\text{b}}=\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{c}}=\hat{\text{k}}+\hat{\text{i}}$, write unit vectors parallel to $\vec{\text{a}}+\vec{\text{b}}-2\vec{\text{c}}$.
AnswerGiven: $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}},\ \vec{\text{b}}=\hat{\text{j}}+\hat{\text{k}},\ \vec{\text{c}}=\hat{\text{k}}+\hat{\text{i}}$Now, $\vec{\text{a}}+\vec{\text{b}}-2\vec{\text{c}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{j}}+\hat{\text{k}}-2\hat{\text{k}}-2\hat{\text{i}}$
$=-\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$ Unit vector parallel to $\vec{\text{a}}+\vec{\text{b}}-2\vec{\text{c}}=\frac{-\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}}{\sqrt{(-1)^2+2^2+(-1)^2}}$ $$$=\frac{-\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}}{\sqrt6}$
View full question & answer→Question 842 Marks
$\text{Find}\ \lambda\ \text{and}\ \mu\ \text{if}\ (2\hat{i}+6\hat{j}+27\hat{k})\times(\hat{i}+\lambda\hat{j}+\mu\hat{k})=\vec{0}.$
Answer$\text{Given:}\ \ (2\hat{i}+6\hat{j}+27\hat{k})\times(\hat{i}+\lambda\hat{j}+\mu\hat{k})=0$
$\Rightarrow\ \ \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&6&27\\1&\lambda&\mu\end{vmatrix}=\vec{0}$
Expanding along first row,
$\hat{i}(6\mu-27\lambda)-\hat{j}(2\mu-27)+\hat{k}(2\lambda-6)$ $=\vec{0}=0\hat{i}+0\hat{j}+0\hat{k}$
Comparing the coefficients of $\hat{i}, \hat{j},\hat{k}$ on both sides, we have
$6\mu-27\lambda=0\ \ \ \ \ .....\text{(i)}$
$2\mu-27=0\ \ \ \ \ \ .....\text{(ii)}$
$\text{And}\ \ 2\lambda-6=0\ \ \ \ \ \ .....\text{(iii)}$
$\text{From eq. (ii),}\ \ 2\mu-27=0 \ \Rightarrow\ \ \mu=\frac{27}{2}$
$\text{From eq. (iii),}\ \ 2\lambda-6=0\ \Rightarrow\ \ \lambda=\frac{6}{2}=3$
Putting the values of $\mu\ \text{and}\ \lambda$ in eq. (i),
$6\bigg(\frac{27}{2}\bigg)-27(3)=0\ $ $ \Rightarrow\ 3(27)-27(3)=0\ \Rightarrow\ \ 0=0$
$\text{Therefore,}\ \ \mu=\frac{27}{2}\ \text{and}\ \lambda=\frac{6}{2}=3$
View full question & answer→Question 852 Marks
If $\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}},\ \vec{\text{b}}=\hat{\text{j}}+2\hat{\text{k}}$, write a unit vector along the vector $3\vec{\text{a}}-2\vec{\text{b}}$.
AnswerGiven: $\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}},\ \vec{\text{b}}=\hat{\text{j}}+2\hat{\text{k}}$
Therefore,
$3\vec{\text{a}}-2\vec{\text{b}}=3\hat{\text{i}}+6\hat{\text{j}}-2\hat{\text{j}}-4\hat{\text{k}}$
$=3\hat{\text{i}}+4\hat{\text{j}}-4\hat{\text{k}}$
Hence, Unit vector along $3\vec{\text{a}}-2\vec{\text{b}}=\frac{3\hat{\text{i}}+4\hat{\text{j}}-4\hat{\text{k}}}{\sqrt{3^2+4^2+(-4)^2}}$
$=\frac{3\hat{\text{i}}+4\hat{\text{j}}-4\hat{\text{k}}}{\sqrt{9+16+16}}$
$=\frac{1}{\sqrt{41}}\big(3\hat{\text{i}}+4\hat{\text{j}}-4\hat{\text{k}}\big)$
View full question & answer→Question 862 Marks
If $\vec{\text{a}},\vec{\text{b}}$ are non-collinear vectors, then find the value of $\big[\vec{\text{a}}\vec{\text{b}}\hat{\text{i}}\big]\hat{\text{i}}+\big[\vec{\text{a}}\vec{\text{b}}\hat{\text{j}}\big]\hat{\text{j}}+\big[\vec{\text{a}}\vec{\text{b}}\hat{\text{k}}\big]\hat{\text{k}}.$
AnswerFor any vector $\vec{\text{r}},$ we have
$\big(\vec{\text{r}}.\hat{\text{i}}\big)\hat{\text{i}}+\big(\vec{\text{r}}.\hat{\text{j}}\big)\hat{\text{j}}+\big(\vec{\text{r}}.\hat{\text{k}}\big)\hat{\text{k}}=\vec{\text{r}}$
Replacing $\vec{\text{r}}$ by $\vec{\text{a}}\times\vec{\text{b}},$ we have
$\big[(\vec{\text{a}}\times\vec{\text{b}}\big).\hat{\text{i}}\big]+\big[\big(\vec{\text{a}}\times\vec{\text{b}}\big).\hat{\text{j}}+\big[\big(\vec{\text{a}}\times\vec{\text{b}}\big).\hat{\text{k}}\big]\hat{\text{k}}=\vec{\text{a}}\times\vec{\text{b}}$
$\therefore\big[\vec{\text{a}}\vec{\text{b}}\hat{\text{i}}\big]\hat{\text{i}}\big[\vec{\text{a}}\vec{\text{b}}\hat{\text{j}}\big]\hat{\text{j}}+\big[\vec{\text{a}}\vec{\text{b}}\hat{\text{k}}\big]\hat{\text{k}}=\vec{\text{a}}\times\vec{\text{b}}$
View full question & answer→Question 872 Marks
Write a unit vector perpendicular to $\hat{\text{i}}+\hat{\text{j}}$ and $\hat{\text{j}}+\hat{\text{k}}.$
AnswerLet $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}};\vec{\text{b}}=0\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&1&0\\0&1&1 \end{vmatrix}$
$=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\sqrt{1+1+1}$
$=\sqrt{3}$
Unit vector perpendicular to $\vec{\text{a}}$ and $\vec{\text{b}}$ is, $\frac{\vec{\text{a}}\times\vec{\text{b}}}{\big|\vec{\text{a}}\times\vec{\text{b}}\big|}=\frac{1}{\sqrt{3}}\big(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$
View full question & answer→Question 882 Marks
If $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ are non-coplanar vectors, prove that the given vectors are non-coplanar:
$2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}},\ \vec{\text{a}}+\vec{\text{b}}-2\vec{\text{c}}$ and $\vec{\text{a}}+\vec{\text{b}}-3\vec{\text{c}}$
AnswerLet if possible the given vectors are coplanar. Then one of the vector is expressible in the terms of the other two.
We have,
$2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}=\text{x}\big(\vec{\text{a}}+\vec{\text{b}}-2\vec{\text{c}}\big)+\text{y}\big(\vec{\text{a}}+\vec{\text{b}}-3\vec{\text{c}}\big)$
$=\vec{\text{a}}(\text{x + y})+\vec{\text{b}}(\text{x + y})+\vec{\text{c}}(-2\text{x}-3\text{y})$
$\Rightarrow\text{x + y}=2,\ \text{x + y}=-1,\ -2\text{x}-3\text{y}=3$
which is not true, as $\text{x + y}=2\neq-1$. Hence, the given vectors are non-coplanar.
View full question & answer→Question 892 Marks
Find a unit vector in the direction of $\vec{\text{a}}=2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}$.
AnswerGiven:
$\vec{\text{a}}=2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}$
$|\vec{\text{a}}|=\sqrt{2^2+(-3)^2+6^2}$
$=\sqrt{4+9+36}$
$=\sqrt{49}$
$=7$
Unit vector $=\frac{\vec{\text{a}}}{|\vec{\text{a}}|}=\frac{2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}}{7}$
$=\frac{2}7\hat{\text{i}}-\frac{3}7\hat{\text{j}}+\frac{6}7\hat{\text{k}}$
View full question & answer→Question 902 Marks
For any two vectors $\vec{\text{a}}$ and $\vec{\text{b}},$ find $\vec{\text{a}}.\big(\vec{\text{b}}\times\vec{\text{a}}\big).$
AnswerLet:
$\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$
$\vec{\text{b}}=\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$
$\vec{\text{b}}\times\vec{\text{a}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{b}_1&\text{b}_2&\text{b}_3\\\text{a}_1&\text{a}_2&\text{a}_3\end{vmatrix}$
$=\hat{\text{i}}(\text{b}_2\text{a}_3-\text{b}_3\text{a}_2)-\hat{\text{j}}(\text{b}_1\text{a}_3-\text{b}_3\text{a}_1)+\hat{\text{k}}(\text{b}_1\text{a}_2-\text{b}_2\text{a}_1)$
Now,
$\vec{\text{a}}.\big(\vec{\text{b}}\times\vec{\text{a}}\big)$
$=\big(\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}\big).\big[\hat{\text{i}}(\text{b}_2\text{a}_3-\text{b}_3\text{a}_2)\\-\hat{\text{j}}(\text{b}_1\text{a}_3-\text{b}_3\text{a}_1)+\hat{\text{k}}(\text{b}_1\text{a}_2-\text{b}_2\text{a}_1)\big]$
$=\text{a}_1(\text{b}_2\text{a}_3-\text{b}_3\text{a}_2)-\text{a}_2(\text{b}_1\text{a}_3-\text{b}_3\text{a}_1)+\text{a}_3(\text{b}_1\text{a}_2-\text{b}_2\text{a}_1)$
$=\text{a}_1\text{b}_2\text{a}_3-\text{a}_1\text{b}_3\text{a}_2-\text{a}_2\text{b}_1\text{a}_3+\text{a}_2\text{b}_3\text{a}_1+\text{a}_3\text{b}_1\text{a}_2-\text{a}_3\text{b}_2\text{a}_1$
$=0$
View full question & answer→Question 912 Marks
If a vector makes angles $\alpha,\beta,\gamma$ with OX, OY and OZ respectively. then write the value of $\sin^2\alpha+\sin^2\beta+\sin^2\gamma$.
AnswerSuppose, a vector $\overrightarrow{\text{OP}}$ makes an angle $\alpha,\beta,\gamma$ with OX, OY and OZ respectively.Then direction consines of the vector are given by $\text{l}=\cos\alpha,\ \text{m}=\cos\beta,\ \text{n}=\cos\gamma$Consider,
$\sin^2\alpha+\sin^2\beta+\sin^2\gamma\\=1-\cos^2\alpha+1-\cos^2\beta+1-\cos^2\gamma$
$=3-(\cos^2\alpha+\cos^2\beta+\cos^2\gamma)$
$=3-(\text{l}^2+\text{m}^2+\text{n}^2)$
$=3-1$ $[\because\ \text{l}^2+\text{m}^2+\text{n}^2=1]$
$=2$
View full question & answer→Question 922 Marks
Find the angle between the vectors $\vec{\text{a}} $ and $\vec{\text{b}},$ where$\vec{\text{a}}=2\hat {\text{i}}-3\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}} =\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$
AnswerLet $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$.$\big|\vec{\text{a}}\big|=\sqrt{(2)^2+(-3)^2+(1)^{2}}=\sqrt{14}$
$\big|\vec{\text{b}}\big|=\sqrt{(1)^2+(1)^2+(-2)^{2}}=\sqrt{6}$
$\vec{\text{a}}.\vec{\text{b}}=2-3-2=-3$
$\cos\theta=\frac{\vec{\text{a }}.\vec{\text{ b}}}{\big|\vec{a}\big|\big|\vec{b}\big|}=\frac{-3}{\sqrt{14}\sqrt{6}}=\frac{-3}{\sqrt{84}}$
$\Rightarrow\theta=\cos^{-1}\Big(\frac{-3}{\sqrt{84}}\Big)$
View full question & answer→Question 932 Marks
Write the value of $\lambda$ so that vectora $\vec{\text{a}}=2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$ are perpendicular to each other.
AnswerWe have
$\vec{\text{a}}=2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$
The given vectors are perpendicular. so, their dot product is zero.
$\big(2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}\big).\big(\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\big)$
$\Rightarrow2-2\lambda+3=0$
$\Rightarrow5-2\lambda=0$
$\Rightarrow-2\lambda=-5$
$\Rightarrow\lambda=\frac{5}{2}$
View full question & answer→Question 942 Marks
Write two different vectors having same direction.
AnswerLet $\vec{\text{p}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ and $\vec{\text{q}}=2\hat{\text{i}}+4\hat{\text{j}}+6\hat{\text{k}}$
Then, direction cosines of $\vec{\text{p}}$ are
$\text{l}=\frac{1}{\sqrt{1^2+2^2+3^2}}=\frac{1}{\sqrt{14}},\text{m}=\frac{2}{\sqrt{1^2+2^2+3^2}}=\frac{2}{\sqrt{14}}$ and $\text{n}=\frac{3}{\sqrt{1^2+2^2+3^2}}=\frac{3}{\sqrt{14}}$
Direction cosines of $\vec{\text{q}}$ are
$\text{l}=\frac{2}{\sqrt{2^2+4^2+6^2}}=\frac{2}{2\sqrt{14}}=\frac{1}{\sqrt{14}},$ $\text{m}=\frac{4}{\sqrt{2^2+4^2+6^2}}=\frac{4}{2\sqrt{14}}=\frac{2}{\sqrt{14}}$ and $\text{n}=\frac{6}{\sqrt{2^2+4^2+6^2}}=\frac{6}{2\sqrt{14}}=\frac{3}{\sqrt{14}}$
The direction cosines of two vectors are same. Hence the two different vectors $\vec{\text{p}},\vec{\text{q}}$ have same directions.
View full question & answer→Question 952 Marks
Show that the vector $\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ is equally inclined with the axes OX, OY and OZ.
AnswerLet $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
Then,
$|\vec{\text{a}}|=\sqrt{1^1+1^1+1^1}=\sqrt{3}$
Therefore, the direction ratios of $\vec{\text{a}}$ are $\Big(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\Big).$
Now, let $\alpha,\beta$ and $\gamma$ be the angles format by $\vec{\text{a}}$ with the positive directions of x, y and z axes.
Then, we have $\cos\alpha=\frac{1}{\sqrt{3}},\cos\beta=\frac{1}{\sqrt{3}},\cos\gamma=\frac{1}{\sqrt{3}}.$
Hence, the given vector is equally inclined to axes OX, OY and OZ.
View full question & answer→Question 962 Marks
Write the direction cosines of the vector $\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ .
AnswerGiven: $\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
Then, direction cosines are
$\frac{1}{\sqrt{1^2+2^2+3^2}},\frac{2}{\sqrt{1^2+2^2+3^2}},\frac{3}{\sqrt{1^2+2^2+3^2}}$ or, $\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}$
View full question & answer→Question 972 Marks
Find the components along the coordinate axis of the position vector of the following point:Q(-5, 1)
AnswerHere, Q = (-5, 1)
Position vector of $\text{Q}=-5\hat{\text{i}}+\hat{\text{j}}$
Component of Q along x-axis $=-5\hat{\text{i}}$
Component of Q along x-axis $=\hat{\text{j}}$
View full question & answer→Question 982 Marks
Write the component of $\vec{\text{b}}$ along $\vec{\text{a}}.$
AnswerComponent of $\vec{\text{b}}$ on $\vec{\text{a}}$ is
$\Big\{\frac{\vec{\text{a}}.\vec{\text{b}}}{|\vec{\text{a}}|}\Big\}\hat{\text{a}}=\Bigg\{\frac{\big(\vec{\text{a}}.\vec{\text{b}}\big)}{|\vec{\text{a}}|^2}\Bigg\}\vec{\text{a}}=\frac{\big(\vec{\text{a}}.\vec{\text{b}}\big)\vec{\text{a}}}{|\vec{\text{a}}|^2}$
View full question & answer→Question 992 Marks
Find $\big|\vec{\text{a}}-\vec{\text{b}}\big|$ if
$|\vec{\text{a}}|=2,\big|\vec{\text{b}}\big|=5$ and $\vec{\text{a}}.\vec{\text{b}}=8$
AnswerGiven that$|\vec{\text{a}}|=2,\big|\vec{\text{b}}\big|=5$ and $\vec{\text{a}}.\vec{\text{b}}=8\dots(1)$
We know that
$\big|\vec{\text{a}}-\vec{\text{b}}\big|^2=|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-2\vec{\text{a}}.\vec{\text{b}}$
$=2^2+5^2-2(8)$ [using (1)]
$=4+25-16$
$=13$
$\therefore \big|\vec{\text{a}}-\vec{\text{b}}\big|=\sqrt{13}$
View full question & answer→Question 1002 Marks
If the $\vec{\text{a}}$ and $\vec{\text{b}}$ are such that $|\vec{\text{a}}|=3,\big|\vec{\text{b}}\big|=\frac{2}{3}$ and $\vec{\text{a}}\times\vec{\text{b}}$ is a unit vector, then the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
AnswerLet the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ be $\theta.$
It is given that $\vec{\text{a}}\times\vec{\text{b}}$ is a unit vector.
