Miscellaneous Exercise (Revised) - Chapter 10 - Vector Algebra - Ncert Solutions class 12 - Maths

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NCERT Class 12 Maths: Chapter 10 - Vector Algebra Solutions

Miscellaneous Exercise Question 1.

Write down a unit vector in XY-plane making an angle of $30^{\circ}$ in anti-clockwise direction with the positive direction of $x$ - axis.

Answer.

Let $\overrightarrow{\mathrm{OP}}$ be the unit vector in $\mathrm{XY}-$ plane such that $\angle \mathrm{XOP}=30^{\circ}$.

Therefore, $|\overrightarrow{\mathrm{OP}}|=1 \Rightarrow \mathrm{OP}=1$
By Triangle Law of Addition of vectors,
In $\Delta \mathrm{OMP}, \overrightarrow{\mathrm{OP}}=\overrightarrow{\mathrm{OM}}+\overrightarrow{\mathrm{MP}}=(\mathrm{OM}) \hat{i}+(\mathrm{MP}) \hat{j}$
[Unit vector along $\mathrm{OX}$ is $\hat{i}$ and that is along $\mathrm{OY}$ is $\hat{j}$ ]
$\Rightarrow \overrightarrow{\mathrm{OP}}=\mathrm{OP} \frac{\mathrm{OM}}{\mathrm{OP}} \hat{i}+\mathrm{OP} \frac{\mathrm{MP}}{\mathrm{OP}} \hat{j}$ [Dividing and multiplying by OP in R.H.S.]
$\Rightarrow \overrightarrow{\mathrm{OP}}=(1) \cos 30^{\circ} \hat{i}+(1) \sin 30^{\circ} \hat{j}$ [Using eq. (i)]

$\Rightarrow \overrightarrow{\mathrm{OP}}=\frac{\sqrt{3}}{2} \hat{i}+\frac{1}{2} \hat{j}$

Miscellaneous Exercise Question 2.

Find the scalar components and magnitude of the vector joining the points $P$ $\left(x_1, y_1, z_1\right)$ and $\mathbf{Q}\left(x_2, y_2, z_2\right)$.

Answer.

Given points are $\mathrm{P}\left(x_1, y_1, z_1\right)$ and $\mathrm{Q}\left(x_2, y_2, z_2\right)$
$\Rightarrow$ Position vector of point $\mathrm{P}=\left(x_1, y_1, z_1\right)=x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k}$
And Position vector of point $\mathrm{Q}=\left(x_2, y_2, z_2\right)=x_2 \hat{i}+y_2 \hat{j}+z_2 \hat{k}$
Now $\overrightarrow{\mathrm{PQ}}=$ Position vector of $\mathrm{Q}-$ Position vector of $\mathrm{P}$
$
\begin{aligned}
& =x_2 \hat{i}+y_2 \hat{j}+z_2 \hat{k}-\left(x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k}\right) \\
& =x_2 \hat{i}+y_2 \hat{j}+z_2 \hat{k}-x_1 \hat{i}-y_1 \hat{j}-z_1 \hat{k}=\left(x_2-x_1\right) \hat{i}+\left(y_2-y_1\right) \hat{j}+\left(z_2-z_1\right) \hat{k}
\end{aligned}
$
$\therefore$ Scalar components of the vector $\overrightarrow{\mathrm{PQ}}$ are the coefficients of $\hat{i}, \hat{j}, \hat{k}$ in $\overrightarrow{\mathrm{PQ}}$, i.e.,
$
\left(x_2-x_1\right),\left(y_2-y_1\right),\left(z_2-z_1\right)
$

And magnitude of vector $\overrightarrow{\mathrm{PQ}}=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}$

Miscellaneous Exercise Question 3.

A girl walks $4 \mathbf{~ k m}$ towards west, then she walks $3 \mathbf{~ k m}$ in a direction $30^{\circ}$ east of north and stops. Determine the girl's displacement from her initial point of departure.

Answer.

Let the initial point of departure is origin $(0,0)$ and the girl walks a distance $\mathrm{OA}=4 \mathrm{~km}$ towards west.

Through the point A, draw a line AQ parallel to a line OP, which is $30^{\circ}$ East of North, i.e., in East-North quadrant making an angle of $30^{\circ}$ with North.

