Question 12 Marks
If $\vec a = \hat i + \hat j + \hat k,\vec b = 2\hat i - \hat j + 3\hat k$ and $\vec c = \hat i - 2\hat j + \hat k$ find a unit vector parallel to the vector $2\vec a - \vec b + 3\vec c$
AnswerGiven: Vectors $\vec a = \hat i + \hat j + \hat k,\vec b = 2\hat i - \hat j + 3\hat k$ and $\vec c = \hat i - 2\hat j + \hat k$
Let $\vec d = 2\vec a - \vec b + 3\vec c$
$= 2\left( {\hat i + \hat j + \hat k} \right) - \left( {2\hat i - \hat j + 3\hat k} \right) + 3\left( {\hat i - 2\hat j + \hat k} \right)$
$= 2\hat i + 2\hat j + 2\hat k - 2\hat i + \hat j - 3\hat k + 3\hat i - 6\hat j + 3\hat k$
$ = 3\hat i - 3\hat j + 2\hat k$
$\therefore$ A unit vector parallel to the vector $\vec d$ is
$\hat d = \frac{{\vec d}}{{\left| {\vec d} \right|}} = \frac{{3\hat i - 3\hat j + 2\hat k}}{{\sqrt {9 + 9 + 4} }}$
$= \frac{{3\hat i - 3\hat j + 2\hat k}}{{\sqrt {22} }}$
$= \frac{3}{{\sqrt {22} }}\hat i - \frac{3}{{\sqrt {22} }}\hat j + \frac{2}{{\sqrt {22} }}\hat k$
View full question & answer→Question 22 Marks
Find the value of x for which $x\left( {\hat i + \hat j + \hat k} \right)$ is a unit vector.
AnswerAs $x$$\left( {\hat i + \hat j + \hat k} \right)$ is a unit vector, therefore, $ \left| {x\left( {\hat i + \hat j + \hat k} \right)} \right| = 1 \Rightarrow x\sqrt {1 + 1 + 1} = 1 \Rightarrow x = \frac{1}{{\sqrt 3 }} \\ $
View full question & answer→Question 32 Marks
If $\vec a = \vec b + \vec c$, then is it true that $\left| {\vec a} \right| = \left| {\vec b} \right| + \left| {\vec c} \right|$? Justify your answer.
AnswerGiven: $\vec a = \vec b + \vec c$
$\therefore$ Either the vectors $\vec a,\vec b,\vec c$ are collinear or form the sides of a triangle.
Case I: Vectors $\vec a,\vec b,\vec c$ are collinear.
Let $\vec a = \overrightarrow {AC} ,\vec b = \overrightarrow {AB} $ and $\vec c = \overrightarrow {BC} $
Then $\vec a = \overrightarrow {AC} = \overrightarrow {AB} + \overrightarrow {BC} = \vec b + \vec c$
Also, $\left| {\vec a} \right| = AC = AB + BC = \left| {\vec b} \right| + \left| {\vec c} \right|$
Case II: Vectors $\vec a,\vec b,\vec c$ form a triangle.
Here also by Triangle Law of vectors, $\vec a = \vec b + \vec c$
But $\left| {\vec a} \right| < \left| {\vec b} \right| + \left| {\vec c} \right|$ [$\because $ Each side of a triangle is less than sum of the other two sides]
$\therefore \,\left| {\vec a = \vec b + \vec c} \right| = \left| {\vec b} \right| + \left| {\vec c} \right|$ is true only when vectors $\vec b$ and $\vec c$are collinear vectors.
View full question & answer→Question 42 Marks
Find the scalar components and magnitude of the vector joining the points $P(x_1, y_1, z_1)$ and $Q (x_2, y_2, z_2)$
AnswerGiven points are $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2).$
The vector obtained by joining the given points $P$ and $Q$ is:
$\vec{\mathrm{PQ}}$ = position vector of $Q -$ position vector of $P$
= $\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right) \hat{\mathrm{i}}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right) \hat{\mathrm{j}}+\left(\mathrm{z}_{2}-\mathrm{z}_{1}\right) \hat{\mathrm{k}}$
$|\vec{\mathrm{PQ}}|=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}+\left(\mathrm{z}_{2}-\mathrm{z}_{1}\right)^{2}}$
Hence, the scalar components and the magnitude of the vector joining the given points $P\ \&\ Q$ are $\{(x_2- x_1), (y_2- y_1), (z_2- z_1)\}$ and $\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}$ respectively.
View full question & answer→Question 52 Marks
Prove that $(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^{2}+|\vec{b}|^{2}$, if and only if $\vec{a}, \vec{b}$ are perpendicular, given $\vec{a} \neq \vec{0}, \vec{b} \neq \vec{0}$.
AnswerGiven: $(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^{2}+|\vec{b}|^{2}$
To prove: vectors $\vec{a}$ and $\vec b$ are mutually perpendicular to each other.
$\Rightarrow \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}=0$
$\because(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^{2}+|\vec{b}|^{2}$
$\Rightarrow(\vec{\mathrm{a}} \cdot \vec{\mathrm{a}}+\vec{\mathrm{a}} \cdot \vec{\mathrm{b}}+\vec{\mathrm{b}} \cdot \vec{\mathrm{a}}+\vec{\mathrm{b}} \cdot \vec{\mathrm{b}})=|\vec{\mathrm{a}}|^{2}+|\vec{\mathrm{b}}|^{2}$
$(\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a})$ scalar product is commutative.