$\therefore\big|\vec{\text{a}}\times\vec{\text{b}\big|}=1$
$\Rightarrow|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta=1$
$\Rightarrow3\times\frac{2}{3}\times\sin\theta=1$
$\Rightarrow\sin\theta=\frac{1}{2}$
$\Rightarrow\theta=\frac{\pi}{6}$
Thus, the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ is $\frac{\pi}{6}.$
View full question & answer→Question 1012 Marks
For any two non-zero vectors, write the value of $\frac{\big|\vec{\text{a}}+\vec{\text{b}}\big|^2+\big|\vec{\text{a}}-\vec{\text{b}}\big|^2}{|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2}.$
AnswerWe have
$\frac{\big|\vec{\text{a}}+\vec{\text{b}}\big|^2+\big|\vec{\text{a}}-\vec{\text{b}}\big|^2}{|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2}$
$=\frac{|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+2\vec{\text{a}}.\vec{\text{b}}+|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-2\vec{\text{a}}.\vec{\text{b}}}{|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2}$
$=\frac{2\big(|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2\big)}{|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2}$
$=2$
View full question & answer→Question 1022 Marks
Define vector product of two vectors.
AnswerIf $\vec{\text{a}}$ and $\vec{\text{b}}$ are two non-zero non-parallel vectors, then the vector product denoted by $\vec{\text{a}}\times\vec{\text{b}}$ is defined as $\vec{\text{a}}\times\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta\eta.$
Here, $\theta$ is the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ and $\eta$ is
the unit vector perpendicular to the plane of $\vec{\text{a}}$ and $\vec{\text{b}}$ such that $\vec{\text{a}},\vec{\text{b}}$ and $\eta$ from a right handed system.
View full question & answer→Question 1032 Marks
Find the magnitude of two vectors $\vec{a}\ \text{and}\ \vec{b},$ having the same magnitude and such that the angle between them is 60° and their scalar product is $\frac{1}{2}\cdot$
Answer$\text{Given:}\ \ \big|\vec{a}\big|=\Big|\vec{b}\Big|, \text{angle}\ \theta$ (say) between $\vec{a}\ \text{and}\ \vec{b},$ is 60° and their scalar (i.e., dot) product $=\frac{1}{2}$ $\Rightarrow\ \ \vec{a}.\vec{b}=\frac{1}{2}\ \ \Rightarrow\ \ \big|\vec{a}.\Big|\vec{b}\Big|\ \text{cos}\ \theta=\frac{1}{2}$ $\text{Putting}\ \big|\vec{a}\big|=\Big|\vec{b}\Big| \ \text{and}\ \theta=60^\circ,\ \text{we have}\ \ \ \big|\vec{a}\big|.\big|\vec{a}\big|\ \text{cos}\ 60^\circ=\frac{1}{2}$ $\Rightarrow\ \ \big|\vec{a}|^2.\Big(\frac{1}{2}\Big)=\frac{1}{2}\ \ \Rightarrow\ \ \big|\vec{a}\big|^2=1\ \ \Rightarrow\ \ \big|\vec{a}\big|=1$$\therefore\ \ \Big|\vec{b}\Big|=\big|\vec{a}\big|=1$
$\therefore\ \ \big|\vec{a}\big|=1\ \text{and}\ \Big|\vec{b}\Big|=1$
View full question & answer→Question 1042 Marks
Find the position vector of the min-point of the line segment AB, where A is the point (3, 4, -2) and B is the point (1, 2, 4).
AnswerGiven: A(3, 4, -2) and B(1, 2, 4) Let C is the mid-point of AB $\therefore$ Position vector of $\text{C}=\frac{3\hat{\text{i}}+4\hat{\text{j}}-2\hat{\text{k}}+\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}}2$$=\frac{4\hat{\text{i}}+6\hat{\text{j}}+2\hat{\text{k}}}2$
$=2\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}$
View full question & answer→Question 1052 Marks
If $\vec{\text{a}}=3\hat{\text{i}}+4\hat{\text{j}}$ and $\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},$ find the value of $\big|\vec{\text{a}}\times\vec{\text{b}}\big|.$
AnswerGiven:
$\vec{\text{a}}=3\hat{\text{i}}+4\hat{\text{j}}+0\hat{\text{k}}$
$\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
$\therefore\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\3&4&0\\1&1&1 \end{vmatrix}$
$=(4-0)\hat{\text{i}}-(3-0)\hat{\text{j}}+(3-4)\hat{\text{k}}$
$=4\hat{\text{i}}-3\hat{\text{j}}-\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\sqrt{16+9+1}$
$=\sqrt{26}$
View full question & answer→Question 1062 Marks
If $\vec{\text{c}}$ is a unit vector perpendicular to the vectors $\vec{\text{a}}$ and $\vec{\text{b}},$ write another unit vector perpendicular to $\vec{\text{a}}$ and $\vec{\text{b}}.$
Answer$\vec{\text{c}}$ is a unit vector perpendicular to both $\vec{\text{a}}$ and $\vec{\text{b}}.$
$\Rightarrow\vec{\text{c}}=\frac{\vec{\text{a}}\times\vec{\text{b}}}{\big|\vec{\text{a}}\times\vec{\text{b}}\big|}$
$\Rightarrow-\vec{\text{c}}=\frac{\vec{\text{b}}\times\vec{\text{a}}}{\big|\vec{\text{a}}\times\vec{\text{b}}\big|}$
Therefore, $-\vec{\text{c}}$ is perpendicular to $\vec{\text{b}}$ and $\vec{\text{a}}.$
Thus, $-\vec{\text{c}}$ is another unit vector perpendicular to $\vec{\text{a}}$ and $\vec{\text{b}}.$
View full question & answer→Question 1072 Marks
Find the angle between the vectors $\vec{\text{a}} $ and $\vec{\text{b}},$ where$\vec{\text{a}}=\hat {\text{i}}-\hat{\text{j}}$ and $\vec{\text{b}} = \hat{\text{j}}+\hat{\text{k}}$
AnswerLet $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$.
$\big|\vec{\text{a}}\big|=\sqrt{(1)^2+(-1)^2}=\sqrt{2}$
$\big|\vec{\text{b}}\big|=\sqrt{(1)^2+(1)^2}=\sqrt{2}$
$\vec{\text{a}}.\vec{\text{b}}=0-1+0=-1$
$\cos\theta=\frac{\vec{\text{a }}.\vec{\text{ b}}}{\big|\vec{a}\big|\big|\vec{b}\big|}=\frac{-1}{\sqrt{2}\sqrt{2}}=\frac{-1}{2}$
$\Rightarrow\theta=\cos^{-1}\big(\frac{-1}{2}\big)=\frac{2\pi}{3}$
View full question & answer→Question 1082 Marks
Find the area of the parallelogram whose adjacent sides are determined by the vectors $\vec{a}=\hat{i}-\hat{j}+3\hat{k}\ \text{and}\ \vec{b}=2\hat{i}-7\hat{j}+\hat{k}.$
Answer Given: Vectors representing two adjacent sides of a parallelogram are $\vec{a}=\hat{i}-\hat{j}+3\hat{k}\ \text{and}\ \vec{b}=2\hat{j}-7\hat{j}+\hat{k}$ $\therefore\ \vec{a}\times\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-1&3\\2&-7&1\end{vmatrix}=\hat{i}(-1+21)-\hat{j}(1-6)+\hat{k}(-7+2)$ $=20\hat{i}+5\hat{j}-5\hat{k}$ $\text{Now}\ \ \text{Area of parallelogram}=\big|\vec{a}\times\vec{b}\big|=\sqrt{400+25+25}$ $=\sqrt{450}=15\sqrt{2}\ \text{sq. units}$
View full question & answer→Question 1092 Marks
Show that the following triads of vectors are coplanar:
$\vec{\text{a}}=-4\hat{\text{i}}-6\hat{\text{j}}-2\hat{\text{k}},\vec{\text{b}}=-\hat{\text{i}}+4\hat{\text{j}}+3\hat{\text{k}},\vec{\text{c}}=-8\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$
AnswerWe know that three vectors $\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}$ are coplanar iff their scalar triple product is zero i.e. $\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=0.$
Here,
$\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=\begin{vmatrix}-4&-6&-2\\-1&4&3\\-8&-1&3 \end{vmatrix}\\$
$=-4(12+3)+6(-3+24)-2(1+32)$
$= -60 + 126 -66$
$= 0$
Hence, the given vectors are coplanar.
View full question & answer→Question 1102 Marks
Write the angle between the vectors $\vec{\text{a}}\times\vec{\text{b}}$ and $\vec{\text{b}}\times\vec{\text{a}}.$
Answer$\vec{\text{b}}\times\vec{\text{a}}=-\vec{\text{a}}\times\vec{\text{b}}$
So, $\vec{\text{a}}\times\vec{\text{b}}$ and $\vec{\text{b}}\times\vec{\text{a}}$ are vectors of same magnitude but opposite in direction.
Thus, the angle between the vectors $\vec{\text{a}}\times\vec{\text{b}}$ and $\vec{\text{b}}\times\vec{\text{a}}$ is $180^\circ.$
View full question & answer→Question 1112 Marks
Write a unit vector making equal acute angles with the coordinates axes.
AnswerSuppose $\vec{\text{r}}$ makes an angle $\alpha$ wuth each of the axis OX, OY and OZ.
Then, its direction cosines are $\text{l}=\cos\alpha,\ \text{m}=\cos\alpha,\ \text{n}=\cos\alpha$.
Now,
$\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Rightarrow\ \text{l}^2+\text{l}^2+\text{l}^2=1$ $[\because\text{l = m = n}]$
$\Rightarrow\ 3\text{l}^2=1$
$\Rightarrow\ \text{l}^2=\frac{1}3$
$\Rightarrow\ \text{l}=\pm\frac{1}{\sqrt3}$
Since the angle is acute Hence, we take only positive value
Therefore, unit vector is $\Big(\frac{1}{\sqrt3}\hat{\text{i}}+\frac{1}{\sqrt3}\hat{\text{j}}+\frac{1}{\sqrt3}\hat{\text{k}}\Big)$.
View full question & answer→Question 1122 Marks
Find the vector in the direction of vector $2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}$ which has magnitude 21 units.
AnswerLet $\vec{\text{a}}=2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}$
$\therefore\ |\vec{\text{a}}|=\sqrt{2^2+(-3)^2+6^2}$
$=\sqrt{4+9+36}$
$=\sqrt{49}$
$=7$
Unit vector in the direction of $\vec{\text{a}}=\frac{\vec{\text{a}}}{|\vec{\text{a}}|}=\frac{2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}}7$
$\therefore$ vector in the direction of vector $\vec{\text{a}}$ which has magnitude 21 units
$=21\times\Big(\frac{2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}}7\Big)$
$=3\big(2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}\big)$
$=6\hat{\text{i}}-9\hat{\text{j}}+18\hat{\text{k}}$
View full question & answer→Question 1132 Marks
Write the length (magnitude) of a vector whose projections on the coordinate axes are 12, 3 and 4 units.
Answer Given: Projections on the coordinate axes are 12, 3, 4 units. Therefore,
Length of vector $=\sqrt{12^2+3^2+4^2}$
$=\sqrt{169}$
$=13$
View full question & answer→Question 1142 Marks
Write the value of $\big(\hat{\text{i}}\times\hat{\text{j}}\big).\hat{\text{k}}+\big(\hat{\text{j}}+\hat{\text{k}}\big).\hat{\text{j}}$
Answer$\big(\hat{\text{i}}\times\hat{\text{j}}\big).\hat{\text{k}}+\big(\hat{\text{j}}+\hat{\text{k}}\big).\hat{\text{j}}$
$=\hat{\text{k}}.\hat{\text{k}}+\hat{\text{j}}.\hat{\text{j}}+\hat{\text{k}}.\hat{\text{j}}$ $\big(\because\hat{\text{i}}\times\hat{\text{j}}=\hat{\text{k}}\big)$
$=|\hat{\text{k}}|^2+|\hat{\text{j}}|^2+0$ $\big(\because\hat{\text{k}}.\hat{\text{j}}=0\big)$
$=1^2+1^2$
$=2$
View full question & answer→Question 1152 Marks
If $|\vec{\text{a}}|=4$ and $-3\leq\lambda\leq2$, then write the range of $|\lambda\vec{\text{a}}|$.
AnswerIt is given that
$-3\leq\lambda\leq2$
$\Rightarrow\ -3\times|\vec{\text{a}}|\leq\lambda|\vec{\text{a}}|\leq2\times|\vec{\text{a}}|$
$\Rightarrow\ -3\times4\leq|\lambda\vec{\text{a}}|\leq2\times4$ $\big(\text{k}|\vec{\text{a}}|=|\text{k}\vec{\text{a}}|,\ \text{k}$ is scalar$\big)$
$\Rightarrow\ -12\leq|\lambda\vec{\text{a}}|\leq8$
Thus, the range of $|\lambda\vec{\text{a}}|$ is [-12, 8]
View full question & answer→Question 1162 Marks
If $|\vec{\text{a}}|=10,\big|\vec{\text{b}}\big|=2$ and $\big|\vec{\text{a}}\times\vec{\text{b}}\big|=16,$ find $\vec{\text{a}}.\vec{\text{b}}.$
AnswerWe know
$\big(\vec{\text{a}}.\vec{\text{b}}\big)^2+\big|\vec{\text{a}}\times\vec{\text{b}}\big|^2=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2$
$\Rightarrow\big(\vec{\text{a}}.\vec{\text{b}}\big)^2+(16)^2=(10)^2\times2^2$ $\big(\because\big|\vec{\text{a}}\times\vec{\text{b}}\big|=16,|\vec{\text{a}}|=10\text{ and }\big|\vec{\text{b}}\big|=2\big)$
$\Rightarrow\big(\vec{\text{a}}.\vec{\text{b}}\big)^2+256=400$
$\Rightarrow\big(\vec{\text{a}}.\vec{\text{b}}\big)^2=144$
$\Rightarrow\big(\vec{\text{a}}.\vec{\text{b}}\big)^2=\pm12$
View full question & answer→Question 1172 Marks
Find the projection of the vector $\hat{i}+3\hat{j}+7\hat{k}$ on the vector $7\hat{i}-\hat{j}+8\hat{k}.$
Answer$\text{Let}\ \ \vec{a}=\hat{i}+3\hat{j}+7\hat{k}\ \text{and}\ \vec{b}=7\hat{i}-\hat{j}+8\hat{k}$
Porjection of vector $\vec{a}\ \text{and}\ \vec{b}=\frac{\vec{a}.\vec{b}}{\big|\vec{b}\big|}=\frac{(1)(7)+(3)(-1)+7(8)}{\sqrt{(7)^2+(-1)^2+(8)^2}}$ $=\frac{7-3+56}{\sqrt{49+1+64}}=\frac{60}{\sqrt{114}}$
View full question & answer→Question 1182 Marks
Find the position vector of the mid point of the vector joining the points P(2, 3, 4) and Q(4, 1, –2).