Again, let the girl walks a distance $\mathrm{AB}=3 \mathrm{~km}$ along this direction $\overrightarrow{\mathrm{OQ}}$

$\therefore \overrightarrow{\mathrm{OA}}=4(-\hat{i})=-4 \hat{i}$ $\qquad$ .(i) $\left[\because\right.$ vector $\overrightarrow{\mathrm{OA}}$ is along $\left.\mathrm{XX}^{\prime}\right]$

Now, draw BM perpendicular to $x$-axis.
In $\triangle \mathrm{AMB}$, by Triangle Law of Addition of vectors,
$
\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{AM}}+\overrightarrow{\mathrm{MB}}=(\mathrm{AM}) \hat{i}+(\mathrm{MB}) \hat{i}
$

Dividing and multiplying by $\mathrm{AB}$ in R.H.S.,
$
\begin{aligned}
& \overrightarrow{\mathrm{AB}}=\mathrm{AB} \frac{\mathrm{AM}}{\mathrm{AB}} \hat{i}+\mathrm{AB} \frac{\mathrm{MB}}{\mathrm{AB}} \hat{j}=3 \cos 60^{\circ} \hat{i}+3 \sin 60^{\circ} \hat{j} \\
& \Rightarrow \overrightarrow{\mathrm{AB}}=3 \frac{1}{2} \hat{i}+3 \frac{\sqrt{3}}{2} \hat{i}=\frac{3}{2} \hat{i}+\frac{3 \sqrt{3}}{2} \hat{j} \quad \ldots \ldots \ldots . \text { (ii) }
\end{aligned}
$
$\therefore$ Girl's displacement from her initial point $\mathrm{O}$ of departure to final point $\mathrm{B}$,
$
\begin{aligned}
& \overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{AB}}=-4 \hat{i}+\left(\frac{3}{2} \hat{i}+\frac{3 \sqrt{2}}{2} \hat{j}\right)=\left(-4+\frac{3}{2}\right) \hat{i}+\frac{3 \sqrt{3}}{2} \hat{j} \\
& \Rightarrow \overrightarrow{\mathrm{OB}}=\frac{-5}{2} \hat{i}+\frac{3 \sqrt{3}}{2} \hat{j}
\end{aligned}
$

Miscellaneous Exercise Question 4.

If $\vec{a}=\vec{b}+\vec{c}$, then is it true that $|\vec{a}|=|\vec{b}|+|\vec{c}|$ ? Justify your answer.

Answer.

Given: $\vec{a}=\vec{b}+\vec{c}$
$\therefore$ 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 $\vec{a}=\overrightarrow{\mathrm{AC}}, \vec{b}=\overrightarrow{\mathrm{AB}}$ and $\vec{c}=\overrightarrow{\mathrm{BC}}$
Then $\vec{a}=\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=\vec{b}+\vec{c}$
Also, $|\vec{a}|=\mathrm{AC}=\mathrm{AB}+\mathrm{BC}=|\vec{b}|+|\vec{c}|$
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 $|\vec{a}|<|\vec{b}|+|\vec{c}|[\because$ Each side of a triangle is less than sum of the other two sides]
$\therefore|\vec{a}=\vec{b}+\vec{c}|=|\vec{b}|+|\vec{c}|$ is true only when vectors $\vec{b}$ and $\vec{c}$ are collinear vectors.

Miscellaneous Exercise Question 5.

Find the value of $x$ for which $x(\hat{i}+\hat{j}+\hat{k})$ is a unit vector.

Answer.

Since $x(\hat{i}+\hat{j}+\hat{k})=x \hat{i}+x \hat{j}+x \hat{k}$ is a unit vector,
Therefore, $|x \hat{i}+x \hat{j}+x \hat{k}|=1$
$\therefore \sqrt{x^2+x^2+x^2}=1 \Rightarrow \sqrt{3 x^2}=1$
Squaring both sides, $3 x^2=1$
$
\Rightarrow x^2=\frac{1}{3}
$

$\Rightarrow x= \pm \frac{1}{\sqrt{3}}$

Miscellaneous Exercise Question 6.