$\Rightarrow\left(|\vec{\mathrm{a}}|^{2}+|\vec{\mathrm{b}}|^{2}+2 \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}\right)=|\vec{\mathrm{a}}|^{2}+|\vec{\mathrm{b}}|^{2}$
$\Rightarrow(2 . \vec{\mathrm{a}} . \vec{\mathrm{b}})=0$
$\Rightarrow \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}=0$
Hence, $\vec a$ and $\vec b$ are mutually perpendicular to each other. as $\vec{a} \neq 0$ and $\overrightarrow{\mathrm{b}} \neq 0$ is given.
View full question & answer→Question 62 Marks
Show that the direction cosines of a vector equally inclined to the axes $OX, OY$ and $OZ$ are $\pm \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$
AnswerLet the required vector be inclined to $OX, OY$ and $OZ$ at angle $\alpha$.
Then the direction cosine of the vectors are $\cos \alpha , \cos \alpha$ and $\cos \alpha$.
We have, $\cos^2 \alpha + \cos^{2 }\alpha + \cos^2 \alpha = 1 [$Sum of the squares of direction cosines is $1]$
$\Rightarrow 3 \cos^2\alpha = 1$
$\Rightarrow \cos^{2 }\alpha = \frac{1}{3}$
$\Rightarrow \cos \alpha=\frac{1}{\sqrt{3}}$
Hence, the direction cosines of the required vector which are equally inclined to the axis are $\pm \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$
View full question & answer→Question 72 Marks
Write down a unit vector in $XY$-plane, making an angle of $30^\circ$ with the positive direction of x-axis.
AnswerLet $\overrightarrow r = x\widehat i + y\widehat j $ be a unit vector in $XY$ plane,making angle $30^\circ$ with positive $X$ axis ,
so we have the vector as $\overrightarrow r = \cos {30^0}\hat i + \sin {30^0}\hat j$
$x = \frac{{\sqrt 3 }}{2};y = \frac{1}{2},\left( {\because \left| {\overrightarrow r } \right| = 1} \right).$
$\therefore \overrightarrow r = \frac{{\sqrt 3 }}{2}\widehat i + \frac{1}{2}\widehat j$ is the required vector.
View full question & answer→Question 82 Marks
If either $\vec{a}=\vec {0}$ or $\vec{b}=\vec{0}$, then $\vec{a} \times \vec{b}=\vec 0$. Is the converse true? Justify your answer with an example.
AnswerFirstly let us consider, Take any parallel non-zero vectors so that
$\vec{a} \times \vec{b}=\vec{0}$
Let $\vec{a}=2 \hat{i}+3 \hat{j}+4 \hat{k}$, $\vec{b}=4 \hat{i}+6 \hat{j}+8 \hat{k}$
Then,
$\vec{a} \times \vec{b}$ = $\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 4 & 6 & 8\end{array}\right|$ = $\hat{i}(24-24)-\hat{j}(16-16)+\hat{k}(12-12)$ = $0 \hat{i}+0 \hat{j}+0 \hat{k}=\vec{0}$
Now, it's seen that $|\vec{a}|=\sqrt{2^{2}+3^{2}+4^{2}}=\sqrt{29}$
$\therefore \vec{a} \neq \vec{0}$
$|\vec{b}|=\sqrt{4^{2}+6^{2}+8^{2}}=\sqrt{116}$
$\therefore \vec{b} \neq \vec{0} $
Thus, the converse of the given statement need not be true.
View full question & answer→Question 92 Marks
Given that $\vec{a} \cdot \vec{b}=0$ and $\vec{a} \times \vec{b}=\overrightarrow{0}$. What can you conclude about the vectors $\vec a $ and $\vec b$?
AnswerGiven that $\vec{a} \cdot \vec{b}=0$
$\Rightarrow|\vec{\mathrm{a}}| \cdot|\vec{\mathrm{b}}| \cdot \cos \theta=0$ (where $\theta$ is the angle between the vectors)
$\Rightarrow|\vec{\mathrm{a}}|=0$ or $|\vec{\mathrm{b}}|=0$ or $\cos \theta=0$
Also given that, $\vec{\mathrm{a}} \times \vec{\mathrm{b}}=0$
$\Rightarrow|\vec{\mathrm{a}}| \cdot|\vec{\mathrm{b}}| \cdot \sin \theta=0$ (where $\theta$ is the angle between the vectors)
$\Rightarrow|\vec{\mathrm{a}}|=0$ or $|\vec{\mathrm{b}}|=0$ or $\sin \theta=0$
As there is no value of $\theta$ for which both sin $\theta$ and cos $\theta$ are zero.