AnswerThe position vector of mid-point R of the vector joining points P(2, 3, 4) and Q(4, 1, -2) is given by,
$\overrightarrow{\text{OR}}=\frac{\big(2\hat{i}+3\hat{j}+4\hat{k}\big)+\big(4\hat{i}+\hat{j}-2\hat{k}\big)}{2}$ $=\frac{(2+4)\hat{i}+(3+1)\hat{j}+(4-2)\hat{k}}{2}$
$=\frac{6\hat{i}+4\hat{j}+2\hat{k}}{2}=3\hat{i}+2\hat{j}+\hat{k}$
View full question & answer→Question 1192 Marks
Find the values of 'a' for which the vectors
$\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}},\vec\beta={\text{a}}\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$ and $\vec\gamma=\hat{\text{i}}+2\hat{\text{j}}+\text{a}\hat{\text{k}}$ are coplanar.
AnswerGiven:
$\vec\alpha=\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$
$\vec\beta={\text{a}}\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$
$\vec\gamma=\hat{\text{i}}+2\hat{\text{j}}+\text{a}\hat{\text{k}}$
We know that three vectors $\vec\alpha,\vec\beta,\vec\gamma$ are coplanar if their scalar product is zero.
$\therefore\big[\vec\alpha \vec \beta \vec\gamma\big]=0$
$\begin{vmatrix}1&2&1\\\text{a}&1&2\\1&2&\text{a} \end{vmatrix}=0$
$\Rightarrow1(\text{a}-4)-2(\text{a}^2-2)+1(2\text{a}-1)=0$
$\Rightarrow-2\text{a}^2+3\text{a}-1=0$
$\Rightarrow2\text{a}^2-3\text{a}+1=0$
$\Rightarrow(\text{a}-1)(2\text{a}-1)=0$
$\Rightarrow\text{a}=1,\frac{1}{2}$
View full question & answer→Question 1202 Marks
For what vaiue of $\lambda$ are the vectors $\vec{\text{a}}=2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$ perpendicular to each other?
AnswerWe know
$\vec{\text{a}}=2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$
Given, $\vec{\text{a}}$ and $\vec{\text{b}}$ are perpendicular. so, their dot product is zero.
$\Rightarrow\big(2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}\big).\big(\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\big)=0$
$\Rightarrow2-2\lambda+3=0$
$\Rightarrow-2\lambda+5=0$
$\therefore\lambda=\frac{5}{2}$
View full question & answer→Question 1212 Marks
Find the unit vector in the direction of vector $\overrightarrow{\text{PQ}}$ where P and Q are the points (1, 2, 3) and (4, 5, 6), respectively.
AnswerThe given points are P (1, 2, 3) and Q (4, 5, 6).$\therefore\overrightarrow{\text{PQ}}=(4-1)\hat{i}+(5-2)\hat{j}+(6-3)\hat{k}=3\hat{i}+3\hat{j}+3\hat{k}$
$\Bigg|\overrightarrow{\text{PQ}}\Bigg|=\sqrt{3^2+3^2+3^2}=\sqrt{9+9+9}=\sqrt{27}=3\sqrt{3} $
Hence, the unit vector in the direction of $\overrightarrow{\text{PQ}}$ is $\frac{\overrightarrow{\text{PQ}}}{\Big|\overrightarrow{\text{PQ}}\Big|}=\frac{3\hat{i}+3\hat{j}+3\hat{k}}{3\sqrt{3}}=\frac{1}{\sqrt{3}}\hat{i}+\frac{1}{\sqrt{3}}\hat{j}+\frac{1}{\sqrt{3}}\hat{k}$
View full question & answer→Question 1222 Marks
Write the unit vector in the direction of $\vec{\text{a}}=3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}$.
AnswerWe have,$\vec{\text{a}}=3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}$
$|\vec{\text{a}}|=\sqrt{3^2+(-2)^2+6^2}$ $=\sqrt{9+4+36}$ $=\sqrt{49}$ $=7$ $\therefore$ Unit vector in the direction of $\vec{\text{a}}=\hat{\text{a}}=\frac{\vec{\text{a}}}{|\vec{\text{a}}|}=\frac{1}7\big(3\hat{\text{i}}-2\hat{\text{j}}+6\hat{\text{k}}\big)=\frac{3}7\hat{\text{i}}-\frac{2}7\hat{\text{j}}+\frac{6}7\hat{\text{k}}$
View full question & answer→Question 1232 Marks
Prove that $\big(\vec{\text{a}}+\vec{\text{b}}\big)\cdot\big(\vec{\text{a}}+\vec{\text{b}}\big)=\big|\vec{\text{a}}\big|^2+\Big|\vec{\text{b}}\Big|^2,$ if and only if $\vec{\text{a}},\vec{\text{b}}$ are perpendicular, given $\vec{\text{a}}\neq\vec{\text{0}},\vec{\text{b}}\neq\vec{\text{0}}.$
Answer$\Big(\vec{\text{a}}+\vec{\text{b}}\Big)\cdot\Big(\vec{\text{a}}+\vec{\text{b}}\Big)=\big|\vec{\text{a}}\big|^2+\Big|\vec{\text{b}}\Big|^2$
$\Leftrightarrow\vec{\text{a}}\cdot\vec{\text{a}}+\vec{\text{a}}\cdot\vec{\text{b}}+\vec{\text{b}}\cdot\vec{\text{a}}+\vec{\text{b}}\cdot\vec{\text{b}}=\big|\vec{\text{a}}\big|^2+\Big|\vec{\text{b}}\Big|^2$ [Distributivity of scalar products over addition]
$\Leftrightarrow\big|\vec{\text{a}}\big|^2+2\vec{\text{a}}\cdot\vec{\text{b}}+\Big|\vec{\text{b}}\Big|^2=\big|\vec{\text{a}}\big|^2+\Big|\vec{\text{b}}\Big|^2\ \ \ $ $\Big[\vec{\text{a}}\cdot\vec{\text{b}}=\vec{\text{b}}\cdot\vec{\text{a}}\ \text{(Scalar product is commutative)}\Big]$
$\Leftrightarrow2\vec{\text{a}}\cdot\vec{\text{b}}=0$
$\Leftrightarrow\vec{\text{a}}\cdot\vec{\text{b}}=0$
$\therefore\vec{\text{a}}\ \text{and}\ \vec{\text{b}} \ \text{are perpendicular.}$ $\ \ \ \Big[\vec{\text{a}}\neq\vec{\text{0}},\ \vec{\text{b}}\neq\vec{\text{0}}\ \text{(Given)}\Big]$
View full question & answer→Question 1242 Marks
Find $\vec{\text{a}}.\big(\vec{\text{b}}\times\vec{\text{c}}\big), $ if $\vec{\text{a}}=2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}},\vec{\text{b}}=-\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{c}}=3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$.
AnswerThe given vectors are $\vec{\text{a}}=2\hat{\text{i}}+\hat{\text{j}}+3\vec{\text{k}},\vec{\text{b}}=-\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{c}}=3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$
Now,
$\vec{\text{b}}\times\vec{\text{c}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\-1&2&1\\3&1&2 \end{vmatrix}=3\hat{\text{i}}+5\hat{\text{j}}-7\hat{\text{k}}$
$\therefore\vec{\text{a}}\big(\vec{\text{b}}\times\vec{\text{c}}\big)=\big(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}\big).\big(3\hat{\text{i}}+5\hat{\text{j}}-7\hat{\text{k}}\big)$
$=2\times3+1\times5=3\times(-7)$
$=6+5-21=-10$
View full question & answer→Question 1252 Marks
Find the values of x and y so that the vectors $2\hat{i}+3\hat{j}\ \text{and}\ x\hat{i}\ +y\hat{j}$ are equal.
AnswerThe two vectors $2\hat{i}+3\hat{j}\ \text{and}\ x\hat{i}\ +y\hat{j}$ will be equal if their corresponding components are equal.
Hence, the required values of x and y are 2 and 3 respectively.
View full question & answer→Question 1262 Marks
Find the value of x for which $\text{x}\Big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\Big)$ is a unit vector.
Answer$\text{x}\Big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\Big)\ \text{is a unit vector if}\ \Big|\text{x}\Big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\Big)\Big|=1$Now,
$\Big|\text{x}\Big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\Big)\Big|=1$ $\Rightarrow{}\sqrt{\text{x}^{2}+\text{x}^{2}+\text{x}^{2}}=1$ $\Rightarrow\sqrt{3\text{x}^2}=1$ $\Rightarrow\sqrt{3\text{x}}=1$ $\Rightarrow\text{x}=\pm\frac{1}{\sqrt{3}}$Hence, the required value of x is $\pm\frac{1}{\sqrt{3}}$
View full question & answer→Question 1272 Marks
If $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ are mutually perpendicular unit vectors, write the value of $\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|.$
AnswerGiven that $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ are unit vectors.
So, $|\vec{\text{a}}|=1,\big|\vec{\text{b}}\big|=1$ and $|\vec{\text{c}}|=1\dots(1)$
Since they are mutually perpendicular,
$\vec{\text{a}}.\vec{\text{b}}=\vec{\text{b}}.\vec{\text{c}}=\vec{\text{c}}.\vec{\text{a}}=0\dots(2)$
Now,
$\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|^2=|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|+|\vec{\text{c}}|^2+2\vec{\text{a}}.\vec{\text{b}}+2\vec{\text{b}}.\vec{\text{c}}+2\vec{\text{c}}.\vec{\text{a}}$
$=1+1+1+0+0+0$ [using (1) and (2)]
$=3$
$\therefore\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|=\sqrt{3}$
View full question & answer→Question 1282 Marks
If $\vec{\text{a}}$ is a unit vector such that $\vec{\text{a}}\times\hat{\text{i}}=\hat{\text{j}},$ find $\vec{\text{a}}.\hat{\text{i}}.$
AnswerWe know
$\hat{\text{k}}\times\hat{\text{i}}=\hat{\text{j}}\dots(1)$
Given: $\vec{\text{a}}\times\hat{\text{i}}=\hat{\text{j}}\dots(2)$
Comparing (1) and (2), we get
$\vec{\text{a}}=\hat{\text{k}}$
Now,
$\vec{\text{a}}.\hat{\text{i}}=\hat{\text{k}}.\hat{\text{i}}$
$=0$
View full question & answer→Question 1292 Marks
$\text{Find}\ |\vec{a}\ \times\vec{b}|,\ \text{if}\ \vec{a}=\hat{i}-7\hat{j}+7\hat{k}\ \text{and}\ \vec{b}=3\hat{i}-2\hat{j}+2\hat{k}.$
Answer$\text{Give:}\ \ \vec{a}=\hat{i}-7\hat{j}+7\hat{k}\ \text{and}\ \vec{b}=3\hat{i}-2\hat{j}+2\hat{k}$
$\vec{a}\times\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-7&7\\3&-2&2\end{vmatrix}$
Expanding along first row,
$\vec{a}\times\vec{b}=\hat{i}\begin{vmatrix}-7&7\\-2&2\end{vmatrix}-\hat{j}\begin{vmatrix}1&7\\3&2\end{vmatrix}+\hat{k}\begin{vmatrix}1&-7\\3&-2\end{vmatrix}$ $=\hat{i}(-14+14)-\hat{j}(2-21)+\hat{k}(-2+21)$
$=0\hat{i}+19\hat{j}+19\hat{k}$
$\therefore\ \ \Big|\vec{a}\times\vec{b}\Big|=\sqrt{(0)^2+(19)^2+(19)^2}=\sqrt{2(19)^2}=19\sqrt{2}$
View full question & answer→Question 1302 Marks
Write the value $\big(\hat{\text{i}}\times\hat{\text{j}}\big).\hat{\text{k}}+\hat{\text{i}}.\hat{\text{j}}$
Answer $\big(\hat{\text{i}}\times\hat{\text{j}}\big).\hat{\text{k}}+\hat{\text{i}}.\hat{\text{j}}$$=\hat{\text{k}}.\hat{\text{k}}+0$
$=|\hat{\text{k}}|^2+0$
$=1^2+0$ $\big(\because|\text{k}|=1\big)$
$=1$
View full question & answer→Question 1312 Marks
If P, Q and R are three collinear points such that $\overrightarrow{\text{PQ}}=\vec{\text{a}}\text{ and }\overrightarrow{\text{QR}}=\vec{\text{b}}$. Find the vector $\overrightarrow{\text{PR}}$.
Answer Given that, P, Q, R are collinear.It also given that, $\overrightarrow{\text{PQ}}=\vec{\text{a}}\text{ and }\overrightarrow{\text{QR}}=\vec{\text{b}}$
$\overrightarrow{\text{PR}}=\overrightarrow{\text{PQ}}+\overrightarrow{\text{QR}}$
$=\vec{\text{a}}+\vec{\text{b}}$
$\overrightarrow{\text{PR}}=\vec{\text{a}}+\vec{\text{b}}$
View full question & answer→Question 1322 Marks
Write the value of $\hat{\text{i}}\times\big(\hat{\text{j}}+\hat{\text{k}}\big)+\hat{\text{j}}\times\big(\hat{\text{k}}+\hat{\text{i}}\big)+\hat{\text{k}}\times\big(\hat{\text{i}}+\hat{\text{j}}\big).$
Answer$\hat{\text{i}}\times\big(\hat{\text{j}}+\hat{\text{k}}\big)+\hat{\text{j}}\times\big(\hat{\text{k}}+\hat{\text{i}}\big)+\hat{\text{k}}\times\big(\hat{\text{i}}+\hat{\text{j}}\big)$
$=\big(\hat{\text{i}}\times\hat{\text{j}}\big)+\big(\hat{\text{i}}\times\hat{\text{k}}\big)+\big(\hat{\text{j}}\times\hat{\text{k}}\big)+\big(\hat{\text{j}}\times\hat{\text{i}}\big)+\big(\hat{\text{k}}\times\hat{\text{i}}\big)+\big(\hat{\text{k}}\times\hat{\text{j}}\big)$
$=\hat{\text{k}}-\hat{\text{j}}+\hat{\text{i}}-\hat{\text{k}}+\hat{\text{j}}-\hat{\text{i}}=\vec{0}$
View full question & answer→Question 1332 Marks
If $|\vec{\text{a}}|=2,\big|\vec{\text{b}}\big|=5$ and $\vec{\text{a}}.\vec{\text{b}}=2,$ find $\big|\hat{\text{a}}-\hat{\text{b}}\big|.$
AnswerWe have $|\vec{\text{a}}|=2,\big|\vec{\text{b}}\big|=5$ and $\vec{\text{a}}.\vec{\text{b}}=2$
Now,$\big|\hat{\text{a}}-\hat{\text{b}}\big|^2=|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-2\vec{\text{a}}.\vec{\text{b}}$
$=2^2+5^2-2(2)$
$=4+25-4$
$=25$
$\therefore\big|\vec{\text{a}}-\vec{\text{b}}\big|=\sqrt{25}=5$
View full question & answer→Question 1342 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are two vectors such that $\big|\vec{\text{a}}.\vec{\text{b}}\big|=\big|\vec{\text{a}}\times\vec{\text{b}}\big|,$ write the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
AnswerLet $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
We know
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\big|\sin\theta\big|$
$\big|\vec{\text{a}}.\vec{\text{b}}\big|=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\big|\cos\theta\big|$
Now,
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\big|\vec{\text{a}}.\vec{\text{b}}\big|$ (Given)
$\Rightarrow|\vec{\text{a}}|\big|\vec{\text{b}}\big||\sin\theta|=|\vec{\text{a}}|\big|\vec{\text{b}}\big||\cos\theta|$
$\Rightarrow|\sin\theta|=|\cos\theta|$
$\Rightarrow\theta=\frac{\pi}{4}$
View full question & answer→Question 1352 Marks
Let the vectors $\vec{a},\vec{b},\vec{c}$ be given as $a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k},\ b_{1}\hat{i}+b_{2}\hat{j}+b_{3}\hat{k},$ $c_{1}\hat{i}+c_{2}\hat{j}+c_{3}\hat{k}.$ Then show that $\vec{a}\times(\vec{b}+\vec{c})=\vec{a}\times\vec{b}+\vec{a}\times\vec{c}.$
AnswerGiven: $\text{Vector}\ \vec{a}=a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k},\ \vec{b}=b_{1}\hat{i}+b_{2}\hat{j}+b_{3}\hat{k}$ $\text{and}\ \vec{c}=c_{1}\hat{i}+c_{2}\hat{j}+c_{3}\hat{k}$
$\therefore\ \ \vec{b}+\vec{c}=(b_1+c_1)\hat{i}+(b_2+c_2)\hat{j}+(b_3+c_3)\hat{k}$
$\text{Now}\ \ \ \ \text{L.H.S}=\vec{a}\times\big(\vec{b}+\vec{c}\big)= \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\ a_1& a_2 & a_3 \\ b_1+c_1& b_2+c_2 & b_3+c_3 \end{vmatrix}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\ a_1& a_2 & a_3 \\ b_1& b_2& b_3\end{vmatrix}+\begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\ a_1& a_2 & a_3 \\ c_1& c_2& c_3\end{vmatrix}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{[By Property of Determinants]}$
$=\vec{a}\times\vec{b}+\vec{a}\times\vec{c}=\text{R.H.S}.$
View full question & answer→Question 1362 Marks
Find the scalar components and magnitude of the vector joining the points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$.