Find a vector of magnitude 5 units and parallel to the resultant of the vectors $\vec{a}=2 \hat{i}+3 \hat{j}-\hat{k}$ and $\vec{b}=\hat{i}-2 \hat{j}+\hat{k}$

Answer.

Given: Vectors $\vec{a}=2 \hat{i}+3 \hat{j}-\hat{k}$ and $\vec{b}=\hat{i}-2 \hat{j}+\hat{k}$
Let vector $\vec{c}$ be the resultant vector of $\vec{a}$ and $\vec{b}$.
$
\begin{aligned}
& \therefore \vec{c}=\vec{a}+\vec{b}=2 \hat{i}+3 \hat{j}-\hat{k}+\hat{i}-2 \hat{j}+\hat{k} \\
& =3 \hat{i}+\hat{j}+0 \hat{k}
\end{aligned}
$
$\therefore$ Required vector pf magnitude 5 units and parallel (or collinear) to resultant vector $\vec{c}=\vec{a}+\vec{b}$ is
$
\begin{aligned}
& 5 \hat{c}=5 \frac{\vec{c}}{|\vec{c}|}=5\left(\frac{3 \hat{i}+\hat{j}+0 \hat{k}}{\sqrt{9+1+0}}\right) \\
= & \frac{5}{\sqrt{10}}(3 \hat{i}+\hat{j}) \\
= & \frac{5}{\sqrt{10}} \times \frac{\sqrt{10}}{\sqrt{10}}(3 \hat{i}+\hat{j}) \\
= & \frac{5 \sqrt{10}}{10}(3 \hat{i}+\hat{j}) \\
\Rightarrow & \hat{c}=\frac{\sqrt{10}}{2}(3 \hat{i}+\hat{j}) \\
= & \frac{3 \sqrt{10}}{2} \hat{i}+\frac{\sqrt{10}}{2} \hat{j}
\end{aligned}
$

Miscellaneous Exercise Question 7.

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 $2 \vec{a}-\vec{b}+3 \vec{c}$.
Answer.

Given: 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 $\vec{d}=2 \vec{a}-\vec{b}+3 \vec{c}$
$
\begin{aligned}
& =2(\hat{i}+\hat{j}+\hat{k})-(2 \hat{i}-\hat{j}+3 \hat{k})+3(\hat{i}-2 \hat{j}+\hat{k}) \\
& =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}
\end{aligned}
$
$\therefore$ A unit vector parallel to the vector $\vec{d}$ is
$
\begin{aligned}
& \hat{d}=\frac{\vec{d}}{|\vec{d}|}=\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}
\end{aligned}
$

Miscellaneous Exercise Question 8.

Show that the points $\mathrm{A}(1,-2,-8), \mathbf{B}(5,0,-2)$ and $\mathrm{C}(11,3,7)$ are collinear and find the ratio in which $\mathrm{B}$ divides $\mathrm{AC}$.

Answer.

Given: Points A $(1,-2,-8), B(5,0,-2)$ and C $(11,3,7)$.
$\therefore$ Position vector of point $\mathrm{A}=(1,-2,-8)=\hat{i}-2 \hat{j}-8 \hat{k}$
Position vector of point $\mathrm{B}=(5,0,-2)=5 \hat{i}+0 \hat{j}-2 \hat{k}$
Position vector of point $\mathrm{C}=(11,3,7)=1 \hat{i}+3 \hat{j}+7 \hat{k}$

Now $\overrightarrow{A B}=$ Position vector of point $B-$ Position vector of point $A$
$
\begin{aligned}
& =5 \hat{i}-2 \hat{k}-(\hat{i}-2 \hat{j}-8 \hat{k})=5 \hat{i}-2 \hat{k}-\hat{i}+2 \hat{j}+8 \hat{k}=4 \hat{i}+2 \hat{j}+6 \hat{k} \\
& \therefore \mathrm{AB}=|\overrightarrow{\mathrm{AB}}|=\sqrt{16+4+36}=\sqrt{56}=2 \sqrt{14}
\end{aligned}
$