$\therefore$ The condition for which $\vec{a} \cdot \vec{b}=0$ and $\vec{\mathrm{a}} \times \vec{\mathrm{b}}=0$ is;
Either $|\vec{\mathrm{a}}|=0$ or $|\vec{\mathrm{b}}|=0$
View full question & answer→Question 102 Marks
Find $\lambda $ and $\mu $ if $\left( {2\hat i + 6\hat j + 27\hat k} \right) \times \left( {\hat i + \lambda \hat j + u\hat k} \right) = \overrightarrow 0 $
AnswerGiven: $\left( {2\hat i + 6\hat j + 27\hat k} \right) \times \left( {\hat i + \lambda \hat j + u\hat k} \right) = 0$ $\Rightarrow \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ 2&6&{27} \\ 1&\lambda &u \end{array}} \right| = \vec 0$
Expanding along first row,
$\hat i\left( {6\mu - 27\lambda } \right) - \hat j\left( {2\mu - 27} \right) + \hat k\left( {2\lambda - 6} \right) $ $= \vec 0 = 0\hat i + 0\hat j + 0\hat k$
Comparing the coefficients of $\hat i,\hat j,\hat k$ on both sides, we have
$6\mu - 27\lambda = 0$ …(i)
$2\mu - 27 = 0$ …(ii)
And $2\lambda - 6 = 0$ ...(iii)
From eq. (ii), $2\mu - 27 = 0 \Rightarrow u = \frac{{27}}{2}$
From eq. (iii), $2\lambda - 6 = 0 \Rightarrow \lambda = \frac{6}{2} = 3$
Putting the values of $\mu $ and $\lambda $ in eq. (i),
$6\left( {\frac{{27}}{2}} \right) - 27\left( 3 \right) = 0 $ $\Rightarrow 3\left( {27} \right) - 27\left( 3 \right) = 0 \Rightarrow 0 =0$
Therefore, $\mu = \frac{{27}}{2}$ and $\lambda = 3.$
View full question & answer→Question 112 Marks
Show that $\left( {\vec a - \vec b} \right) \times \left( {\vec a + \vec b} \right) = 2\left( {\vec a \times \vec b} \right)$
AnswerL.H.S $= \left( {\vec a - \vec b} \right) \times \left( {\vec a + \vec b} \right)$ $= \vec a \times \vec a + \vec a \times \vec b - \vec b \times \vec a - \vec b \times \vec b$
$= 0 + \vec a \times \vec b + \vec a \times \vec b - 0$ $[as \ \vec a×\vec b=-\vec b×\ \vec a]$
$=2(\vec a×\vec b)$
View full question & answer→Question 122 Marks
Find $|\vec{a} \times \vec{b}|$, if $\vec{a}=\hat{i}-7 \hat{j}+7 \hat{k}$ and $\vec{b}=3 \hat{i}-2 \hat{j}+2 \hat{k}$.
AnswerGiven that $\vec{a}=\hat{1}-7 \hat{\jmath}+7 \hat{k}$ and $\vec{\mathrm{b}}=3 \hat{1}-2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
$\vec{a} \times \vec{b}=\left|\begin{array}{ccc} {\hat{1}} & {\hat{\jmath}} & {\hat{k}} \\ {1} & {-7} & {7} \\ {3} & {-2} & {2} \end{array}\right|$
Expanding along first row,
$\vec{a} \times \vec{b}$ = $\hat{\mathrm{i}}[(-7 \times 2)-(-2 \times 7)]-\hat{\mathrm{j}}[(1 \times 2)-(7 \times 3)]+\hat{\mathrm{k}}[(-2 \times 1)-(-7 \times 3)]$
$\Rightarrow \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=(0) \hat{\mathrm{i}}-(-19) \hat{\mathrm{j}}+(19) \hat{\mathrm{k}}$
$\Rightarrow \vec{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=0 \hat{i}+19 \hat{\jmath}+19 \hat{\mathrm{k}}$
$\therefore|\vec{\mathrm{a}} \times \vec{\mathrm{b}}|=\sqrt{0^{2}+19^{2}+19^{2}}=19 \sqrt{2}$
View full question & answer→Question 132 Marks
Find $|\vec{x}|$, if for a unit vector $\vec{a},(\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})=12$
AnswerGiven, vector $\vec a $ is a unit vector
$\therefore|\vec{\mathrm{a}}|=1$
$(\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})=12$
$\Rightarrow \vec{\mathrm{x}} \cdot \vec{\mathrm{x}}+\vec{\mathrm{x}} \cdot \vec{\mathrm{a}}-\vec{\mathrm{a}} \cdot \vec{\mathrm{x}}-\vec{\mathrm{a}} \cdot \vec{\mathrm{a}}=12$
$\Rightarrow|\vec{\mathrm{x}}|^{2}+\vec{\mathrm{x}} \cdot \vec{\mathrm{a}}-\vec{\mathrm{x}} \cdot \vec{\mathrm{a}}-|\vec{\mathrm{a}}|^{2}=12$ [Since $\vec{\mathrm{x}} \cdot \vec{\mathrm{a}}=\vec{\mathrm{a}} \cdot \vec{\mathrm{x}}$]
$\Rightarrow|\vec{\mathrm{x}}|^{2}-|\vec{\mathrm{a}}|^{2}=12$
$\Rightarrow|\vec{\mathrm{x}}|^{2}-1^{2}=12$
$\Rightarrow|\vec{\mathrm{x}}|^{2}=12+1=13$
$\Rightarrow|\vec{\mathrm{x}}|=\sqrt{13}$
$\therefore|\vec{\mathrm{x}}|=\sqrt{13}$
View full question & answer→Question 142 Marks
Find the magnitude of two vectors $\vec a$ and $\vec b$, having the same magnitude and such that the angle between them is $60^\circ$ and their scalar product is $\frac{1}{2}$.
