Question 512 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 522 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 $2\vec{\text{a}}-\vec{\text{b}}+3\vec{\text{c}}$.
AnswerWe 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 532 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)$$=\hat{\text{i}}.\hat{\text{i}}+\hat{\text{j}}.\hat{\text{j}}+\hat{\text{k}}\big(-\hat{\text{k}}\big)$
$=|\hat{\text{i}}|^2+|\hat{\text{j}}|^2-|\hat{\text{k}}|^2$
$=1+1-1$ $\big(\because|\hat{\text{i}}|=1,|\hat{\text{j}}|=1\text{ and } |\hat{\text{k}}|=1\big)$
$=1$
View full question & answer→Question 542 Marks
Write the value of $\hat{\text{i}}\times\big(\hat{\text{j}}\times\hat{\text{k}}\big).$
Answer$\hat{\text{i}}\times\big(\hat{\text{j}}\times\hat{\text{k}}\big)$$=\hat{\text{i}}\times\hat{\text{i}}$
$=0$
View full question & answer→Question 552 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 562 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 572 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 582 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 592 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 602 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 612 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 622 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 632 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 642 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 652 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 662 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 672 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 682 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 692 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 702 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 712 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 722 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 732 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 742 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 752 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 762 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 772 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 782 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 792 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 802 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}$
View full question & answer→Question 812 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 822 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 832 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 842 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}}$.
AnswerGiven 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 852 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 862 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 872 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 882 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 892 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 902 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 912 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:$\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$
$\vec{\text{b}}=\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{a}_1&\text{a}_2&\text{a}_3\\\text{b}_1&\text{b}_2&\text{b}_3\end{vmatrix}$
$=\hat{\text{i}}(\text{a}_2\text{b}_3-\text{a}_3\text{b}_2)-\hat{\text{j}}(\text{a}_1\text{b}_3-\text{a}_3\text{b}_1)+\hat{\text{k}}(\text{a}_1\text{b}_2-\text{a}_2\text{b}_1)$
$\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{b}}$
$=\big[\hat{\text{i}}(\text{a}_2\text{b}_3-\text{a}_3\text{b}_2)-\hat{\text{j}}(\text{a}_1\text{b}_3-\text{a}_3\text{b}_1)+\hat{\text{k}}(\text{a}_1\text{b}_2-\text{a}_2\text{b}_1)\big]\\.\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big)$
$=\text{a}_2\text{b}_1\text{b}_3-\text{a}_3\text{b}_1\text{b}_2-\text{a}_1\text{b}_2\text{b}_3+\text{a}_3\text{b}_1\text{b}_2+\text{a}_1\text{b}_2\text{b}_3-\text{a}_2\text{b}_1\text{b}_3$
$=\text{b}_1(\text{a}_2\text{b}_3-\text{a}_3\text{b}_2)-\text{b}_2(\text{a}_1\text{b}_3-\text{a}_3\text{b}_1)+\text{b}_3(\text{a}_1\text{b}_2-\text{a}_2\text{b}_1)$
$=\text{a}_2\text{b}_1\text{b}_3-\text{a}_3\text{b}_1\text{b}_2-\text{a}_1\text{b}_2\text{b}_3+\text{a}_3\text{b}_1\text{b}_2+\text{a}_1\text{b}_2\text{b}_3-\text{a}_2\text{b}_1\text{b}_3$
$=0$
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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 932 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 942 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 952 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 962 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 972 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 982 Marks
Write the value of $\hat{\text{i}}.\big(\hat{\text{j}}\times\hat{\text{k}}\big)+\hat{\text{j}}.\big(\hat{\text{k}}\times\hat{\text{i}}\big)+\hat{\text{k}}.\big(\hat{\text{i}}\times\hat{\text{j}}\big).$
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 992 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 1002 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
$\cos\theta=\frac{\vec{\text{a}}.\vec{\text{b}}}{|\vec{\text{a}}|\big|\vec{\text{b}}\big|}$
$=\frac{1}{3.3}$
$\cos\theta=\frac{1}{9}$
$\theta=\cos^{-1}\big(\frac{1}{9}\big)$
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