Again $\overrightarrow{\mathrm{BC}}=$ Position vector of point $\mathrm{C}-$ Position vector of point $\mathrm{B}$
$
\begin{aligned}
& =11 \hat{i}+3 \hat{j}+7 \hat{k}-(5 \hat{i}-2 \hat{k})=1 \hat{i}+3 \hat{i}+7 \hat{k}-5 \hat{i}+2 \hat{k}=6 \hat{i}+3 \hat{j}+9 \hat{k} \\
& \therefore \mathrm{BC}=|\overrightarrow{\mathrm{BC}}|=\sqrt{36+9+81}=\sqrt{126}=3 \sqrt{14}
\end{aligned}
$

Again $\overrightarrow{\mathrm{AC}}=$ Position vector of point $\mathrm{C}-$ Position vector of point $\mathrm{A}$
$
\begin{aligned}
& =1 \hat{i}+3 \hat{j}+7 \hat{k}-(\hat{i}-2 \hat{j}-8 \hat{k})=1 \hat{i}+3 \hat{j}+7 \hat{k}-\hat{i}+2 \hat{j}+8 \hat{k}=10 \hat{i}+5 \hat{j}+15 \hat{k} \\
& \therefore \mathrm{AC}=|\overrightarrow{\mathrm{AC}}|=\sqrt{100+25+225}=\sqrt{350}=5 \sqrt{14}
\end{aligned}
$

Now $\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AC}}=4 \hat{i}+2 \hat{j}+6 \hat{k}+6 \hat{i}+3 \hat{j}+9 \hat{k}=10 \hat{i}+5 \hat{j}+15 \hat{k}=\overrightarrow{\mathrm{AC}}$
Therefore, points A, B, C are either collinear or are the vertices of a triangle $\mathrm{ABC}$.
$
\text { Again } \mathrm{AB}+\mathrm{BC}=2 \sqrt{14}+3 \sqrt{14}=5 \sqrt{14}=\mathrm{AC}
$

Now to find ratio in which $B$ divides $A C$
Let the point $\mathrm{B}$ divides $\mathrm{AC}$ in the ratio $\lambda: 1$.
Therefore, using section formula, Position vector of point $\mathrm{B}(5,0,-2)$ is $\frac{\lambda \vec{c}+1 \vec{a}}{\lambda+1}$

$\begin{aligned}
& \Rightarrow(5,0,-2)=\frac{\lambda(11,3,7)+1(1,-2,-8)}{\lambda+1} \\
& \Rightarrow(5,0,-2)(\lambda+1)=\lambda(11,3,7)+(1,-2,-8)
\end{aligned}$

$
\begin{aligned}
& \Rightarrow(\lambda+1)(5 \hat{i}+0 \hat{j}-2 \hat{k})=\lambda(11 \hat{i}+3 \hat{j}+7 \hat{k})+(\hat{i}-2 \hat{j}-8 \hat{k}) \\
& \Rightarrow 5(\lambda+1) \hat{i}-2(\lambda+1) \hat{k}=11 \hat{i}+3 \lambda \hat{j}+7 \lambda \hat{k}+\hat{i}-2 \hat{j}-8 \hat{k}
\end{aligned}
$

Comparing coefficients of $\hat{i}, \hat{j}, \hat{k}$ both sides, we get
$
\begin{aligned}
& 5 \lambda+5=11 \lambda+1, \quad 0=3 \lambda-2, \quad-(2 \lambda+2)=7 \lambda-8 \\
& \Rightarrow-6 \lambda=-4, \quad 3 \lambda=2, \quad-2 \lambda-2=7 \lambda-8 \\
& \Rightarrow \lambda=\frac{4}{6}=\frac{2}{3}, \quad \lambda=\frac{2}{3}, \quad \lambda=\frac{6}{9}=\frac{2}{3} \\
& \text { Therefore, required ratio }=\lambda: 1=\frac{2}{3}: 1=2: 3
\end{aligned}
$

Miscellaneous Exercise Question 9.

Find the position vector of a point $R$ which divides the line joining the two points $P$ and $Q$ whose position vectors are $(2 \vec{a}+\vec{b})$ and $(\vec{a}-3 \vec{b})$ externally in the ratio $1: 2$. Also, show that $P$ is the middle point of line segment RQ.

Answer.