AnswerLet $\theta$ be the angle between $\vec a$ and $\vec b$
Given, magnitude of vector $\vec a$ = magnitude of vector $\vec b$
$|\vec{\mathrm{a}}|=|\vec{\mathrm{b}}|$ and $\theta=60^{\circ}$
Scalar product of $\vec a$ and $\vec b$ = $\frac{1}{2}$
i.e. $\vec{a} \cdot \vec{b}=\frac{1}{2}$
We know, $\vec{\mathrm{a}} \cdot \vec{\mathrm{b}}=|\vec{\mathrm{a}}||\vec{\mathrm{b}}| \cos \theta$
$\Rightarrow \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}=|\vec{\mathrm{a}}|^{2} \cos \theta$
$\Rightarrow \frac{1}{2}=|\vec{a}|^{2} \cos 60^{\circ}$
$\Rightarrow \frac{1}{2}=|\vec{a}|^{2} \times \frac{1}{2}$
$\Rightarrow|\overrightarrow{\mathrm{a}}|^{2}=1$
$\therefore|\overrightarrow{\mathrm{a}}|=1$ [magnitude of vector is positive]
So, $|\vec{\mathrm{b}}|=1$
Hence, $|\overrightarrow{\mathrm{a}}|=|\overrightarrow{\mathrm{b}}|=1$
View full question & answer→Question 152 Marks
Evaluate the product $\left( {3\vec a - 5\vec b} \right).\left( {2\vec a + 7\vec b} \right)$
Answer$\left( {3\vec a - 5\vec b} \right).\left( {2\vec a + 7\vec b} \right)$ $=6{\left| {\vec a} \right|^2} + 11\vec a.\vec b - 35{\left| {\vec b} \right|^2}$
$\left[ {\because \vec a.\vec a = {{\left| {\vec a} \right|}^2},\vec b.\vec b = {{\left| {\vec b} \right|}^2},\vec a.\vec b = \vec b.\vec a} \right]$
View full question & answer→Question 162 Marks
Find the projection of the vector $\hat i + 3\hat j + 7\hat k$ on the vector $7\hat i - \hat j + 8\hat k$
AnswerLet $\vec a = \hat i + 3\hat j + 7\hat k$ and $\vec b = 7\hat i - \hat j + 8\hat k$ Projection of vector $\vec a$ on $\vec b = \frac{{\vec a.\vec b}}{{\left| {\vec b} \right|}}$
$= \frac{{\left( 1 \right)\left( 7 \right) + \left( 3 \right)\left( { - 1} \right) + 7\left( 8 \right)}}{{\sqrt {{{\left( 7 \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( 8 \right)}^2}} }}$
$= \frac{{7 - 3 + 56}}{{\sqrt {49 + 61 + 64} }} = \frac{{60}}{{\sqrt {114} }}$
View full question & answer→Question 172 Marks
Find the projection of the vector $\hat i - \hat j$ on the vector $\hat i + \hat j$.
AnswerLet $\vec {a} = \hat i - \hat j = \hat i - \hat j + 0\hat k$ and $\vec{b}=\hat{i}+\vec{j}+0 . \hat{k}$
$\therefore$ Angle between $\vec a$ and $\vec b$ is
cos $\theta$ $=\frac{\vec {a}\cdot \vec{b}}{|\vec{a}| \cdot|\vec{b}|}$
$=\frac{(\hat i- \hat j+ 0. \hat k) \cdot\left(\hat i+\hat j+0. \hat {k}\right) }{\sqrt{1^{2}+1^{2}} \cdot \sqrt{1^{2}+1^{2}}}$
$=\frac{1-1}{\sqrt{2} \times \sqrt{2}}=0$
$\therefore $ cos $\theta$ = 0 = cos $\pi \over 2$
$\theta$ = $\pi \over 2$
View full question & answer→Question 182 Marks
If either vector $\vec{a}=\vec{0}$ or $\vec{b}=\vec{0}$, then $\vec{a} \cdot \vec{b}$ = 0. But the converse need not be true. Justify your answer with an example.
AnswerIf either $\vec{a}=0$ or $\vec{\mathrm{b}}=0$ then $\vec{a} . \vec{b}=0$
Now let $\vec{\mathrm{a}}=2 \hat{1}+3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}} $ and $\vec{\mathrm{b}}=3 \hat{1}-6 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
$\vec{\mathrm{a}} \cdot \vec{\mathrm{b}}=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}) \cdot(3 \hat{1}-6 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})$
$\Rightarrow \vec{\mathrm{a}} \cdot \vec{\mathrm{b}}=2 \hat{1} .3 \hat{1}-2 \hat{1} \cdot 6 \hat{\mathrm{j}}+2 \hat{1} \cdot 2 \hat{\mathrm{k}}+3 \hat{\mathrm{j}} \cdot 3 \hat{1}$ - $3 \hat{0} \cdot 6 \hat{\jmath}+3 \hat{\jmath} .2 \hat{\mathrm{k}}+6 \hat{\mathrm{k}} .3 \hat{1}-6 \hat{\mathrm{k}} .6 \hat{\mathrm{j}}+6 \hat{\mathrm{k}} .2 \hat{\mathrm{k}}$
$\Rightarrow \vec{a} \cdot \vec{b}$ = 6 - 0 + 0 + 0 - 18 + 0 + 0 - 0 + 12 [$\hat{ {1}} \hat{ {j}}=\hat{ {j}} \cdot \hat{ {k}}=\hat{ {k}} \cdot \hat{ {\imath}}=0$]
$\Rightarrow \vec{a} \cdot \vec{b}$ = 18 - 18 = 0 [$\text { i. }\hat{\mathrm{i}}=\hat{\mathrm{j}} \cdot \hat{\mathrm{j}}=\hat{\mathrm{k}}. \hat{\mathrm{k}}=1$]
$\Rightarrow \vec{a} \cdot \vec{b}=0~ for ~\vec{\mathrm{a}} \neq 0, \vec{\mathrm{b}} \neq 0$
Hence, the converse of given statement need not be true.
View full question & answer→Question 192 Marks
If $\vec { a } \cdot \vec { a } = 0$ and $\vec { a } \cdot \vec { b } = 0,$ then what can be concluded about the vector $\vec { b }$?