AnswerThe vector joining the points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ can be obtained by,
$\overrightarrow{\text{PQ}}$ = Position vector of Q - Position vector of P
$=(\text{x}_{2}-\text{x}_{1})\hat{\text{i}}+(\text{y}_{2}-\text{y}_{1})\hat{\text{j}}+(\text{z}_{2}-\text{z}_{1})\hat{\text{k}}$
$\bigg|\overrightarrow{\text{PQ}}\bigg|=\sqrt{(\text{x}_{2}-\text{x}_{1})^{2}+(\text{y}_{2}-\text{y}_{1})^{2}+(\text{z}_{2}-\text{z}_{1})^{2}}$
Hence, the scalar components and the magnitude of the vector joining the given points are respectively $ \{(\text{x}_{2}-\text{x}_{1}),(\text{y}_{2}-\text{y}_{1}),(\text{z}_{2}-\text{z}_{1})\}\ \text{and}$ $\sqrt{(\text{x}_{2}-\text{x}_{1})^{2}+(\text{y}_{2}-\text{y}_{1})^{2}+(\text{z}_{2}-\text{z}_{1})^{2}}$
View full question & answer→Question 1372 Marks
Write the value of $\hat{\text{i}}.\big(\hat{\text{j}}\times\hat{\text{k}}\big)+\hat{\text{j}}.\big(\hat{\text{k}}\times\hat{\text{i}}\big)+\hat{\text{k}}.\big(\hat{\text{j}}\times\hat{\text{i}}\big).$
Answer$\hat{\text{i}}.\big(\hat{\text{j}}\times\hat{\text{k}}\big)+\hat{\text{j}}.\big(\hat{\text{k}}\times\hat{\text{i}}\big)+\hat{\text{k}}.\big(\hat{\text{j}}\times\hat{\text{i}}\big)$$=\hat{\text{i}}.\hat{\text{i}}+\hat{\text{j}}.\hat{\text{j}}+\hat{\text{k}}\big(-\hat{\text{k}}\big)$
$=|\hat{\text{i}}|^2+|\hat{\text{j}}|^2-|\hat{\text{k}}|^2$
$=1+1-1$ $\big(\because|\hat{\text{i}}|=1,|\hat{\text{j}}|=1\text{ and } |\hat{\text{k}}|=1\big)$
$=1$
View full question & answer→Question 1382 Marks
Show that the following triads of vectors are coplanar:
$\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}},\vec{\text{b}}=3\hat{\text{i}}+2\hat{\text{j}}+7\hat{\text{k}},\vec{\text{c}}=5\hat{\text{i}}+6\hat{\text{j}}+5\hat{\text{k}}$
AnswerWe know that three vectors $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are coplanar iff their scalar triple product is zero i.e. $\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=0.$
Here,
$\Big[\vec{\text{a}}\ \vec{\text{b}}\ \vec{\text{c}}\Big]=\begin{vmatrix}1&2&-1\\3&2&7\\5&6&5 \end{vmatrix}$
$=1(1-42)-2(15-35)-1(18-10)$
$=-60+126-66$
$=0$
Hence, the given vectors are coplanar.
View full question & answer→Question 1392 Marks
If $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},\ \vec{\text{b}}=2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$and $\vec{\text{c}}=\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$, find a unit vector parallel to $2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}$.
Answer We have, $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},\ \vec{\text{b}}=2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$and $\vec{\text{c}}=\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$
$\therefore\ 2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}=2\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)\\-\big(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}\big)+3\big(\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\big)$
$=3\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}$
A unit vector parallel to $2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}$ is given by $\frac{2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}}{\big|2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}\big|}=\frac{(3\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}})}{\sqrt{3^2+(-3)^2+2^2}}$
$=\frac{(3\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}})}{\sqrt{22}}$
$=\frac{3}{\sqrt{22}}\hat{\text{i}}-\frac{3}{\sqrt{22}}\hat{\text{j}}+\frac{2}{\sqrt{22}}\hat{\text{k}}$
View full question & answer→Question 1402 Marks
Write the value of $\hat{\text{i}}\times\big(\hat{\text{j}}\times\hat{\text{k}}\big).$
Answer$\hat{\text{i}}\times\big(\hat{\text{j}}\times\hat{\text{k}}\big)$$=\hat{\text{i}}\times\hat{\text{i}}$
$=0$
View full question & answer→Question 1412 Marks
Evaluate the following:$\big[2\hat{\text{i}}\hat{\text{j}}\hat{\text{k}}\big]+\big[\hat{\text{i}}\hat{\text{k}}\hat{\text{j}}\big]+\big[\hat{\text{k}}\hat{\text{j}}2\hat{\text{i}}\big]$
Answer We have,
$\big[2\hat{\text{i}}\hat{\text{j}}\hat{\text{k}}\big]+\big[\hat {\text{i}}\hat{\text{k}}\hat{\text{j}}\big]+\big[\hat{\text{k}}\hat{\text{j}}2\hat{\text{i}}\big]\\=(2\hat{\text{i}}\times\hat{\text{j}}).\hat{\text{k}}+(\hat{\text{i}}\times\hat{\text{k}}).2\hat{\text{i}}$
$=2\hat{\text{k}}.\hat{\text{k}}+(-\hat{\text{j}}).\hat{\text{j}}+(-\hat{\text{i}}).2\hat{\text{i}}$
$=2-1-2$
$=-1$
Therefore, $\big[2\hat{\text{i}}\hat{\text{j}}\hat{\text{k}}\big]+\big[\hat{\text{i}}\hat{\text{k}}\hat{\text{j}}\big]+\big[\hat{\text{k}}\hat{\text{j}}2\hat{\text{i}}\big]=-1$
View full question & answer→Question 1422 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are two vectors such that $\vec{\text{a}}.\vec{\text{b}}=6,|\vec{\text{a}}|=3$ and $\big|\vec{\text{b}}\big|=4.$ write the projection of $\vec{\text{a}}$ on $\vec{\text{b}}.$
AnswerWe have
$\vec{\text{a}}.\vec{\text{b}}=6$ and $\big|\vec{\text{b}}\big|=4$
The projection of $\vec{\text{a}}$ on $\vec{\text{b}}$ is
$\frac{\vec{\text{a}}.\vec{\text{b}}}{\big|\vec{\text{b}}\big|}$
$=\frac{6}{4}$
$=\frac{3}{2}$
View full question & answer→Question 1432 Marks
Write a vector of magnitude 12 units which makes 45º angle with x-axis, 60º angle with y-axis and an obtuse angle with z-axis.
AnswerSuppose a vector $\vec{\text{r}}$ makes an angle 45º with OX, 60º with OY and having magnitude 12 units. $\text{l}=\cos45^{\circ}=\frac{1}{\sqrt2}$ and $\text{m}=\cos60^{\circ}=\frac{1}2$Now, $\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Rightarrow\ \frac{1}2+\frac{1}4+\text{n}^2=1$
$\Rightarrow\ \text{n}^2=\frac{1}4$
$\Rightarrow\ =-\frac{1}2$ $[\because$ The angle with the z-axis is obtuse$]$
Therefore,
$\vec{\text{r}}=|\vec{\text{r}}|\big(\text{l}\hat{\text{i}}+\text{m}\hat{\text{j}}+\text{n}\hat{\text{k}}\big)$
$=12\Big(\frac{1}{\sqrt2}\hat{\text{i}}+\frac{1}2\hat{\text{j}}-\frac{1}2\hat{\text{k}}\Big)$
$=6\big(\sqrt2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$
View full question & answer→Question 1442 Marks
If G denots the centroid of $\triangle\text{ABC}$, then write the value of $\overrightarrow{\text{GA}}+\overrightarrow{\text{GB}}+\overrightarrow{\text{GC}}$.
AnswerLet $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ be the position vectors of the vertices A, B, C respectively. Then, the position vector of the centroid G is $\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3$
Thus,
$\overrightarrow{\text{GA}}+\overrightarrow{\text{GB}}+\overrightarrow{\text{GC}}$
$=\vec{\text{a}}-\Big(\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3\Big)+\vec{\text{b}}-\Big(\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3\Big)+\vec{\text{c}}-\Big(\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3\Big)$
$=\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big)-3\Big(\frac{\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}}3\Big)$
$=\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big)-\big(\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big)$
$=\vec0$
View full question & answer→Question 1452 Marks
Find a unit vector perpendicular to each of the vector $\vec{a}+\vec{b}\ \text{and}\ \vec{a}-\vec{b},\ \text{where}$ $\vec{a}=3\hat{i}+2\hat{j}+2\hat{k}\ \text{and}\ \vec{b}=\hat{i}+2\hat{j}-2\hat{k}.$
Answer$\text{Given:}\ \ \vec{a}=3\hat{i}+2\hat{j}+2\hat{k}\ \text{and}\ \vec{b}=\hat{i}+2\hat{j}-2\hat{k}$
$\text{On Adding}\ \vec{c}=\vec{a}+\vec{b}=3\hat{i}+2\hat{j+2\hat{k}}\ +\ \hat{i}+2\hat{j}-2\hat{k}$ $=4\hat{i}+4\hat{j}+0\hat{k}$
$\text{On Subtracting}\ \ \ \vec{d}=\vec{a}-\vec{b}=3\hat{i}+2\hat{j}+2\hat{k}\ - \ \hat{i}-2\hat{j}+2\hat{k}$ $=2\hat{i}+0\hat{j}+4\hat{k}$
$\text{Therefore,}\ \ \ \vec{n}=\vec{c}\times\vec{d}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\4&4&0\\2&0&4 \end{vmatrix}$
Expanding along first row $=\hat{i}(16-0)-\hat{i}(16-0)+\hat{k}(0-8)$
$\Rightarrow\ \ \vec{n}=16\hat{i}-16\hat{j}-8\hat{k}$
$\therefore\ \big|\vec{n}\big|=\sqrt{(16)^2+(-16)^2+(-8)^2}$ $\sqrt{256+256+64}=\sqrt{576}=24$
Therefore, a unit vector perpendicular to both $\vec{a}\ \text{and}\ \vec{b}\ \text{is}$
$\hat{n}=\pm\frac{\vec{n}}{|\vec{n}|}=\pm\frac{(16\hat{i}-16\hat{j}-8\hat{k})}{24}$
$=\pm\Big(\frac{16}{24}\hat{i}-\frac{16}{24}\hat{j}-\frac{8}{24}\hat{k}\Big)=\pm\Big(\frac{2}{3}\hat{i}-\frac{2}{3} \hat{j}-\frac{1}{3}\hat{k}\Big)$
View full question & answer→Question 1462 Marks
Write the angle between two vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ with magnitudes $\sqrt{3}$ and 2 respectively if $\vec{\text{a}}.\vec{\text{b}}=\sqrt{6}.$
AnswerLet $\theta$ be the angle between$\vec{\text{a}}$ and$\vec{\text{b}}.$
Given,
$|\vec{\text{a}}|=\sqrt{3};\big|\vec{\text{b}}\big|=2;\vec{\text{a}}.\vec{\text{b}}=\sqrt{6}$
We know that
$\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta$
$\Rightarrow\sqrt{6}=(\sqrt{3})(2)\cos\theta$
$\Rightarrow\cos\theta=\frac{\sqrt{6}}{2\sqrt{3}}=\frac{1}{\sqrt{2}}$
$\Rightarrow\theta=\cos^{-1}\Big(\frac{1}{\sqrt{2}}\Big)=\frac{\pi}{4}$
View full question & answer→Question 1472 Marks
Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (–5, 7).
AnswerThe vector with the initial point P (2, 1) and terminal point Q (-5, 7) can be given by, $\overrightarrow{\text{PQ}}=(-5-2)\hat{i}+(7-1)\hat{j}$ $\Rightarrow\ \overrightarrow{\text{PQ}}=-7\hat{i}+6\hat{j}$ Hence, the required scalar components are -7 and 6 while the vector component are,$-7\hat{i}\ \text{and}\ 6\hat{j}$
View full question & answer→Question 1482 Marks
$\text{Given that}\ \vec{a}\cdot\vec{b}=0\ \text{and}\ \vec{a}\times\vec{b}=\vec{0}.$ What can you conclude about the vectors $\vec{a}\ \text{and}\ \vec{b}$?
Answer$\text{Given:}\ \ \vec{a}.\vec{b}=0\ \Rightarrow\ \ \big|\vec{a}\big|.\big|\vec{b}\big|\cos\theta=0$ $\therefore\ \ \big|\vec{a}\big|=0\ \ \text{or} \ \big|\vec{b}\big|=0$$\ \text{or}\ \cos\theta=0\ \ \Rightarrow\ \ \theta=90^\circ$ $\Rightarrow \ \ \vec{a}=0\ \ \text{or}\ \ \vec{b}=0\ \ \text{or}$ $\ \ \text{vector}\ \vec{a}\ \text{is perpendicular to}\ \vec{b}.\ \ \ \ ......\text{(i)}$ $\text{Again, given}\ \vec{a}\times\vec{b}=0\ \Rightarrow\ \big|\vec{a}\times\vec{b}\big|=0$ $\ \Rightarrow\ \ \big|\vec{a}\big|.\big|\vec{b}\big|\sin\theta=0$ $\therefore\ \ \big|\vec{a}\big|=0\ \ \text{or}\ \ \big|\vec{b}\big|=0\ \ \text{or}\ \ $ $\sin\theta=0\ \ \Rightarrow\ \theta=0^\circ$ $\Rightarrow\ \ \vec{a}=0\ \ \text{or}\ \ \vec{b}=0 \ \ \text{or}\ \ $ $\text{vector}\ \vec{a}\ \text{and}\ \vec{b}\ \text{are collinear or parallel.}\ \ \ \ \ ...\text{(ii)}$ Since, vectors $\vec{a}\ \&\ \vec{b}$ are perpendicular to each other as well as parallel are not possible. ...(iii) Therefore, form eq. (i), (ii) and (iii), $\ \text{either}\ \vec{a}=\vec{0}\ \ \ \text{or}\ \vec{b}=\vec{0}$ $\therefore\ \ \vec{a}.\vec{b}=0 \ \ \text{and}\ \ \vec{a}\times\vec{b}=0$
View full question & answer→Question 1492 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are unit vectors, then write the value of $\big|\vec{\text{a}}\times\vec{\text{b}}\big|^2+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2.$
AnswerIt is given that $\vec{\text{a}}$ and $\vec{\text{b}}$ are unit vectors. $\Rightarrow|\vec{\text{a}}|=\big|\vec{\text{b}}\big|=1\dots(1)$ Now, $\big(\vec{\text{a}}.\vec{\text{b}}\big)^2+\big|\vec{\text{a}}\times\vec{\text{b}}\big|^2$ $=\big(|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta\big)^2+\big(|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta\big)^2$ $=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2(\cos^2\theta+\sin^2\theta)$ $=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2$ (1) $=|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2$ $=1^21^2$ [From (1)]$=1$
View full question & answer→Question 1502 Marks
Find the position vector of a point R which divides the line segment joining points $\text{P}\big(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big)$ and $\text{Q}\big(-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$ in the ratio 2 : 1.Externally
AnswerGiven: R divides the line segment joining the points $\text{P}\big(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big),\text{Q}\big(-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$ in the ratio 2 : 1 externally.