Since position vector of point $\mathrm{R}$ dividing the join of $\mathrm{P}$ and $\mathrm{Q}$ externally in the ratio $1: 2=$ $m: n$ is given by $\vec{c}=\frac{m \vec{b}-n \vec{a}}{m-n}$
$
\therefore \vec{c}=\frac{1(\vec{a}-3 \vec{b})-2(2 \vec{a}+\vec{b})}{1-2} \Rightarrow \vec{c}=\frac{\vec{a}-3 \vec{b}-4 \vec{a}-2 \vec{b}}{-1}=\frac{-3 \vec{a}-5 \vec{b}}{-1}=3 \vec{a}+5 \vec{b}
$

Again position vector of the middle point of the line segment $R Q$
$
\begin{aligned}
& =\frac{1}{2} \text { (Position vector of point } \mathrm{R}-\text { Position vector of point } \mathrm{Q} \text { ) } \\
& =\frac{1}{2}(3 \vec{a}+5 \vec{b}+\vec{a}-3 \vec{b})=\frac{1}{2}(4 \vec{a}+2 \vec{b})=2 \vec{a}+\vec{b}=\text { Position vector of point } \mathrm{P} \text { (given) }
\end{aligned}
$

Therefore, $\mathrm{P}$ is the middle point of the line segment $\mathrm{RQ}$.

Miscellaneous Exercise Question 10.

Two adjacent sides of a parallelogram are $2 \hat{i}-4 \hat{j}+5 \hat{k}$ and $\hat{i}-2 \hat{j}-3 \hat{k}$. Find the unit vector parallel to its diagonal. Also, find its area.

Answer.

Let ABCD is a parallelogram.

Given: The vectors representing two adjacent sides of this parallelogram say,
$
\vec{a}=2 \hat{i}-4 \hat{j}+5 \hat{k} \text { and } \vec{b}=\hat{i}-2 \hat{j}-3 \hat{k}
$

Now vectors along the diagonals $\overrightarrow{\mathrm{AC}}$ and $\overrightarrow{\mathrm{DB}}$ of the parallelogram are
$
\begin{aligned}
& \vec{a}+\vec{b} \text { and } \vec{a}-\vec{b} \\
& \therefore \vec{a}+\vec{b}=2 \hat{i}-4 \hat{j}+5 \hat{k}+\hat{i}-2 \hat{j}-3 \hat{k}=3 \hat{i}-6 \hat{j}+2 \hat{k}
\end{aligned}
$

And $\vec{a}-\vec{b}=2 \hat{i}-4 \hat{j}+5 \hat{k}-\hat{i}+2 \hat{j}+3 \hat{k}=\hat{i}-2 \hat{j}+8 \hat{k}$
Therefore, Unit vectors parallel to (or along) diagonals are
$
\begin{aligned}
& \frac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|} \text { and } \frac{\vec{a}-\vec{b}}{|\vec{a}-\vec{b}|} \\
& \Rightarrow \frac{3 \hat{i}-6 \hat{j}+2 \hat{k}}{\sqrt{9+36+4}} \text { and } \frac{\hat{i}-2 \hat{j}+8 \hat{k}}{\sqrt{1+4+64}} \Rightarrow \frac{3 \hat{i}-6 \hat{j}+2 \hat{k}}{\sqrt{49}} \text { and } \frac{\hat{i}-2 \hat{j}+8 \hat{k}}{\sqrt{69}} \\
& \Rightarrow \frac{3 \hat{i}-6 \hat{j}+2 \hat{k}}{7} \text { and } \frac{\hat{i}-2 \hat{j}+8 \hat{k}}{\sqrt{69}}
\end{aligned}
$

$\begin{aligned}
& \text { Now Area of parallelogram }=\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & -4 & 5 \\
1 & -2 & -3
\end{array}\right| \\
& =(12+10) \hat{i}-(-6-5) \hat{j}+(-4+4) \hat{k}=22 \hat{i}+11 \hat{j}+0 \hat{k} \\
& \Rightarrow|\vec{a} \times \vec{b}|=\sqrt{(22)^2+(11)^2+(0)^2} \\
& =\sqrt{484+121}=\sqrt{605}=11 \sqrt{5} \text { sq. units }
\end{aligned}$

Miscellaneous Exercise Question 11.