AnswerHere ,we are given that $\vec { a } \cdot \vec { a } = 0 \Rightarrow | \vec { a } | ^ { 2 } = 0$
$\Rightarrow \quad | \vec { a } | = 0$ ......(1)
and $\vec { a } \cdot \vec { b } = 0$ ........(2)
$\Rightarrow \quad | \vec { a } | | \vec { b } | \cos \theta = 0$
From Eqs. (1) and (2), it may be concluded that $\vec { b }$ is either zero or non-zero perpendicular vector.
View full question & answer→Question 202 Marks
Find the angle between two vectors $\vec a$ and $\vec b$ with magnitudes $\sqrt 3 $ and 2, respectively, having $\vec a\;.\vec b = \sqrt 6 $
Answer$ \left| {\overrightarrow a } \right| = \sqrt 3 ,\left| {\overrightarrow b } \right| = 2,\overrightarrow a .\overrightarrow b = \sqrt 6 ,$ $\Rightarrow \overrightarrow a .\overrightarrow b = \left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|\cos \theta \Rightarrow \sqrt 6$
$ = 2\sqrt 3 \cos \theta$
$\Rightarrow \cos \theta = \frac{1}{{\sqrt 2 }} \Rightarrow \theta = \frac{\pi }{4} $
View full question & answer→Question 212 Marks
For given vectors, $\vec a = 2\hat i - \hat j + 2\hat k$ and $\vec b = - \hat i + \hat j - \hat k$, find the unit vector in the direction of the vector $\vec a + \vec b$.
AnswerGiven: Vectors $\vec a = 2\hat i - \hat j + 2\hat k$ and $\vec b = - \hat i + \hat j - \hat k$
$\therefore \vec a + \vec b = 2\hat i - \hat j + 2\hat k - \hat i + \hat j - \hat k $ $= \hat i + 0\hat j + \hat k$
Therefore, unit vector in the direction of $ {(\vec a + \vec b)} = \frac{{\hat i + 0\hat j + \hat k}}{{\sqrt {{{\left( 1 \right)}^2} + {{\left( 0 \right)}^2} + {{\left( 1 \right)}^2}} }}$
$= \frac{{\hat i + 0\hat j + \hat k}}{{\sqrt {1 + 0 + 1} }} = \frac{{\hat i + \hat k}}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}\hat i + \frac{1}{{\sqrt 2 }}\hat k$
View full question & answer→Question 222 Marks
Find the unit vector in the direction of vector $\overrightarrow {PQ} $ , where P and Q are the points (1, 2, 3) and (4, 5, 6), respectively
Answer$\overrightarrow{PQ}=3\widehat{i}+3\widehat{j}+3\widehat{k}$, then $\hat{PQ}=\frac{\vec{PQ}}{|\vec {PQ}|}=\frac{3\hat i +3 \hat j +3 \hat k}{\sqrt{3^2+3^2+3^2}}$
$=\frac{3\hat i +3 \hat j +3 \hat k}{\sqrt{27}}=\frac{3(\hat i +\hat j +\hat j)}{3\sqrt3}$
$=\frac{1}{\sqrt3}\hat i+\frac{1}{\sqrt3}\hat j +\frac{1}{\sqrt 3}\hat k$
View full question & answer→Question 232 Marks
Find the unit vector in the direction of the vector $\vec a = \hat i + \hat j + 2\hat k$
AnswerWe know that a unit vector in the direction of the vector $\vec a = \hat i + \hat j + 2\hat k$is $\hat a = \frac{{\vec a}}{{\left| {\vec a} \right|}} = \frac{{\hat i + \hat j + 2\hat k}}{{\sqrt {{1^2} + {1^2} + {2^2}} }} = \frac{{\hat i + \hat j + 2\hat k}}{{\sqrt {1 + 1 + 4} }} = \frac{{\hat i + \hat j + 2\hat k}}{{\sqrt 6 }}$
$\Rightarrow \hat a = \frac{1}{{\sqrt 6 }}\hat i + \frac{1}{{\sqrt 6 }}\hat j + \frac{2}{{\sqrt 6 }}\hat k$
View full question & answer→Question 242 Marks
Find the sum of the vectors: $\vec a = \hat i - 2\hat j + \hat k,\vec b = - 2\hat i + 4\hat j + 5\hat k$ and $\vec c = \hat i - 6\hat j - 7\hat k$
AnswerGiven: $\vec a = \hat i - 2\hat j + \hat k,\vec b = - 2\hat i + 4\hat j + 5\hat k$ and $\vec c = \hat i - 6\hat j - 7\hat k$ Adding, $\vec a + \vec b + \vec c $ $= \hat i - 2\hat j + \hat k - 2\hat i + 4\hat j + 5\hat k + \hat i - 6\hat j - 7\hat k$
$ = 0\hat i - 4\hat j - \hat k = - 4\hat j - \hat k $
View full question & answer→Question 252 Marks
Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (– 5, 7).
AnswerThe scalar and vector components of the vector with initial point (2, 1) and terminal point (– 5, 7) is given by (-5 – 2) i.e. –7 and (7 – 1) i.e. 6. Therefore, the scalar components are – 7 and 6, and vector components are - $7\hat i$ and $6\hat j$
View full question & answer→Question 262 Marks
Find the values of x and y so that the vectors $2 \hat{i}+3 \hat{j}$ and $x \hat{i}+y \hat{j}$ are equal.
Answer$\Rightarrow 2\hat i + 3\hat j\; = \;x\hat i + y\hat j \Rightarrow x = 2,y = 3 \\\\\\ $
View full question & answer→Question 272 Marks
Write two different vectors having same direction.