Therefore position vector of $\text{R}=\frac{2\big(-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)+1\big(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big)}{2+1}$
$=-3\hat{\text{i}}+\hat{\text{k}}$
View full question & answer→Question 1512 Marks
If $\vec{\text{a}}=3\hat{\text{i}}-\hat{\text{j}}-4\hat{\text{k}},\ \vec{\text{b}}=-2\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}}$ and $\vec{\text{c}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$, find $|3\vec{\text{a}}-2\vec{\text{b}}+4\vec{\text{c}}|$.
AnswerGiven $\vec{\text{a}}=3\hat{\text{i}}-\hat{\text{j}}-4\hat{\text{k}},\ \vec{\text{b}}=-2\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}},\ \vec{\text{c}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$
Now, $3\vec{\text{a}}-2\vec{\text{b}}+4\vec{\text{c}}=3\big(3\hat{\text{i}}-\hat{\text{j}}-4\hat{\text{k}}\big)-2\big(-2\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}}\big)+4\big(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}\big)$
$=9\hat{\text{i}}-3\hat{\text{j}}-12\hat{\text{k}}+4\hat{\text{i}}-8\hat{\text{j}}+6\hat{\text{k}}+4\hat{\text{i}}+8\hat{\text{j}}-4\hat{\text{k}}$
$=17\hat{\text{i}}-3\hat{\text{j}}-10\hat{\text{k}}$
$\therefore\ |3\vec{\text{a}}-2\vec{\text{b}}+4\vec{\text{c}}|=\sqrt{17^2+(-3)^2+(-10)^2}$
$=\sqrt{289+9+100}$
$=\sqrt{398}$
View full question & answer→Question 1522 Marks
If $\vec{\text{a}}=\text{x}\hat{\text{i}}+2\hat{\text{j}}-\text{z}\hat{\text{k}}$ and $\vec{\text{b}}=3\hat{\text{i}}-\text{y}\hat{\text{j}}+\hat{\text{k}}$ are two equal vectors, then write the value of x + y + z.
AnswerGiven: $\vec{\text{a}}=\text{x}\hat{\text{i}}+2\hat{\text{j}}-\text{z}\hat{\text{k}}$ and $\vec{\text{b}}=3\hat{\text{i}}-\text{y}\hat{\text{j}}+\hat{\text{k}}$ Since the two vectors are equal. We have,$\vec{\text{a}}=\text{x}\hat{\text{i}}+2\hat{\text{j}}-\text{z}\hat{\text{k}}=\vec{\text{b}}=3\hat{\text{i}}-\text{y}\hat{\text{j}}+\hat{\text{k}}$
$\Rightarrow\ \text{x}=3,\ \text{y}=-2,\ \text{z}=-1$
$\therefore\ \text{x + y + z}=3-2-1=0$
View full question & answer→Question 1532 Marks
If $\vec{\text{a}},\vec{\text{b}}$ are two vectors, then write the truth value of the following statement:$\vec{\text{a}}=-\vec{\text{b}}\Rightarrow\big|\vec{\text{a}}\big|=\big|\vec{\text{b}}\big|$
AnswerTrue
$\vec{\text{a}}=-\vec{\text{b}}$
Take modulus both sides
$\big|\vec{\text{a}}\big|=\big|-\vec{\text{b}}\big|$
$\Rightarrow\big|\vec{\text{a}}\big|=\big|\vec{\text{b}}\big|$ $\Big[\therefore\ \big|-\vec{\text{b}}\big|=\big|\vec{\text{b}}\big|\Big]$
View full question & answer→Question 1542 Marks
$\text{Find}\ \big|\vec{x}\big|,$ if for a unit unit vector $\vec{a},\ (\vec{x}-\vec{a})\cdot(\vec{x}+\vec{a})=12.$
Answer$\text{Given:}\ \vec{a}\ \text{is a unit vector}\ \Rightarrow\ \big|\vec{a}\big|=1\ \ .......(\text{i})$
$\big(\vec{x}-\vec{a}\big)\big(\vec{a}+\vec{a}\big)=12\ $ $\Rightarrow\ \vec{x}.\vec{x}+\vec{x}.\vec{a}-\vec{a}.\vec{x}-\vec{a}.\vec{a}=12$
$\Rightarrow\ \ \big|\vec{x}\big|^2+\vec{x}.\vec{a}-\vec{a}.\vec{x}-\big|\vec{a}\big|^2=12\ $ $\Rightarrow\ \big|\vec{x}\big|^2+\vec{a}.\vec{x}-\vec{a}.\vec{x}-\big|\vec{a}\big|^2=12$
$ \Rightarrow\ \ \big|\vec{x}\big|^2-\big|\vec{a}\big|^2=12$
$\text{Putting}\ \big|\vec{a}\big|=1\ \text{from eq. (i),}\ \ \ \ \big|\vec{x}\big|^2-1=12$
$\Rightarrow\ \big|\vec{x}\big|^2=13\ \ \Rightarrow\ \ \big|\vec{x}\big|=13$
View full question & answer→Question 1552 Marks
If $\vec{\text{a}}\text{ and }\vec{\text{b}}$ denote the position vectors of points A and B respectively and C is a point on AB such that 3AC = 2AB, then write the position vector of C.
AnswerGiven: $\vec{\text{a}}\text{ and }\vec{\text{b}}$ denote the position vectors of points A and B respectively and C is a point on AB such that 3AC = 2AB.Let $\vec{\text{c}}$ is the position vector of C.
Now, $\overrightarrow{\text{AB}}=\vec{\text{b}}-\vec{\text{a}}$ $\overrightarrow{\text{AC}}=\vec{\text{c}}-\vec{\text{a}}$ Consider, $3\text{AC}=2\text{AB}$ $\Rightarrow\ 3\big(\vec{\text{c}}-\vec{\text{a}}\big)=2\big(\vec{\text{b}}-\vec{\text{a}}\big)$ $\Rightarrow\ 3\vec{\text{c}}-3\vec{\text{a}}=2\vec{\text{b}}-2\vec{\text{a}}$ $\Rightarrow\ 3\vec{\text{c}}=2\vec{\text{b}}+\vec{\text{a}}$ $\Rightarrow\ \vec{\text{c}}=\frac{1}3\big(2\vec{\text{b}}+\vec{\text{a}}\big)$ $\Rightarrow\ \vec{\text{c}}=\frac{1}3\big(\vec{\text{a}}+2\vec{\text{b}}\big)$ Hence, the position vector of C is $\frac{1}3\big(\vec{\text{a}}+2\vec{\text{b}}\big)$
View full question & answer→Question 1562 Marks
Find the angle between the vectors $\vec{\text{a}} $ and $\vec{\text{b}},$ where$\vec{\text{a}}=3\hat {\text{i}}-2\hat{\text{j}}-6\hat{\text{k}}$ and $\vec{\text{b}} =4\hat{\text{i}}-\hat{\text{j}}+8\hat{\text{k}}$
AnswerLet $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$.
$\big|\vec{\text{a}}\big|=\sqrt{(3)^2+(-2)^2+(-6)^{2}}=\sqrt{49}=7$
$\big|\vec{\text{b}}\big|=\sqrt{(4)^2+(1)^2+(8)^{2}}=\sqrt{81}=9$
$\vec{\text{a}}.\vec{\text{b}}=12+2-48=-34$
$\cos\theta=\frac{\vec{\text{a }}.\vec{\text{ b}}}{\big|\vec{a}\big|\big|\vec{b}\big|}=\frac{-34}{(7)(9)}=\frac{-34}{63}$
$\Rightarrow\theta=\cos^{-1}\big(\frac{-34}{63}\big)$
View full question & answer→Question 1572 Marks
A vector $\vec{\text{r}}$ is inclined at equal acute angles to x-axis, y-axis and z-axis. If $|\vec{\text{r}}|=6$ units, find $\vec{\text{r}}$.
AnswerHere, $\alpha=\beta=\gamma$
$\Rightarrow\ \cos\alpha=\cos\beta=\cos\gamma$
$\Rightarrow\ \text{l}=\text{m}=\text{n}=\text{x}$ (say)
We know that,
$\text{l}^2+\text{m}^2+\text{n}^2=1$
$\text{x}^2+\text{x}^2+\text{x}^2=1$
$3\text{x}^2=1$
$\text{x}^2=\frac{1}3$
$\text{x}=\pm\frac{1}{\sqrt3}$
$\text{l}=\pm\frac{1}{\sqrt3},\ \text{m}=\pm\frac{1}{\sqrt3},\ \text{n}=\pm\frac{1}{\sqrt3}$
View full question & answer→Question 1582 Marks
Write the projection of the vector $\hat{\text{i}}+3\hat{\text{j}}+7\hat{\text{k}}$ on the vector $2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}.$
AnswerWe know that projection of $\vec{\text{a}}$ on $\vec{\text{b}}=\frac{\vec{\text{a}}.\vec{\text{b}}}{\big|\vec{\text{b}}\big|}.$
Let $\vec{\text{a}}=\hat{\text{i}}+3\hat{\text{j}}+7\hat{\text{k}}$ and $\vec{\text{b}}=2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}.$
$\therefore$ projection of $\vec{\text{a}}$ on $\vec{\text{b}}$
$=\frac{\big(\hat{\text{i}}+3\hat{\text{j}}+7\vec{\text{k}}\big).\big(2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}.\big)}{\big|2\hat{\text{i}}-3\hat{\text{j}}+6\hat{\text{k}}.\big|}$
$=\frac{1\times2+3\times(-3)+7\times6}{\sqrt{2^2+(-3)^2+6^2}}$
$=\frac{2-9+42}{\sqrt{49}}$
$=\frac{35}{7}$
$=5$
View full question & answer→Question 1592 Marks
If two vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ are such that $|\vec{\text{a}}|=2,\big|\vec{\text{b}}\big|=1$ and $\vec{\text{a}}.\vec{\text{b}}=1,$ then find the value of $\big(3\vec{\text{a}}-5\vec{\text{b}}\big).\big(2\vec{\text{a}}+7\vec{\text{b}}\big).$
Answer$\big(3\vec{\text{a}}-5\vec{\text{b}}\big).\big(2\vec{\text{a}}+7\vec{\text{b}}\big)$
$=3\vec{\text{a}}.2\vec{\text{a}}+3\vec{\text{a}}.7\vec{\text{b}}-5\vec{\text{b}}.2\vec{\text{a}}-5\vec{\text{b}}.7\vec{\text{b}}$
$=6\vec{\text{a}}.\vec{\text{a}}+21\vec{\text{a}}.\vec{\text{b}}-10\vec{\text{a}}.\vec{\text{b}}-35\vec{\text{b}}.\vec{\text{b}}$
$=6|\vec{\text{a}}|^2+11\vec{\text{a}}.\vec{\text{b}}-35\big|\vec{\text{b}}\big|^2$
$=6\times2^2+11\times1-35\times1^2$
$=35-35$
$=0$
View full question & answer→Question 1602 Marks
Find the manitude of two vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ that are of the same magnitude, are inclined at 60° and whose scalar product is $\frac{1}{2}.$
AnswerGiven that the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$ is $30^{\circ}.$
Also,
$|\vec{\text{a}}|=\big|\vec{\text{b}}\big|;\vec{\text{a}}.\vec{\text{b}}=\frac{1}{2}$
We know that
$\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta$
$\Rightarrow\frac{1}2{}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos60$
$\Rightarrow\frac{1}{2}=|\vec{\text{a}}|^2\big(\frac{1}{2}\big)$
$\Rightarrow|\vec{\text{a}}|^2=1$
$\Rightarrow|\vec{\text{a}}|=1$
$\therefore|\vec{\text{a}}|=\big|\vec{\text{b}\big|}=1$
View full question & answer→Question 1612 Marks
A unit vector $\vec{\text{r}}$ makes angles $\frac{\pi}3\text{ and }\frac{\pi}2$ with $\hat{\text{j}}\text{ and } \hat{\text{k}}$ respectively and an acute angle $\theta$ with $\hat{\text{i}}$. Find $\theta$.
AnswerA unit vector makes an angle $\frac{\pi}3\text{ and }\frac{\pi}2$ with $\hat{\text{j}}\text{ and } \hat{\text{k}}$Let l, m, n be its direction cosines
$\therefore\ \text{l}=\cos\theta,\ \text{m}=\cos\big(\frac{\pi}3\big)=\frac{1}2,\ \text{n}=\cos\big(\frac{\pi}2\big)=0$
Now,
$\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Rightarrow\ \text{l}^2+\frac{1}4+0=1$
$\Rightarrow\ \text{l}^2=1-\frac{1}4=\frac{3}4$
$\Rightarrow\ \text{l}=\pm\frac{\sqrt3}2$
$\therefore\ \vec{\text{r}}$ makes an acute angle 30º, 150º with $\hat{\text{i}}$
Since, angle $\theta$ is acute.
$\therefore\ \theta=30^{\circ}$
View full question & answer→Question 1622 Marks
Write the position vector of a point dividing the line segment joining points having position vectors $\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$ and $2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}$ externally in the ratio 2 : 3.
AnswerLet A and B be the points with position vectors $\vec{\text{a}} = \hat{\text{i}} + \hat{\text{j}} - 2\hat{\text{k}},\vec{\text{b}} = 2\hat{\text{i}} - \hat{\text{j}} + 3\hat{\text{k}} $ respectively.
Let C divide AB externally in the ratio 2 : 3 such that AC : CB = 2 : 3
$\therefore $ Postion vector of C $ = \frac{2\big(2\hat{\text{i}} - \hat{\text{j}} + 3\hat{\text{k}}\big) - 3 \big(\hat{\text{i}} + \hat{\text{j}} - 2\hat{\text{k}}\big)}{2 - 3}$
$ = \frac{4\hat{\text{i}} - 2\hat{\text{j}} + 6\hat{\text{k}} - 3\hat{\text{i}} - 3\hat{\text{j}} + 6\hat{\text{k}}}{-1}$
$= \frac{\hat{\text{i}} - 5\hat{\text{j}} + 12\hat{\text{k}}}{-1}$
$= -\hat{\text{i}} + 5\hat{\text{j}} - 12\hat{\text{k}}$
View full question & answer→Question 1632 Marks
Write a unit vector in the direction of $\overrightarrow{\text{PQ}}$, where P and Q are the points (1, 3, 0) and (4, 5, 6) respectively.
AnswerP(1, 3, 0) and Q(4, 5, 6) are the given points.