Show that the direction cosines of a vector equally inclined $\frac{1}{\sqrt{3}}$ to the axes ox, oy and $\mathrm{Oz}$ are $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$.

Answer.

Let $l, m, n$ be the direction cosines of a vector equally inclined to axes $\mathrm{OX}, \mathrm{OY}$ and $\mathrm{OZ}$ respectively.
$\therefore$ A unit vector along the given vector is
$
\begin{aligned}
& \hat{l}+m \hat{j}+n \hat{k} \text { and }|\hat{a}|=1 \\
& \Rightarrow \sqrt{l^2+m^2+n^2}=1 \Rightarrow l^2+m^2+n^2=1
\end{aligned}
$

Let the given vector (for which unit vector is $\hat{a}$ ) make equal angle (given) $\theta, \theta, \theta$ (say) with $\mathrm{OX}(\Rightarrow \hat{i}), \mathrm{OY}(\Rightarrow \hat{j})$ and $\mathrm{OZ}(\Rightarrow \hat{k})$.
$\therefore$ The given vector is in positive octant OXYZ and hence $\theta$ is acute. $\qquad$ (ii)

Now angle $\theta$ between $\hat{a}$ and $\hat{i} \cos \theta=\frac{\hat{a} \hat{i}}{|\hat{a}| \cdot \hat{i} \mid}$

$
\begin{aligned}
& \Rightarrow \cos \theta=\frac{(\hat{i}+m \hat{j}+n \hat{k}) \cdot(\hat{i}+0 \hat{j}+0 \hat{k})}{(1)(1)} \\
& \Rightarrow \cos \theta=l(1)+m(0)+n(0) \\
& \Rightarrow l=\cos \theta
\end{aligned}
$

Similarly, angle $\theta$ between $\hat{a}$ and $\hat{j}, m=\cos \theta$
And angle $\theta$ between $\hat{a}$ and $\hat{k}, n=\cos \theta$
Putting the values of $l, m, n$ in eq. (i), we get
$
\begin{aligned}
& \cos ^2 \theta+\cos ^2 \theta+\cos ^2 \theta=1 \\
& \Rightarrow 3 \cos ^2 \theta=1 \\
& \Rightarrow \cos ^2 \theta=\frac{1}{3} \\
& \Rightarrow \cos \theta= \pm \frac{1}{\sqrt{3}}
\end{aligned}
$

But $\cos \theta=\frac{1}{\sqrt{3}}[\because \theta$ is acute and hence $\cos \theta$ is positive $]$
Therefore, required vectors $l, m, n$ are $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$ and $\frac{1}{\sqrt{3}}$.

Miscellaneous Exercise Question 12.

Let $\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k}, \vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k}$ and $\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}$. Find a vector $\vec{b}$ which is perpendicular to both $\vec{a}$ and $\vec{b}$ and $\vec{c} \cdot \vec{d}=15$.

Answer.

Given: Vectors $\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k}$ and $\vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k}$
We know that the cross-product of two vectors, $\vec{a} \times \vec{b}$ is a vector perpendicular to both $\vec{a}$ and $\vec{b}$.

Hence, vector $\vec{a}$ which is also perpendicular to both $\vec{a}$ and $\vec{b}$ is $\vec{d}=\lambda(\vec{a} \times \vec{b})$ where $\lambda=1$ or some other scalar.

Therefore, $\vec{d}=\lambda\left|\begin{array}{rrr}i & \hat{j} & \hat{k} \\ 1 & 4 & 2 \\ 3 & -2 & 7\end{array}\right|$
$
\begin{aligned}
& =\lambda[\hat{i}(28+4)-\hat{j}(7-6)+\hat{k}(-2-12)] \\
& \Rightarrow \vec{d}=\lambda[32 \hat{i}-\hat{j}-14 \hat{k}] \\
& \Rightarrow \vec{d}=32 \lambda \hat{i}-\lambda \hat{j}-14 \lambda \hat{k} \\
&
\end{aligned}
$

Now given $\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}$ and $\vec{c} \cdot \vec{d}=15$
$
\vec{c} \cdot \vec{d}=15
$