Answer$\vec a = \hat i + 2\hat j + 3\hat k$ $\vec b = 2\hat i + 4\hat j + 6\hat k$
$\vec b = 2\left( {\hat i + 2\hat j + 3\hat k} \right)$
$\vec b = 2\vec a$
View full question & answer→Question 282 Marks
Write two different vectors having same magnitude.
Answer$\vec a = \hat i + 2\hat j + 3\hat k$ $\left| {\vec a} \right| = \sqrt {1 + 4 + 9} = \sqrt {14} $
$\vec b = 3\hat i + 2\hat j + 1\hat k$
$\left| {\vec b} \right| = \sqrt {9 + 4 + 1} = \sqrt {14} $
View full question & answer→Question 292 Marks
Find the position vector of the mid-point of the vector joining the points P (2, 3, 4) and Q(4, 1, -2).
AnswerGiven: Point P (2, 3, 4) and Q(4,1, - 2)
$\therefore$ Position vector of point P is $\vec a = 2\hat i + 3\hat j + 4\hat k$
And Position vector of point Q is $\vec b = 4\hat i + \hat j - 2\hat k$
And Position vector of mid-point R of PQ is $\frac{{\vec a + \vec b}}{2} = \frac{{2\hat i + 3\hat j + 4\hat k + 4\hat i + \hat j - 2\hat k}}{2}$
$=\frac{{6\hat i + 4\hat j + 2\hat k}}{2} = 3\hat i + 2\hat j + \hat k$
View full question & answer→Question 302 Marks
Find the position vector (externally) of a point R which divides the line joining two points P and Q whose position vectors are $\hat{i}+2 \hat{j}-\hat{k}$ and $-\hat{i}+\hat{j}+\hat{k}$ respectively, in the ratio 2 : 1.
AnswerGiven that $\vec{{P}}=\hat{1}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $\vec{\mathrm{Q}}=-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$
The $\vec{\mathrm{R}}$ does not lie on the segment PQ (external division). If $\mathrm{m}: n$ is the ratio in which $\vec{\mathrm{R}}$ divides PQ, then
$\vec{\mathrm{R}}=\frac{m \vec{\mathrm{Q}}-n \vec{\mathrm{P}}}{m-n}$
Given m : n = 2 : 1, m = 2 and n = 1
$\Rightarrow \vec{\mathrm{R}}=\frac{2(-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})-1(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}})}{2-1}$ = $\frac{-3 \hat{1}+0 \hat{\jmath}+3 \hat{k}}{1}=-3 \hat{1}+3 \hat{k}$
View full question & answer→Question 312 Marks
Find the position vector (internally) of a point R which divides the line joining two points P and Q whose position vectors are $\hat{i}+2 \hat{j}-\hat{k}$ and $-\hat{i}+\hat{j}+\hat{k}$ respectively, in the ratio 2 : 1.
AnswerGiven that $\vec{{P}}=\hat{1}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $\vec{\mathrm{Q}}=-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$
The $\vec{\mathrm{R}}$ lies on the segment PQ(internal division). If $\mathrm{m}: \mathrm{n}$ is the ratio in which $\vec{\mathrm{R}}$ divides $\mathrm{PQ}$ then
$\vec{\mathrm{R}}=\frac{\mathrm{m} \vec{\mathrm{Q}}+\mathrm{n} \vec{\mathrm{P}}}{\mathrm{m}+\mathrm{n}}$
Given m : n = 2 : 1, m = 2 and n = 1
$\Rightarrow \vec{\mathrm{R}}=\frac{2(-\hat{\imath}+\hat{\jmath}+\hat{\mathrm{k}})+1(\hat{\imath}+2 \hat{\jmath}-\hat{\mathrm{k}})}{2+1}$ = $\frac{-1 \hat{1}+4 \hat{\jmath}+\hat{k}}{3}=\frac{-1}{3} \hat{i}+\frac{4}{3} \hat{\jmath}+\frac{1}{3} \hat{k}$
View full question & answer→Question 322 Marks
Find the direction cosines of the vector $\hat i + 2\hat j + 3\hat k$
AnswerLet
$\overrightarrow a =\hat i + 2\hat j + 3\hat k $, Then ,
$\widehat a = \frac{{\overrightarrow a }}{{\left| {\overrightarrow a } \right|}} = \frac{{\widehat i + 2\widehat j + 3\widehat k}}{{\sqrt {{1^2} + {2^2} + {3^2}} }} = \frac{{\widehat i + 2\widehat j + 3\widehat k}}{{\sqrt {14} }}$
Therefore , the D.C.’s of vector a are :
$\frac{1}{{\sqrt {14} }},\;\frac{2}{{\sqrt {14} }},\;\frac{3}{{\sqrt {14} }}.$
View full question & answer→Question 332 Marks
Show that the vectors 2$\hat i$ - 3$\hat j$ + 4$\hat k$ and -4$\hat i$ + 6$\hat j$ - 8$\hat k$ are collinear.
AnswerLet $\vec a$ = 2$\hat i$ - 3$\hat j$ + 4$\hat k$ and $\vec b$ = -4$\hat i$ + 6$\hat j$ - 8$\hat k$.
Now, $\vec b$ = -4$\hat i$ + 6$\hat j$ - 8$\hat k$ = -2(2$\hat i$ - 3$\hat j$ + 4$\hat k$) = -2$\vec a$
Thus, $\vec b$ = -2$\vec a$
Hence, $\vec a$ and $\vec b$ are collinear.