$\therefore\ \overrightarrow{\text{PQ}}=\big(4\hat{\text{i}}+5\hat{\text{j}}+6\hat{\text{k}}\big)-\big(\hat{\text{i}}+3\hat{\text{j}}+0\hat{\text{k}}\big)$
$=3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}$
$\Rightarrow\ \Big|\overrightarrow{\text{PQ}}\Big|=\sqrt{3^2+2^2+6^2}$
$=\sqrt{9+4+36}$
$=\sqrt{49}$
$=7$
$\therefore$ Unit vector in the direction of $\overrightarrow{\text{PQ}}=\frac{\overrightarrow{\text{PQ}}}{\Big|\overrightarrow{\text{PQ}}\Big|}=\frac{3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}}7=\frac{1}7\big(3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}\big)$
View full question & answer→Question 1642 Marks
Write a vector satisfying $\vec{\text{a}}.\hat{\text{i}}=\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}\big)=\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)=1.$
AnswerLet $\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$
$\vec{\text{a}}.\hat{\text{i}}=\text{a}_1$
$\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}\big)=\text{a}_1+\text{a}_2$
$\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)=\text{a}_1+\text{a}_2+\text{a}_3$
Given that
$\vec{\text{a}}.\hat{\text{i}}=\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}\big)=\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)=1$
$\Rightarrow\text{a}_1=\text{a}_1+\text{a}_2=\text{a}_1+\text{a}_2+\text{a}_3=1$
$\Rightarrow\text{a}_1=1;\text{a}_1+\text{a}_2=1;\text{a}_1+\text{a}_2+\text{a}_3=1$
$\Rightarrow\text{a}_1=1;1+\text{a}_2=1;1+\text{a}_2+\text{a}_3=1$
$\Rightarrow\text{a}_1=1;\text{a}_2=0;1+0+\text{a}_3=1$
$\Rightarrow\text{a}_1=1;\text{a}_2=0;\text{a}_3=0$
So, $\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}=1\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}=\hat{\text{i}}$
View full question & answer→Question 1652 Marks
If $\vec{\text{a}}.\vec{\text{a}}=0$ and $\vec{\text{a}}.\vec{\text{b}}=0,$ what can you conclude about the vector $\vec{\text{b}}?$
AnswerGiven that $\vec{\text{a}}.\vec{\text{a}}=0$
$\Rightarrow|\vec{\text{a}}|^2=0$
$\Rightarrow|\vec{\text{a}}|=0\dots(1)$
Also, given that
$\vec{\text{a}}.\vec{\text{b}}=0$
$\Rightarrow|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta=0$ (where $\theta$ is the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$)
$\Rightarrow0\big|\vec{\text{b}}\big|\cos\theta=0$ [From (1)]
$\Rightarrow0=0$
So, it means that for any vector $\vec{\text{b}},$ the given equation $\vec{\text{a}}.\vec{\text{b}}=0$ is satisfid.
View full question & answer→Question 1662 Marks
For given vectors, $\vec{a}=2\hat{i}-\hat{j}+2\hat{k}\ \ \text{and}\ \vec{b}=-\hat{i}+\hat{j}-\hat{k},$ find the unit vector in the direction of the vector $\vec{a}+\vec{b}.$
AnswerThe given vectors are $\vec{a}=2\hat{i}-\hat{j}+2\hat{k}\ \ \text{and}\ \vec{b}=-\hat{i}+\hat{j}-\hat{k}$ $\vec{a}=2\hat{i}-\hat{j}+2\hat{k}$ $\vec{b}=-\hat{i}+\hat{j}-\hat{k}$ $\therefore\vec{a}+\vec{b}=(2-1)\hat{i}+(-1+1)\hat{j}+(2-1)\hat{k}$ $=1\hat{i}+0\hat{j}+1\hat{k}=\hat{i}+\hat{k}$ $\big|\vec{a}+\vec{b}\big|=\sqrt{1^2+1^2}=\sqrt{2}$Hence, the unit vector in the direction of $\big(\vec{a}+\vec{b}\big)$ is
$\frac{\big(\vec{a}+\vec{b}\big)}{\big|\vec{a}+\vec{b}\big|}=\frac{\hat{i}+\hat{k}}{\sqrt{2}}=\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{k}$
View full question & answer→Question 1672 Marks
For any two vectors $\vec{\text{a}}$ and $\vec{\text{b}},$ write when $\big|\vec{\text{a}}+\vec{\text{b}}\big|=\big|\vec{\text{a}}-\vec{\text{b}}\big|$ holds.
AnswerGiven that
$\big|\vec{\text{a}}+\vec{\text{b}}\big|=\big|\vec{\text{a}}-\vec{\text{b}}\big|$
Squaring both sides, we get
$\big|\vec{\text{a}}+\vec{\text{b}}\big|^2=\big|\vec{\text{a}}-\vec{\text{b}}\big|^2$
$\Rightarrow|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+2\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-2\vec{\text{a}}.\vec{\text{b}}$
$\Rightarrow4\vec{\text{a}}.\vec{\text{b}}=0$
$\Rightarrow\vec{\text{a}}.\vec{\text{b}}=0$
⇒ $\vec{\text{a}}$ and $\vec{\text{b}}$ are perpendiculalr.
View full question & answer→Question 1682 Marks
If $\vec{\text{a}},\vec{\text{b}}$ are two vectors, then write the truth value of the following statement:$\big|\vec{\text{a}}\big|=\big|\vec{\text{b}}\big|\Rightarrow\vec{\text{a}}=\vec{\text{b}}$
AnswerFalse
$\big|\vec{\text{a}}\big|=\big|\vec{\text{b}}\big|\Rightarrow\vec{\text{a}}=\vec{\text{b}}$
Consider an example,
$\vec{\text{a}}=\text{i}+\sqrt3\text{j}$ and $\vec{\text{b}}=\sqrt2\text{i}+\sqrt2\text{j}$
$\big|\vec{\text{a}}\big|=\sqrt{1^2+\big(\sqrt3\big)^2}=2$ and $\big|\vec{\text{b}}\big|=\sqrt{\big(\sqrt2\big)^2+\big(\sqrt2\big)^2}=2$
Thus, $\big|\vec{\text{a}}\big|=\big|\vec{\text{b}}\big|$ but $\vec{\text{a}}\neq\vec{\text{b}}$
View full question & answer→Question 1692 Marks
Find the position vector of a point R which divides the line segment joining points $\text{P}\big(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big)$ and $\text{Q}\big(-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$ in the ratio 2 : 1.Internally
AnswerGiven: R divides the line segment joining the points $\text{P}\big(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big),\text{Q}\big(-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$ in the ratio 2 : 1 internally.
Therefore position vector of $\text{R}=\frac{2\big(-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)+1\big(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\big)}{2+1}$
$=\frac{1}3\big(-\hat{\text{i}}+4\hat{\text{j}}+3\hat{\text{k}}\big)$
View full question & answer→Question 1702 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are unit vectors, find the angle between $\vec{\text{a}}+\vec{\text{b}}$ and $\vec{\text{a}}-\vec{\text{b}}.$
AnswerWe have
$|\vec{\text{a}}|=1$ and $\big|\vec{\text{b}}\big|=1\dots(1)$
Now,
$\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2$
$=1^2-1^2$ [using (1)]
$=0$
$\Rightarrow\vec{\text{a}}+\vec{\text{b}}$ and $\vec{\text{a}}-\vec{\text{b}}$ are perpendicular.
$\therefore$ angle between $\big(\vec{\text{a}}+\vec{\text{b}}\big)$ and $\big(\vec{\text{a}}-\vec{\text{b}}\big)$ is 90°.
View full question & answer→Question 1712 Marks
Write the value of $\big(\vec{\text{a}}.\hat{\text{i}}\big)\hat{\text{i}}+\big(\vec{\text{a}}.\hat{\text{j}}\big)\hat{\text{j}}+\big(\vec{\text{a}}.\hat{\text{k}}\big)\hat{\text{k}},$ where $\vec{\text{a}}$ is any vector.
AnswerLet $\vec{\text{a}}=\text{a}_1\vec{\text{i}}+\text{a}_2\vec{\text{j}}+\text{a}_3\vec{\text{k}}$
Now,
$\big(\vec{\text{a}}.\hat{\text{i}}\big)\hat{\text{i}}+\big(\vec{\text{a}}.\hat{\text{j}}\big)\hat{\text{j}}+\big(\vec{\text{a}}.\hat{\text{k}}\big)\hat{\text{k}}$
$=\text{a}_1\vec{\text{i}}+\text{a}_2\vec{\text{j}}+\text{a}_3\vec{\text{k}}$
$=\vec{\text{a}}$
View full question & answer→Question 1722 Marks
If $\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}},$ then write the value of $\big|\vec{\text{r}}\times\hat{\text{i}}\big|^2.$
AnswerGiven: $\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$
Now,
$\vec{\text{i}}=\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$
$\vec{\text{r}}\times\vec{\text{i}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{x}&\text{y}&\text{z}\\1&0&0 \end{vmatrix}$
$=0\hat{\text{i}}+\text{z}\hat{\text{j}}-\text{y}\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{r}}\times\vec{\text{i}}\big|=\sqrt{\text{x}^2+\text{y}^2}$
$\Rightarrow\big|\vec{\text{r}}\times\vec{\text{i}}\big|^2=\text{x}^2+\text{y}^2$
View full question & answer→Question 1732 Marks
Find the projection of $\vec{\text{a}}$ on $\vec{\text{b}}$ if $\vec{\text{a}}.\vec{\text{b}}=8$ and $\vec{\text{b}}=2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}.$
AnswerWe have
$\vec{\text{a}}.\vec{\text{b}}=8$ and $\vec{\text{b}}=2\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}$
The projection of $\vec{\text{a}}$ on $\vec{\text{b}}$ is
$\frac{\vec{\text{a}}.\vec{\text{b}}}{|\vec{\text{b}}|}$
$=\frac{8}{\sqrt{4+36+9}}$
$=\frac{8}{7}$
View full question & answer→Question 1742 Marks
Find a vector $\vec{\text{a}}$ of magnitude $5\sqrt2$, making an angle of $\frac{\pi}4$ with x-axis, $\frac{\pi}2$ with y-axis and an acute angle $\theta$ with z-axis.
AnswerIt is given that vector $\vec{\text{a}}$ makes an angle of $\frac{\pi}4$ with x-axis, $\frac{\pi}2$ with y-axis and an acute angle $\theta$ with z-axis.$\therefore\ \text{l}=\cos\frac{\pi}4=\frac{1}{\sqrt2},\text{m}=\cos\frac{\pi}2=0,\text{n}=\cos\theta$
Now,
$\text{l}^2+\text{m}^2+\text{n}^2=1$
$\Rightarrow\ \frac{1}2+0+\cos^2\theta=1$
$\Rightarrow\ \cos^2\theta=1-\frac{1}2=\frac{1}2$
$\Rightarrow\ \cos\theta=\frac{1}{\sqrt2}$ ($\theta$ is acute)
We know that
$\vec{\text{a}}=|\vec{\text{a}}|\big(\text{l}\hat{\text{i}}+\text{m}\hat{\text{j}}+\text{n}\hat{\text{k}}\big)$
$\Rightarrow\ \vec{\text{a}}=5\sqrt2\Big(\frac{1}{\sqrt2}\hat{\text{i}}+0\hat{\text{j}}+\frac{1}{\sqrt2}\hat{\text{k}}\Big)$
$\Rightarrow\ \vec{\text{a}}=5\big(\hat{\text{i}}+0\hat{\text{j}}+\hat{\text{k}}\big)$
View full question & answer→Question 1752 Marks
Find $\lambda,$ if $\big(2\hat{\text{i}}+6\hat{\text{j}}+14\hat{\text{k}}\big)\times\big(\hat{\text{i}}-\lambda\hat{\text{j}}+7\hat{\text{k}}\big)=\vec{0}.$
AnswerGiven: $\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&6&14\\1&-\lambda&7\end{vmatrix}=\vec{0}$
$\Rightarrow\hat{\text{i}}(42+14\lambda)-0\hat{\text{j}}+\hat{\text{k}}(-2\lambda-6)=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$
$\Rightarrow42+14\lambda=0;-2\lambda-6=0$
$\Rightarrow\lambda=-3$ (This satisfies the above equations)
View full question & answer→Question 1762 Marks
Write the position vector of a point dividing the line segment joining points A and B with position vectors $\vec{\text{a}}\text{ and }\vec{\text{b}}$ externally in the ratio 1 : 4, where $\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}$ and $\vec{\text{b}}=-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$.
AnswerThe position vectors of A and B are
$\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}$
$\vec{\text{b}}=-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
Let C divides AB in the ratio such that AB : CB = 1 : 4
Position vector of $\text{C}=\frac{1\big(-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)-4\big(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}\big)}{1-4}$
$=\frac{-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}-8\hat{\text{i}}-12\hat{\text{j}}-16\hat{\text{k}}}{-3}$
$=\frac{-9\hat{\text{i}}-11\hat{\text{j}}-15\hat{\text{k}}}{-3}$
$=3\hat{\text{i}}+\frac{11\hat{\text{j}}}3+5\hat{\text{k}}$
View full question & answer→Question 1772 Marks
Find $\vec{\text{a}}.\vec{\text{b}}$ when
$\vec{\text{a}}=\hat{\text{j}}+2\hat{\text{k}}$ and $\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{k}}$
Answer$\vec{\text{a}}.\vec{\text{b}}=(\hat{\text{j}}+2\hat{\text{k}}).(2\hat{\text{i}}+\hat{\text{k}})$
$=(0\times\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}})(2\hat{\text{i}}+0\times\hat{\text{j}}+\hat{\text{k}})$
$=(0)(2)+(1)(0)+(2)(1)$
$=0+0+2$
$\vec{\text{a}}.\vec{\text{b}}=2$
View full question & answer→Question 1782 Marks
Find the unit vector in the direction of the vector $\vec{a}=\hat{i}+\hat{j}+2\hat{k}.$
AnswerThe unit vector $\hat{a}$ in the direction of vector $\vec{a}=\hat{i}+\hat{j}+2\hat{k}$ is given by $\hat{a}=\frac{\vec{a}}{|a|}.$
$\Big|\vec{a}\Big|=\sqrt{1^2+1^2+2^2}=\sqrt{1+1+4}=\sqrt{6}$
$\therefore{\hat{a}}=\frac{\vec{a}}{\big|\vec{a}\big|}=\frac{\hat{i}+\hat{j}+2\hat{k}}{\sqrt{6}}=\frac{1}{\sqrt{6}}\hat{i}+\frac{1}{\sqrt{6}}\hat{j}+\frac{2}{\sqrt{6}}\hat{k}$
View full question & answer→Question 1792 Marks
For any two vectore $\vec{\text{a}}$ and $\vec{\text{b}}$, show that $\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=0\Leftrightarrow|\vec{\text{a}}|=\big|\vec{\text{b}}\big|.$
AnswerWe have
$\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=0$
$\Rightarrow|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2=0$
$\Rightarrow|\vec{\text{a}}|^2=\big|\vec{\text{b}}\big|^2$
$\Rightarrow|\vec{\text{a}}|=\big|\vec{\text{b}}\big|$
View full question & answer→Question 1802 Marks
Find a unit vector in the direction of the vector $\vec{\text{a}}=3\hat{\text{i}}-2\hat{\text{j}}+6\hat{\text{k}}$.
AnswerGiven: $\vec{\text{a}}=3\hat{\text{i}}-2\hat{\text{j}}+6\hat{\text{k}}$ Then, $|\vec{\text{a}}|=\sqrt{3^2+(-2)^2+6^2}$ $=\sqrt{9+4+36}$ $=\sqrt{49}$ $=7$ $\therefore$ Unit vector $=\frac{\vec{\text{a}}}{|\vec{\text{a}}|}=\frac{3\hat{\text{i}}-2\hat{\text{j}}+6\hat{\text{k}}}7$$=\frac{3}7\hat{\text{i}}-\frac{2}7\hat{\text{j}}+\frac{6}7\hat{\text{k}}$
View full question & answer→Question 1812 Marks
For what value of $\lambda$ are the vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ perpendicular to each other if
$\vec{\text{a}}=\lambda\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=5\hat{\text{i}}-9\hat{\text{j}}+2\hat{\text{k}}$
AnswerIf the vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ are perpendicular to each other, then
$\vec{\text{a}}.\vec{\text{b}}=0$
$\Rightarrow\Big(\lambda\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}\Big).\Big(5\hat{\text{i}}-9\hat{\text{j}}+2\hat{\text{k}}\Big)=0$
$\Rightarrow5\lambda-18+2=0$
$\Rightarrow5\lambda-16=0$
$\Rightarrow5\lambda=16$
$\Rightarrow\lambda=\frac{16}{5}$
View full question & answer→Question 1822 Marks
What is the angle between vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ with magnitudes 2 and $\sqrt{3}$ respectively? Given $\vec{\text{a}}.\vec{\text{b}}=\sqrt{3}.$
AnswerLet $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
Given that
$|\vec{\text{a}}|=2,\big|\vec{\text{b}}\big|=\sqrt{3}$ and $\vec{\text{a}}.\vec{\text{b}}=\sqrt{3}$
We know that
$\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta$
$\Rightarrow\sqrt{3}=(2)(\sqrt{3})\cos\theta$
$\Rightarrow\cos\theta=\frac{\sqrt{3}}{2\sqrt{3}}$
$\Rightarrow\cos\theta=\frac{1}2{}$
$\Rightarrow\theta=\cos^{-1}\big(\frac{1}2{}\big)=\frac{\pi}{3}$
View full question & answer→Question 1832 Marks
If the vectors $3\hat{\text{i}}+\text{m}\hat{\text{j}}+\hat{\text{k}}$ and $2\hat{\text{i}}-\hat{\text{j}}-8\hat{\text{k}}$ are orthonal, find m.