$
\begin{aligned}
& \Rightarrow 2(32 \lambda)+(-1)(-\lambda)+4(-14 \lambda)=15 \\
& \Rightarrow 64 \lambda+\lambda-56 \lambda=15 \\
& \Rightarrow 9 \lambda=15 \\
& \Rightarrow \lambda=\frac{15}{9} \\
& \Rightarrow \lambda=\frac{5}{3}
\end{aligned}
$

Putting $\lambda=\frac{5}{3}$ in eq. (i), we get
$
\vec{d}=\frac{5}{3}[32 \hat{i}-\hat{j}-14 \hat{k}]
$

$\Rightarrow \vec{d}=\frac{1}{3}[160 \hat{i}-5 \hat{j}-70 \hat{k}]$

Miscellaneous Exercise Question 13.

The scalar product of the vector $\hat{i}+\hat{j}+\hat{k}$ with a unit vector along the sum of vectors $2 \hat{i}+4 \hat{j}-5 \hat{k}$ and $\lambda \hat{i}+2 \hat{j}+3 \hat{k}$ is equal to one. Find the value of $\lambda$.

Answer.

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}+4 \hat{j}-5 \hat{k}$ and $\vec{c}=\lambda \hat{i}+2 \hat{j}+3 \hat{k}$
Now $\vec{b}+\vec{c}=\vec{d}$ (say) $=(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}$
$\therefore \hat{d}$ a unit vector along $\vec{b}+\vec{c}=\vec{d}$ is
$
\begin{aligned}
& \hat{d}=\frac{\vec{d}}{|\vec{d}|}=\frac{(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{(2+\lambda)^2+(6)^2+(2)^2}} \\
& =\frac{(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{4+\lambda^2+4 \lambda+36+4}} \\
& =\frac{(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{4+\lambda^2+4 \lambda+40}} \\
& \Rightarrow \hat{d}=\frac{(2+\lambda) \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{\lambda^2+4 \lambda+44}} \\
& =\frac{(2+\lambda)}{\sqrt{\lambda^2+4 \lambda+44}} \hat{i}+\frac{6}{\sqrt{\lambda^2+4 \lambda+44}} \hat{j}-\frac{2}{\sqrt{\lambda^2+4 \lambda+44}} \hat{k} \\
&
\end{aligned}
$

Also given Dot product of $\vec{a}$ and $\vec{d}$ is 1 .
$
\begin{aligned}
& \Rightarrow \vec{a} \cdot \vec{d}=1 \\
& \Rightarrow \frac{1(2+\lambda)}{\sqrt{\lambda^2+4 \lambda+44}}+\frac{1(6)}{\sqrt{\lambda^2+4 \lambda+44}}+\frac{1(-2)}{\sqrt{\lambda^2+4 \lambda+44}}=1
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow 2+\lambda+6-2=\sqrt{\lambda^2+4 \lambda+44} \\
& \Rightarrow \lambda+6=\sqrt{\lambda^2+4 \lambda+44}
\end{aligned}
$

Squaring both sides,
$
\begin{aligned}
& (\lambda+6)^2=\lambda^2+4 \lambda+44 \\
& \Rightarrow(\lambda+6)^2=\lambda^2+4 \lambda+44 \\
& \Rightarrow \lambda^2+12 \lambda+36=\lambda^2+4 \lambda+44 \\
& \Rightarrow 8 \lambda=8 \Rightarrow \lambda=1
\end{aligned}
$

Miscellaneous Exercise Question 14.

If $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular vectors of equal magnitudes, show that the vector $\vec{a}+\vec{b}+\vec{c}$ is equally inclined to $\vec{a}, \vec{b}, \vec{c}$.

Answer.