View full question & answer→Question 342 Marks
Find a vector in the direction of vector $5\hat i - \hat j + 2\hat k$ which has a magnitude of 8 units.
AnswerLet
$\overrightarrow{a}=5\widehat{i}-\widehat{j}+2\widehat{k}$
then,
$8\hat a = 8\frac{\vec a}{|\vec a|}$
=$8.\frac{5 \hat i-\hat j +2 \hat k}{\sqrt{5^2+(-1)^2+2^2}}$
$=\frac{8(5 \hat i-\hat j +2 \hat k)}{\sqrt{30}}$
$=\frac{40}{\sqrt{30}}\hat i -\frac{8}{\sqrt{30}}\hat j +\frac{16}{\sqrt{30}}\hat k$
View full question & answer→Question 352 Marks
Write the direction ratio’s of the vector $\vec{a}=\hat{i}+\hat{j}-2 \hat{k}$ and hence calculate its direction cosines.
AnswerDirection Ratio's of $\vec r$ are 1, 1, -2
$\left| {\vec r} \right| = \sqrt {1 + 1 + 4} = \sqrt 6 $
Direction Ratio's of $\vec r$ are $\frac{1}{{\sqrt 6 }}.\frac{1}{{\sqrt 6 }},\frac{{ - 2}}{{\sqrt 6 }}$
View full question & answer→Question 362 Marks
Find the unit vector in the direction of the sum of the vectors $\vec a = 2\hat i + 2\hat j - 5\hat k$ and $\vec b = 2\hat i + \hat j + 3\hat k$.
AnswerLet $\vec c = \vec a + \vec b$
$ = \left( {2\hat i + 2\hat j - 5\hat k} \right) + \left( {2\hat i + \hat j + 3\hat k} \right)$
$ = 4\hat i + 3\hat j - 2\hat k$
$\left| {\vec c} \right| = \sqrt {16 + 9 + 4} $
$ = \sqrt {29} $
The required unit vector is
$\hat c = \frac{{\vec c}}{{\left| {\vec c} \right|}}$
$ = \frac{{4\vec i + 3\vec j - 2\vec k}}{{\sqrt {29} }}$
$ = \frac{{4\hat i + 3\hat j - 2\hat k}}{{\sqrt {29} }}$
$ = \frac{4}{{\sqrt {29} }}\hat i + \frac{3}{{\sqrt {29} }}\hat j - \frac{2}{{\sqrt {29} }}\hat k$
View full question & answer→Question 372 Marks
Find a vector in the direction of vector $\vec a = \hat i - 2\hat j$ that has magnitude 7 units.
Answer$\hat a = \frac{{\vec a}}{{\left| {\vec a} \right|}}$ $= \frac{{\left( {\hat i - 2\hat j} \right)}}{{\sqrt 5 }}$
Vector having magnitude equal to 7 and in the direction of $\vec a$is
$7\hat a = 7\left( {\frac{{\hat i - 2\hat j}}{{\sqrt 5 }}} \right)$
$= \frac{7}{{\sqrt 5 }}\left( {\hat i - 2\hat j} \right)$
View full question & answer→Question 382 Marks
Find unit vector in the direction of vector $\vec{a}=2 \hat{i}+3 \hat{j}+\hat{k}$
AnswerThe unit vector in the direction of a vector $\vec a$ is given by $\hat{a}=\frac{1}{|\vec{a}|} \vec{a}$
Now $|\vec{a}|=\sqrt{2^{2}+3^{2}+1^{2}}=\sqrt{14}$
Therefore $\hat{a}=\frac{1}{\sqrt{14}}(2 \hat{i}+3 \hat{j}+\hat{k})=\frac{2}{\sqrt{14}} \hat{i}+\frac{3}{\sqrt{14}} \hat{j}+\frac{1}{\sqrt{14}} \hat{k}$
View full question & answer→Question 392 Marks
Let $\vec{a}=\hat{i}+2 \hat{j}$ and $\vec{b}=2 \hat{i}+\hat{j}$. Is $|\vec{a}|=|\vec{b}|$? Are the vectors $\vec a $ and $\vec b$ equal?
AnswerWe have $|\vec{a}|=\sqrt{1^{2}+2^{2}}=\sqrt{5}$ and $|\overrightarrow{\mathrm{b}}|=\sqrt{2^{2}+1^{2}}=\sqrt{5}$
So, $|\vec{a}|=|\vec{b}|$. But, the two vectors are not equal since their corresponding components are distinct.
View full question & answer→Question 402 Marks
Find the values of x, y and z so that the vectors $\vec{a}=x \hat{i}+2 \hat{j}+z \hat{k}$ and $\vec{b}=2 \hat{i}+y \hat{j}+\hat{k}$ are equal.
AnswerNote that two vectors are equal if and only if their corresponding components are equal. Thus, the given vectors $\vec a$ and $\vec b$ will be equal if and only if
x = 2, y = 2, z = 1
View full question & answer→Question 412 Marks
Find $|\vec{a} \times \vec{b}|$ if $\vec{a}=2 \hat{i}+\hat{j}+3 \hat{k}$ and $\vec{b}=3 \hat{i}+5 \hat{j}-2 \hat{k}$
AnswerWe have,
$\vec{a} \times \vec{b}=\left|\begin{array}{ccc} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ {2} & {1} & {3} \\ {3} & {5} & {-2} \end{array}\right|$
= $\hat{i}(-2-15)-(-4-9) \hat{j}+(10-3) \hat{k}=-17 \hat{i}+13 \hat{j}+7 \hat{k}$
Hence, $|\vec{a} \times \vec{b}|=\sqrt{(-17)^{2}+(13)^{2}+(7)^{2}}=\sqrt{507}$
View full question & answer→Question 422 Marks
Prove that for any two vectors $\vec{a}$ and $\vec{b}$, we always have $|\vec{a} \cdot \vec{b}| \leq|\vec{a}||\vec{b}|$ (CauchySchwartz inequality).