AnswerIt is given that the vectors are othgonal. so, their dot product is zero.
$\big(3\hat{\text{i}}+\text{m}\hat{\text{j}}+\hat{\text{k}}\big).\big(2\hat{\text{i}}-\hat{\text{j}}-8\hat{\text{k}}\big)=0$
$\Rightarrow6-\text{m}-8=0$
$\Rightarrow-\text{m}-2=0$
$\Rightarrow\text{m}=-2$
View full question & answer→Question 1842 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are mutually perpendicular unit vectors, write the value of $\big|\vec{\text{a}}+\vec{\text{b}}\big|.$
Answer$\vec{\text{a}}$ and $\vec{\text{b}}$ are unit vectors and they are perpendicular.
$\Rightarrow|\vec{\text{a}}|=\big|\vec{\text{b}}\big|=1;\vec{\text{a}}.\vec{\text{b}}=0\dots(1)$
Now,
$\big|\vec{\text{a}}+\vec{\text{b}}\big|^2=|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+2\vec{\text{a}}.\vec{\text{b}}$
$=1+1+2(0)$ [using (1)]
$=2$
$\therefore\big|\vec{\text{a}}+\vec{\text{b}}\big|=\sqrt{20}$
View full question & answer→Question 1852 Marks
Evaluate the following:
$\big[\hat{\text{i}}\hat{\text{j}}\hat{\text{k}}\big]+\big[\hat{\text{j}}\hat{\text{k}}\hat{\text{i}}\big]+\big[\hat{\text{k}}\hat{\text{i}}\hat{\text{j}}\big]$
AnswerWe have,
$\big[\hat{\text{i}}\hat{\text{ j}}\hat{\text{ k}}\big]+\big[\hat{\text{j}}\hat{\text{ k}}\hat{\text{ i}}\big]+\big[\hat{\text{k}}\hat{\text{i}}\hat{\text{j}}\big]$
$=(\hat{\text{i}}\times\hat{\text{j}}).\hat{\text{k}}(\hat{\text{j}}\times\hat{\text{k}}).\hat{\text{i}}+(\hat{\text{k}}\times\hat{\text{i}}).\hat{\text{j}}$
$=\hat{\text{k}}.\hat{\text{k}}+\hat{\text{i}}.\hat{\text{i}}+\hat{\text{j}}.\hat{\text{j}}$
$=1+1+1$
$=3$
Therefore, $\big[\hat{\text{i}}\hat{\text{j}}\hat{\text{k}}\big]+\big[\hat{\text{j}}\hat{\text{k}}\hat{\text{i}}\big]+\big[\hat{\text{k}}\hat{\text{i}}\hat{\text{j}}\big]=3$
View full question & answer→Question 1862 Marks
Can a vector have direction angles 45º, 60º, 120º?
AnswerYes,
Let a vector makes an angle $\alpha=45^{\circ},\ \beta=60^{\circ},\ \gamma=120^{\circ}$ with OX, OY, OZ respectively.
Let l, m, n be the direction cosines of the vector. Then,
$\text{l}=\cos45^{\circ}=\frac{1}{\sqrt{2}}$
$\text{m}=\cos60^{\circ}=\frac{1}2$
$\text{n}=\cos120^{\circ}=-\frac{1}2$
So,
$\text{l}^2+\text{m}^2+\text{n}^2$
$=\frac{1}2+\frac{1}4+\frac{1}4$
$=1$
Since, the vector has direction cosines such that $\text{l}^2+\text{m}^2+\text{n}^2=1$
Hence, a vector can have direction angles 45º, 60º, 120º
View full question & answer→Question 1872 Marks
If $\overrightarrow{\text{AO}}+\overrightarrow{\text{OB}}=\overrightarrow{\text{BO}}+\overrightarrow{\text{OC}}$, prove that A, B, C are collinear points.
AnswerHere, $\overrightarrow{\text{AO}}+\overrightarrow{\text{OB}}=\overrightarrow{\text{BO}}+\overrightarrow{\text{OC}}$
$\overrightarrow{\text{OA}}-\overrightarrow{\text{BO}}=\overrightarrow{\text{OB}}-\overrightarrow{\text{CO}}$
$\overrightarrow{\text{AB}}=\overrightarrow{\text{BC}}$
So, $\overrightarrow{\text{AB}}$ is parallel to $\overrightarrow{\text{BC}}$ but $\vec{\text{B}}$ is a common vector. Hence,
A, B, C are collinear.
View full question & answer→Question 1882 Marks
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are two vectors such that $\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big).=0,$ find the relation betwen the magnitudes of $\vec{\text{a}}$ and $\vec{\text{b}}.$
AnswerGiven that$\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=0$
$\Rightarrow|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2=0$
$\Rightarrow|\vec{\text{a}}|^2=\big|\vec{\text{b}}\big|^2$
$\therefore|\vec{\text{a}}|=\big|\vec{\text{b}}\big|$
View full question & answer→Question 1892 Marks
Find the angle at which the following vectors are inclined to each of the coordinate axes:
$\hat{\text{j}}-\hat{\text{k}}$
AnswerLet $\vec{\text{r}}$ be the given vector, and let it make an angle $\alpha,\beta,\gamma$ with OX, OY, OZ respectively. Then, its direction cosines are $\cos\alpha,\cos\beta,\cos\gamma$.So direction ratios of $\vec{\text{r}}=\hat{\text{j}}-\hat{\text{k}}$ are proportional to 0, 1, -1. Therefore,
Direction cosine of $\vec{\text{r}}$ are $\frac{0}{\sqrt{0+1^2+(-1^2)}},\frac{1}{\sqrt{0+1^2+(-1^2)}},\frac{-1}{\sqrt{0+1^2+(-1^2)}}$ or $0,\frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}}$
$\therefore\cos\alpha=0,\cos\beta=\frac{1}{\sqrt{2}},\cos\gamma=\frac{-1}{\sqrt{2}}$
$\Rightarrow\alpha=\cos^{-1}=(0),\beta=\cos^{-1}=\Big(\frac{-1}{\sqrt{2}}\Big),\gamma=\cos^{-1}=\Big(\frac{-1}{\sqrt{2}}\Big)$
$\Rightarrow\alpha=\frac{\pi}{2},\beta=\frac{\pi}{4},\gamma=\frac{3\pi}{4}$
View full question & answer→Question 1902 Marks
Find the direction cosines of the vector joining the points A (1, 2, -3) and B (-1, -2, 1), directed from A to B.
AnswerThe given points are A (1, 2 -3) and B (-1, -2, 1).
$\therefore{\overrightarrow{\text{AB}}}=(-1-1)\hat{i}+(-2-2)\hat{j}+\{1-(-3)\}\hat{k}$
$\Rightarrow\overrightarrow{\text{AB}}=-2\hat{i}-4\hat{j}+4\hat{k}$
$\therefore\bigg|\overrightarrow{\text{AB}}\bigg|=\sqrt{(-2)^2+(-4)^2+4^2}=\sqrt{4+16+16}=\sqrt{36}=6$
Hence, the direction cosines of $\overrightarrow{\text{AB}}\ \text{are}\ \bigg(-\frac{2}{6},-\frac{4}{6},\frac{4}{6}\bigg)=\bigg(-\frac{1}{3},\frac{2}{3},\frac{2}{3}\bigg).$
View full question & answer→Question 1912 Marks
For any two vectors $\vec{\text{a}} $ and $\vec{\text{b}},$ write when $\big|\vec{\text{a}}+\vec{\text{b}}\big|=|\vec{\text{a}}|+\big|\vec{\text{b}}\big|$ holds.
AnswerGiven that $\big|\vec{\text{a}}+\vec{\text{b}}\big|=|\vec{\text{a}}|+\big|\vec{\text{b}}\big|$ Squaring both sides, we get $\big|\vec{\text{a}}+\vec{\text{b}}\big|^2=\big(|\vec{\text{a}}|+\big|\vec{\text{b}}\big|\big)^2$ $\Rightarrow|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+2\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+2|\vec{\text{a}}|\big|\vec{\text{b}}\big|$ $\Rightarrow\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|$ $\Rightarrow|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta=|\vec{\text{a}}|\big|\vec{\text{b}}\big|$ (where $\theta$ is the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$) $\Rightarrow\cos\theta=1$ $\Rightarrow\theta=0^{\circ}$$\Rightarrow\vec{\text{a}}$ and $\vec{\text{b}}$ are parallel.
View full question & answer→Question 1922 Marks
Write a vector in the direction of vector $5\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$ which has magnitude of 8 unit.
AnswerGiven:
$\vec{\text{a}}=5\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$
$|\vec{\text{a}}|=\sqrt{5^2+(-1)^2+2^2}$
$=\sqrt{25+1+4}$
$=\sqrt{30}$
$\therefore$ Position vector in the direction of vector $=8\times\frac{\vec{\text{a}}}{|\vec{\text{a}}|}$
$=\frac{8}{\sqrt{30}}\big(5\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\big)$
View full question & answer→Question 1932 Marks
Find $\vec{\text{a}}.\vec{\text{b}}$ when
$\vec{\text{a}}=\hat{\text{j}}-\hat{\text{k}}$ and $\vec{\text{b}}=2\hat{\text{i}}+3\hat{\text{j}}-2\hat{\text{k}}$
Answer$\vec{\text{a}}.\vec{\text{b}}=(\hat{\text{j}}-\hat{\text{k}}).(2\hat{\text{i}}+3\hat{\text{j}}-2\hat{\text{k}})$
$=(0\times\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})(2\hat{\text{i}}+3\hat{\text{j}}-2\hat{\text{k}})$
$=(0)(2)+(1)(3)+(-1)(-2)$
$=0+3+2$
$\vec{\text{a}}.\vec{\text{b}}=5$
View full question & answer→Question 1942 Marks
If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are position vectors of the points A, B and C respectively, write the value of$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{AC}}$.
AnswerGiven: $\vec{\text{a}},\vec{\text{b}}\text{ and }\vec{\text{c}}$ are the position vectors of A, B, C respectively. Then,
$\overrightarrow{\text{AB}}=\vec{\text{b}}-\vec{\text{a}}$
$\overrightarrow{\text{BC}}=\vec{\text{c}}-\vec{\text{b}}$
$\overrightarrow{\text{AC}}=\vec{\text{c}}-\vec{\text{a}}$
Therefore,
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{AC}}\\=\vec{\text{b}}-\vec{\text{a}}+\vec{\text{c}}-\vec{\text{b}}+\vec{\text{c}}-\vec{\text{a}}$
$=2\big(\vec{\text{c}}-\vec{\text{a}}\big)$
View full question & answer→Question 1952 Marks
Write the value of $\big[\hat{\text{i}}+\hat{\text{j}}\hat{\text{j}}+\hat{\text{k}}\hat{\text{k}}+\hat{\text{i}}\big].$
AnswerWe have
$\big[\hat{\text{i}}+\hat{\text{j}}\hat{\text{j}}+\hat{\text{k}}\hat{\text{k}}+\hat{\text{i}}\big]=\big[\big(\hat{\text{i}}+\hat{\text{j}}\big)\times\big(\hat{\text{j}}+\hat{\text{k}}\big)\big].\big(\hat{\text{k}}+\hat{\text{i}}\big)$ $\big(\therefore\big[\vec{\text{a}}\vec{\text{ b }}\vec{\text{c}}\big]=\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}\big)$
$=\big[\hat{\text{i}}\times\hat{\text{j}}+\hat{\text{i}}\times\hat{\text{k}}+\hat{\text{j}}\times\hat{\text{j}}+\hat{\text{j}}\times\hat{\text{k}}\big].\big(\hat{\text{k}}+\hat{\text{i}}\big)$
$=\big[\hat{\text{k}}-\hat{\text{j}}+\hat{\text{i}}\big].\big(\hat{\text{k}}+\hat{\text{i}}\big)$
$=\big[\big(\hat{\text{k}}.\hat{\text{k}}\big)+\big(\hat{\text{k}}.\hat{\text{i}}\big)-\big(\hat{\text{j}}.\hat{\text{k}}\big)-\big(\hat{\text{j}}.\hat{\text{i}}\big)+\big(\hat{\text{i}}.\hat{\text{k}}\big)+\big(\hat{\text{i}}.\hat{\text{i}}\big)\big]$
$=1+0-0-0+0+1=2$
View full question & answer→Question 1962 Marks
Find the projection of the vector $\hat{i}-\hat{j}$ on the vector $\hat{i}+\hat{j}.$
Answer$\text{Let}\ \ \vec{a}=\hat{i}-\hat{j}=\hat{i}-\hat{j}+0\hat {k}\ \text{and}\ \vec{b}=\hat{i}+\hat{j}$ $=\hat{i}+\hat{j}+0\hat {k}$
Projection of vector $\vec{a}\ \text{and}\ \vec{b}=\frac{\vec{a}.\vec{b}}{\big|\vec{b}\big|}=\frac{(1)(1)+(-1)(1)+0(0)}{\sqrt{(1)^2+(1)^2+(0)^2}}$ $=\frac{1-1+0}{\sqrt{2}}=\frac{0}{\sqrt{2}}=0$
If projection of vector $\vec{a}\ \text{and}\ \vec{b}$ is zero, then vector $\vec{a}$ is perpendicular to $\vec{b}.$
View full question & answer→Question 1972 Marks
Write the value of the area of the parallelogram determined by the vectors $2\hat{\text{i}}$ and $3\hat{\text{j}}.$
AnswerLet:
$\vec{\text{a}}=2\hat{\text{i}}$
$\vec{\text{b}}=3\hat{\text{j}}$
$\vec{\text{a}}\times\vec{\text{b}}=6\big(\hat{\text{i}}\times\hat{\text{j}}\big)$
$=6\hat{\text{k}}$
Area of the parallelogram $=\big|\vec{\text{a}}\times\vec{\text{b}}\big|$
$=6|\hat{\text{k}}|$
$=6(1)$
$=6\text{ sq. units}$
View full question & answer→Question 1982 Marks
Find the unit vector in the direction of $3\hat{\text{i}}+4\hat{\text{j}}-12\hat{\text{k}}$.
AnswerLet $\vec{\text{a}}=3\hat{\text{i}}+4\hat{\text{j}}-12\hat{\text{k}}$Then, $\big|\vec{\text{a}}\big|=\sqrt{3^2+4^2+(-12)^2}$
$=\sqrt{9+16+144}$
$=\sqrt{169}$
$=13$
So, a unit vector in the direction of $\vec{\text{a}}$ is given by
$\hat{\text{a}}=\frac{\vec{\text{a}}}{|\vec{\text{a}}|}=\frac{1}{13}\big(3\hat{\text{i}}+4\hat{\text{j}}-12\hat{\text{k}}\big)$
$=\frac{3}{13}\hat{\text{i}}+\frac{4}{13}\hat{\text{j}}-\frac{12}{13}\hat{\text{k}}$
View full question & answer→Question 1992 Marks
Write a unit vector in the direction of $\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$.
