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 $\vec{\text{a}}$ 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 52 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 62 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 72 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 82 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 92 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 102 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 112 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 122 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 132 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 142 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 152 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 162 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 172 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 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 $\hat{\text{a}},\hat{\text{b}}$ 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 202 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 212 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 222 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 232 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 242 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 252 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 262 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 272 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 282 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 292 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 302 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 312 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 322 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 332 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 342 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 352 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 362 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 372 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 382 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 392 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 402 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 412 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 422 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 432 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 442 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 452 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 462 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 472 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 482 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 492 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 502 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)
$\Rightarrow|\vec{\text{a}}|\big|\vec{\text{b}}\big||\sin\theta|=|\vec{\text{a}}|\big|\vec{\text{b}}\big||\cos\theta|$
$\Rightarrow|\sin\theta|=|\cos\theta|$
$\Rightarrow\theta=\frac{\pi}{4}$
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