AnswerThe inequality holds trivially when either $\vec{a}=\vec{0}$ or $\vec{b}=\vec{0}$. Actually, in such a situation we have $|\vec{a} \cdot \vec{b}|=0=|\vec{a}||\vec{b}|$. So, let us assume that $|\vec{a}| \neq 0 \neq|\vec{b}|$.
Then, we have $\frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}||\vec{b}|}=|\cos \theta| \leq 1$
Therefore $|\vec{a} \cdot \vec{b}| \leq|\vec{a}||\vec{b}|$
View full question & answer→Question 432 Marks
If ${\vec a}$ is a unit vector and $(\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})=8$, then find $\left| {\vec x} \right|$.
Answer$\left| {\vec a} \right| = 1$
$\left( {\vec x - \vec a} \right).\left( {\vec x + \vec a} \right) = 8$
${\left| {\vec x} \right|^2} - |\vec a|^2 = 8$
$|\vec x|^2-1=8$
${\left| {\vec x} \right|^2} = 9$
$\left| {\vec x} \right| = 3$
View full question & answer→Question 442 Marks
Find $|\vec{a}-\vec{b}|$, if two vectors $\vec a$ and $\vec b$ are such that $|\vec{a}|=2,|\vec{b}|=3$ and $\vec{a} \cdot \vec{b}=4$.
AnswerWe have
$|\vec{a}-\vec{b}|^{2}=(\vec{a}-\vec{b}) \cdot(\vec{a}-\vec{b})$
= $\vec{a} \cdot \vec{a}-\vec{a} \cdot \vec{b}-\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}$
= $|\vec{a}|^{2}-2(\vec{a} \cdot \vec{b})+|\vec{b}|^{2}$
= $(2)^{2}-2(4)+(3)^{2}$
Therefore, $|\vec{a}-\vec{b}|=\sqrt{5}$
View full question & answer→Question 452 Marks
Find the projection of the vector $\vec{a}=2 \hat{i}+3 \hat{j}+2 \hat{k}$ on the vector $\vec{b}=\hat{i}+2 \hat{j}+\hat{k}$.
AnswerThe projection of vector $\vec a$ on the vector $\vec b$ is given by
$\frac{1}{|\vec{b}|}(\vec{a} \cdot \vec{b})=\frac{(2 \times 1+3 \times 2+2 \times 1)}{\sqrt{(1)^{2}+(2)^{2}+(1)^{2}}}=\frac{10}{\sqrt{6}}=\frac{5}{3} \sqrt{6}$
View full question & answer→Question 462 Marks
If $\vec{a}=5 \hat{i}-\hat{j}-3 \hat{k}$ and $\vec{b}=\hat{i}+3 \hat{j}-5 \hat{k}$, then show that the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are perpendicular.
AnswerWe know that , two non-zero vectors are perpendicular if their scalar product is zero.
According to the Question , $\vec{a}+\vec{b}=(5 \hat{i}-\hat{j}-3 \hat{k})+(\hat{i}+3 \hat{j}-5 \hat{k})=6 \hat{i}+2 \hat{j}-8 \hat{k}$
and $\vec{a}-\vec{b}=(5 \hat{i}-\hat{j}-3 \hat{k})-(\hat{i}+3 \hat{j}-5 \hat{k})=4 \hat{i}-4 \hat{j}+2 \hat{k}$
So , Dot Product of two vectors :
$(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=(6 \hat{i}+2 \hat{j}-8 \hat{k}) \cdot(4 \hat{i}-4 \hat{j}+2 \hat{k})=24-8-16=0$
Hence, $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are perpendicular vectors.
View full question & answer→Question 472 Marks
Find angle $\theta$ between the vectors $\vec{a}=\hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$
AnswerThe angle $\theta$ between two vectors $\vec a$ and $\vec b$ is given by
$\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$
Now $\vec{a} \cdot \vec{b}=(\hat{i}+\hat{j}-\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})=1-1-1=-1$
Therefore, we have $\cos \theta=\frac{-1}{3}$
hence the required angle is $\theta=\cos ^{-1}\left(-\frac{1}{3}\right)$
View full question & answer→Question 482 Marks
Find the angle between two vectors $\vec a $ and $\vec b$ with magnitudes 1 and 2 respectively and when $\vec{a} \cdot \vec{b}=1$.
AnswerGiven $\vec{a} \cdot \vec{b}=1,|\vec{a}|=1$ and $|\vec{b}|=2$. We have
$\theta=\cos ^{-1}\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a} \| \vec{b}|}\right)=\cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}$
View full question & answer→Question 492 Marks
Find the vector joining the points P(2, 3, 0) and Q(-1, -2, -4) directed from P to Q.
AnswerSince the vector is to be directed from P to Q, clearly P is the initial point and Q is the terminal point. So, the required vector joining P and Q is the vector $\vec {PQ}$, given by
$\vec{\mathrm{PQ}}=(-1-2) \hat{i}+(-2-3) \hat{j}+(-4-0) \hat{k}$
i.e. $\overrightarrow{\mathrm{PQ}}=-3 \hat{i}-5 \hat{j}-4 \hat{k}$
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