AnswerGiven:$\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$
$\big|\vec{\text{b}}\big|=\sqrt{2^2+1^2+2^2}$ $=\sqrt{4+1+4}$ $=\sqrt9$ $=3$ $\therefore$ Unit vector $=\frac{\vec{\text{b}}}{\big|\vec{\text{b}}\big|}=\frac{1}3\big(2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)$ $=\frac{2}3\hat{\text{i}}+\frac{1}3\hat{\text{j}}+\frac{2}3\hat{\text{k}}$
View full question & answer→Question 2002 Marks
Find the angle between two vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ with magnitudes 1 and 2 respectively and when $\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\sqrt{3.}$
AnswerLet $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
We know $\big|\vec{\text{a}}\times\vec{\text{b}}\big|=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta$
$\Rightarrow\sqrt{3}=(1)(2)\sin\theta$
$\Rightarrow\sin\theta=\frac{\sqrt{3}}{2}$
$\Rightarrow\theta=\frac{\pi}{3}$
View full question & answer→Question 2012 Marks
If $\vec{\text{a}}=2\hat{\text{i}}+\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},$ find the magnitude of $\vec{\text{a}}\times\vec{\text{b}}.$
AnswerGiven:$\vec{\text{a}}=2\hat{\text{i}}+0\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&0&1\\1&1&1 \end{vmatrix}$
$=(0-1)\hat{\text{i}}-(2-1)\hat{\text{j}}+(2-0)\hat{\text{k}}$
$=-\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\sqrt{(-1)^2+(1-)^2+2^2}$
$=\sqrt{6}$
View full question & answer→Question 2022 Marks
Write down a unit vector in XY-plane, making an angle of 30° with the positive direction of x-axis.
AnswerIf $\vec{\text{r}}$ is a unit vector in the XY-plane, then $\vec{\text{r}}=\cos\theta\hat{\text{i}}+\text{sin}\theta\hat{\text{j}}.$
Here, $\theta$ is the angle made by the unit vector with the positive direction of the x-axis.
Therefore, for $\theta$ = 30°:
$\vec{\text{r}}=\cos30^{\circ}\hat{\text{i}}+\text{sin}30^{\circ}\hat{\text{j}}=\frac{\sqrt{3}}{2}\hat{\text{i}}+\frac{1}{2}\hat{\text{j}}$
Hence, the required unit vector is $\frac{\sqrt{3}}{2}\hat{\text{i}}+\frac{1}{2}\hat{\text{j}}$
View full question & answer→Question 2032 Marks
If a unit vector $\vec{a}$ makes angles $\frac{\pi}{3}\ \text{with}\ \hat{i},\frac{\pi}{4}\ \text{with}\ \hat{j}$ and an acute angle $\theta$ with $\hat{k},$ then find $\theta$ and hence, the components of $\vec{a}.$
Answer$\text{Let}\ \ \ \hat{a}=\text{x}\hat{i}+\text{y }\hat{j}+\text{z}\hat{k}\ \text{be a unit vector.}\ \ \ \ ......\text{(i)}$ $\Rightarrow\ \ |\hat{a}|=1\ \ \Rightarrow\ \ \sqrt{\text{x}^2+\text{y}^2+\text{z}^2}=1$ $\text{Squaring both sides,}\ \ \ \ \ \text{x}^2+\text{y}^2+\text{z}^2=1\ \ \ .....\text{(ii)}$ $\text{Given:}\ \ \text{Angle between vectors}\ \hat{a}\ \text{and}\ \hat{i}=\hat{i}+0\hat{j}+0\hat{k}\ \text{is}\ \frac{\pi}{3}.$ $\therefore\ \cos\frac{\pi}{3}=\frac{\hat{a}.\hat{i}}{\big|\hat{a}\big|.\big|\hat{i}\big|}\ $ $\Rightarrow\ \ \frac{1}{2}=\frac{\text{x}(1)+\text{y}(0)+\text{z}(0)}{(1)(1)}$ $\Rightarrow\ \ \ \frac{1}{2}=\text{x}\ \ \ ....\text{(iii)}$Again, given $\text{Angel between vectors}\ \hat{a}\ \text{and}\ \hat{j}=0\hat{i}+\hat{j}+0\hat{k}\ \text{is}\ \frac{\pi}{4}.$
$\therefore\ \cos\frac{\pi}{4}=\frac{\hat{a}.{\hat{j}}}{|\hat{a}|.|\hat{j}|}\ \Rightarrow\ \frac{1}{\sqrt{2}}=\frac{\text{x}(0)+\text{y}(1)+\text{z}(0)}{(1)(1)}$
$\Rightarrow\ \frac{1}{\sqrt{2}}=\text{y}\ \ ......\text{(iv)}$
Again, given $\text{Angel between vectors}\ \hat{a}\ \text{and}\ \hat{k}=0\hat{i}+0\hat{j}+\hat{k}\ \text{is}\ \theta,\ \text{where}\ \theta\ \text{is acute angle.}$
$\therefore\ \cos\theta=\frac{\hat{a}.{\hat{k}}}{|\hat{a}|.|\hat{k}|}\ $ $\Rightarrow\ \ \cos\theta=\frac{\text{x}(0)+\text{y}(0)+\text{z}(1)}{(1)(1)}$ $\Rightarrow\ \ \cos\theta=\text{z}\ \ \ .......\text{(v)}$ Putting the values of x, y and z in eq. (ii),$\frac{1}{4}+\frac{1}{2}+\cos^2\theta=1$ $\ \Rightarrow\ \ \cos^2\theta=1-\frac{1}{4}-\frac{1}{2}$
$\Rightarrow\ \ \ \ \ \ \ \cos^2\theta=\frac{4-1-2}{4}=\frac{1}{4}$ $\ \Rightarrow\ \ \cos\theta=\pm\frac{1}{2}$ Since $\theta$ is acute angle, therefore cos $\theta$ is positive and hence $\frac{1}{2}=\cos\frac{\pi}{3}\ \Rightarrow\ \ \theta=\frac{\pi}{3}$ From eq. (v), $\ \ \text{z}=\cos\theta=\frac{1}{2}$ Putting values of x, y and z in eq. (i), $\ \ \hat{a}=\frac{1}{2}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{2}\hat{k}$$\therefore\ \ \text{Components of}\ \hat{a}\ \text{are coefficients of}\ \hat{i},\hat{j},\hat{k}\ \text{in}\ \hat{a}$
$\Rightarrow\ \ \frac{1}{2},\frac{1}{\sqrt{2}},\frac{1}{2}\ \text{and angle}\ \theta=\frac{\pi}{3}$
View full question & answer→Question 2042 Marks
Evaluate the product $\big(3\vec{a}-5\vec{b})\cdot\big(2\vec{a}+7\vec{b}).$
Answer$\text{Given:}\ \ \ \ \big(3\vec{a}-5\vec{b}\big).\big(2\vec{a}+7\vec{b}\big)$ $=\big({3\vec{a}\big)}.\big(2\vec{a}\big)+\big(3\vec{a}\big).\big(7\vec{b}\big)-\big(5\vec{b}\big).\big(2\vec{a}\big)-\big(5\vec{b}\big).\big(7\vec{b}\big)$
$=6\vec{a}.\vec{a}+21\vec{a}.\vec{b}-10\vec{b}.\vec{a}-35\vec{b}.\vec{b}$
$=6\big|\vec{a}\big|^2+21\vec{a}.\vec{b}-10\vec{a}.\vec{b}-35\Big|\vec{b}\Big|^2$
$=6\big|\vec{a}\big|^2+11\vec{a}.\vec{b}-35\Big|\vec{b}\Big|^2$
View full question & answer→Question 2052 Marks
If $\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}$ and $\vec{\text{b}}=-\hat{\text{j}}+2\hat{\text{k}},$ find $\big(\vec{\text{a}}-2\vec{\text{b}}\big).\big(\vec{\text{a}}+\vec{\text{b}}\big).$
AnswerWe have
$\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}$ and $\vec{\text{b}}=-\hat{\text{j}}+2\hat{\text{k}}$
$\vec{\text{a}}-2\vec{\text{b}}=\big(\hat{\text{i}}-\hat{\text{j}}\big)-2\big(-\hat{\text{j}}+2\hat{\text{k}}\big)\\=\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{j}}-4\hat{\text{k}}=\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}}$
$\vec{\text{a}}+\vec{\text{b}}=\hat{\text{i}}-\hat{\text{j}}-\hat{\text{j}}+2\hat{\text{k}}=\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}$
$\big(\vec{\text{a}}-2\vec{\text{b}}\big).\big(\vec{\text{a}}+\vec{\text{b}}\big)$
$=\big(\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}}\big).\big(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}\big)$
$=1-2-8$
$=-9$
View full question & answer→Question 2062 Marks
Prove that 1, 1, 1 cannot be direction cosines of a straight line.
AnswerLet 1, 1, 1 be the direction cosines of a straight line. Then,
$1^2+1^2+1^2=3\neq1$
Since direction cosines of a line which makes equal angle with the axes must satisfy
$\text{l}^2+\text{m}^2+\text{n}^2=1$
Hence 1, 1, 1 cannot be the direction cosines of a straight line.
View full question & answer→Question 2072 Marks
Find the angle betwwen two vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ if$|\vec{\text{a}}|=\sqrt{3},\big|\vec{\text{b}}\big|=2$ and $\vec{\text{a}}.\vec{\text{b}}=\sqrt{6}$
AnswerWe have,
$|\vec{\text{a}}|=\sqrt{3},\big|\vec{\text{b}}\big|=2$ and $\vec{\text{a}}.\vec{\text{b}}=\sqrt{6}$
Let $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$. then
$\cos\theta=\frac{\vec{\text{a}}.\vec{\text{b}}}{|\vec{\text{a}}|\big|\vec{\text{b}}\big|}$
$=\frac{\sqrt{6}}{\sqrt{3}\times2}$
$=\frac{1}{\sqrt{2}}$
$\theta=\cos^{-1}\Big(\frac{1}{\sqrt{2}}\Big)$
$=\frac{\pi}{4}$
View full question & answer→Question 2082 Marks
Show that
$(\vec{a}-\vec{b})\times(\vec{a}+\vec{b})=2(\vec{a}\times{\vec{b}})$
Answer$\text{L.H.S}=(\vec{a}-\vec{b})\times(\vec{a}+\vec{b})=\vec{a}\times\vec{a}+\vec{a}\times\vec{b}-\vec{b}\times\vec{a}-\vec{b}\times\vec{b}$ $\ \Big[\because\vec{a}\times\vec{a}=\vec{0},\vec{b}\times\vec{b}=\vec{0},\vec{b}\times\vec{a}=-\vec{a}\times\vec{b}\Big]$
$=\vec{0}+\vec{a}\times\vec{b}+\vec{a}\times\vec{b}-\vec{0}=2\vec{a}\times\vec{b}=\text{R.H.S.}$
View full question & answer→Question 2092 Marks
If $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}},\ \vec{\text{b}}=\hat{\text{j}}+\hat{\text{k}},\ \vec{\text{c}}=\hat{\text{k}}+\hat{\text{i}}$, find the unit vector in the direction of $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}$.
AnswerLet $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}},\ \vec{\text{b}}=\hat{\text{j}}+\hat{\text{k}},\ \vec{\text{c}}=\hat{\text{k}}+\hat{\text{i}}$
Then, $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{j}}+\hat{\text{k}}+\hat{\text{k}}+\hat{\text{i}}$
$=2\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$
$\therefore\ |\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}|=\sqrt{2^2+2^2+2^2}$
$=\sqrt{12}$
$=2\sqrt3$
Therefore, unit vector in the direction of $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=\frac{2\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)}{2\sqrt3}=\frac{1}{\sqrt3}\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$
View full question & answer→Question 2102 Marks
Find the angle at which the following vectors are inclined to each of the coordinate axes: $\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
AnswerLet $\vec{\text{r}}$ be the given vector, and let it make an angle $\alpha,\beta,\gamma$ with OX, OY, OZ respectively. Then, its direction cosines are $\cos\alpha,\cos\beta,\cos\gamma$.So direction ratios of $\vec{\text{r}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ are proportional to 1, -1, 1. Therefore, Direction cosine of $\vec{\text{r}}$ are $\frac{1}{\sqrt{1^2+(-1)^2+1^2}},\frac{-1}{\sqrt{1^2+(-1)^2+1^2}},\frac{1}{\sqrt{1^2+(-1)^2+1^2}}$ or, $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}.$ $\therefore\cos\alpha=\frac{1}{\sqrt{3}},\cos\beta=\frac{1}{\sqrt{3}},\cos\gamma=\frac{1}{\sqrt{3}}$ $\alpha=\cos^{-1}=\Big(\frac{1}{\sqrt{3}}\Big),\beta=\cos^{-1}=\Big(\frac{-1}{\sqrt{3}}\Big),\gamma=\cos^{-1}=\Big(\frac{1}{\sqrt{3}}\Big)$
View full question & answer→Question 2112 Marks
What is the cosine of the angle with the vector $\sqrt2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ makes with y-axis?
AnswerGiven $\sqrt2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
Therefore, direction cosines are $\frac{\sqrt2}{\sqrt{(\sqrt2)^2+1^2+1^2}},\frac{1}{\sqrt{(\sqrt2)^2+1^2+1^2}},\frac{1}{\sqrt{(\sqrt2)^2+1^2+1^2}}$ or $\frac{1}{\sqrt2},\frac{1}2,\frac{1}2$
So, cosine angle with respect to y-axis is $\frac{1}2$
View full question & answer→Question 2122 Marks
Find the angle between two vectors $\vec{a}\ \text{and}\ \vec{b}$ with magnitudes $\sqrt{3}\ \text{and}\ 2,$ respectively having $\vec{a}\cdot\vec{b}=\sqrt{6}.$
Answer$\text{Given:}\ \ \big|\vec{a}\big|=\sqrt{3},\Big|\vec{b}\Big|=2\ {\text{and}}\ \vec{a}.\vec{b}=\sqrt{6}$ Let $\theta$ be the angle between the vector $\vec{a}\ \text{and}\ \vec{b}.$ We know that $\text{cos}\ \theta=\frac{\vec{a}.\vec{b}}{\big|\vec{a}\big|.\big|\vec{b}\big|}$$\Rightarrow\ \ \text{cos}\ \theta=\frac{\sqrt{6}}{\sqrt{3}.2}=\frac{\sqrt{3}.\sqrt{2}}{\sqrt{3}.\sqrt{2}.\sqrt{2}}=\frac{1}{\sqrt{2}}$ $\Rightarrow\ \ \text{cos}\ \theta=\text{cos}\frac{\pi}{4}$
$\Rightarrow\ \theta= \frac{\pi}{4}$
View full question & answer→Question 2132 Marks
Show that the direction cosines of a vector equally inclined to the axes $OX, OY$ and $OZ$ are $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}.$
AnswerLet a vector be equally inclined to axes OX, OY, and OZ at angle a.
Then, the direction cosines of the vector are cos a, cos a, and cos a.
Now,
$\cos^{2}\alpha+\cos^{2}\alpha+\cos^{2}\alpha=1$
$\Rightarrow3\cos^{2}\alpha=1$
$\Rightarrow\cos\alpha=\frac{1}{\sqrt{3}}$
Hence, the direction cosines of the vector which are equally inclined to the axes are $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}.$
View full question & answer→Question 2142 Marks
Find $\big|\vec{\text{a}}-\vec{\text{b}}\big|$ if
$|\vec{\text{a}}|=2,\big|\vec{\text{b}}\big|=3$ and $\vec{\text{a}}.\vec{\text{b}}=4$
AnswerGiven that$|\vec{\text{a}}|=2,\big|\vec{\text{b}}\big|=3$ and $\vec{\text{a}}.\vec{\text{b}}=4\dots(1)$
We know that
$\big|\vec{\text{a}}-\vec{\text{b}}\big|^2=|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-2\vec{\text{a}}.\vec{\text{b}}$
$=2^2+3^2-2(4)$ [using (1)]
$=4+9-8$
$=5$
$\therefore \big|\vec{\text{a}}-\vec{\text{b}}\big|=\sqrt{5}$
View full question & answer→Question 2152 Marks
For what of $\lambda$ are the vectors $\vec{\text{a}}=2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$ perpendicular to each other?
AnswerWe have
$\vec{\text{a}}=2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$
Given that $\vec{\text{a}}$ and $\vec{\text{b}}$ are perpendicular.
$\Rightarrow\vec{\text{a}}.\vec{\text{b}}=0$
$\Rightarrow\big(2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}\big).\big(\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\big)=0$
$\Rightarrow2-2\lambda+3=0$
$\Rightarrow5-2\lambda=0$
$\therefore\lambda=\frac{5}{2}$
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