Given: $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular vectors of equal magnitude.
$
\therefore \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}=0, \vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{b}=0, \vec{c} \cdot \vec{a}=\vec{a} \cdot \vec{c}=0
$

And $|\vec{a}|=|\vec{b}|=|\vec{c}|=\lambda$ (say)
Let vector $\vec{d}=\vec{a}+\vec{b}+\vec{c}$ make angles $\theta_1, \theta_2, \theta_3$ with vectors $\vec{a}, \vec{b}, \vec{c}$ respectively.
$
\begin{aligned}
& \therefore \cos \theta_1=\frac{\vec{d} \cdot \vec{a}}{|\vec{d}| \cdot|\vec{a}|}=\frac{(\vec{a}+\vec{b}+\vec{c}) \cdot \vec{a}}{|\vec{a}+\vec{b}+\vec{c}||\vec{a}|} \\
& =\frac{\vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c}}{|\vec{a}+\vec{b}+\vec{c}| \cdot|\vec{a}|} \\
& =\frac{\vec{a} \cdot \vec{a}+0+0}{|\vec{a}+\vec{b}+\vec{c}| \cdot|\vec{a}|} \text { [From eq. (i)] }
\end{aligned}
$

$
\Rightarrow \cos \theta_1=\frac{|\vec{a}|^2}{|\vec{a}+\vec{b}+\vec{c}| \cdot|\vec{a}|}=\frac{|\vec{a}|}{|\vec{a}+\vec{b}+\vec{c}|}
$

We know that $|\vec{a}+\vec{b}+\vec{c}|^2=(\vec{a}+\vec{b}+\vec{c})^2$
$
=(\vec{a})^2+(\vec{b})^2+(\vec{c})^2+2 \vec{b} \cdot \vec{c}+2 \vec{a} \cdot \vec{b}+2 \vec{a} \cdot \vec{c}
$

Putting the values from eq. (i) and (ii),
$
\begin{aligned}
& |\vec{a}+\vec{b}+\vec{c}|^2=(\lambda)^2+(\lambda)^2+(\lambda)^2+0+0+0=3 \lambda^2 \\
& \Rightarrow|\vec{a}+\vec{b}+\vec{c}|=\lambda \sqrt{3}
\end{aligned}
$

Now $\cos \theta_1=\frac{|\vec{a}|}{|\vec{a}+\vec{b}+\vec{c}|}$
$
\begin{aligned}
& =\frac{\lambda}{\lambda \sqrt{3}}=\frac{1}{\sqrt{3}} \\
& \Rightarrow \theta_1=\cos ^{-1} \frac{1}{\sqrt{3}}
\end{aligned}
$

Similarly, $\theta_2=\cos ^{-1} \frac{1}{\sqrt{3}}$ and $\theta_3=\cos ^{-1} \frac{1}{\sqrt{3}}$
$
\begin{aligned}
& \therefore \theta_1=\theta_2=\theta_3 \\
& =\cos ^{-1} \frac{1}{\sqrt{3}}
\end{aligned}
$

$\text { Therefore, } \vec{a}+\vec{b}+\vec{c} \text { is equally inclined to the vectors } \vec{a}, \vec{b} \text { and } \vec{c} \text {. }$

Miscellaneous Exercise Question 15.

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 $\vec{a} \neq \overrightarrow{0}, \vec{b} \neq \overrightarrow{0}$

Answer.

We know that $(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=\vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}=|\vec{a}|^2+\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{b}+|\vec{b}|^2$
$
\begin{aligned}
& \Rightarrow(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b}) \\
& =|\vec{a}|^2+|\vec{b}|^2+2|\vec{a}| \cdot|\vec{b}|
\end{aligned}
$

Now if $\vec{a}$ and $\vec{b}$ are perpendicular $\Rightarrow \vec{a} \cdot \vec{b}=0$
Putting $\vec{a} \cdot \vec{b}=0$ in $(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})$
$
\begin{aligned}
& =|\vec{a}|^2+|\vec{b}|^2+2|\vec{a}| \cdot|\vec{b}| \\
& \Rightarrow(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b}) \\
& =|\vec{a}|^2+|\vec{b}|^2 \quad \ldots \ldots \ldots . .(\mathrm{ii})
\end{aligned}
$
$
\Rightarrow|\vec{a}|^2+|\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2+2|\vec{a}| \cdot|\vec{b}|
$
$=|\vec{a}|^2+|\vec{b}|^2$ [Putting value of $(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})$ in eq. (i)]
$
\begin{aligned}
& \Rightarrow 0=2 \vec{a} \cdot \vec{b} \\
& \Rightarrow \vec{a} \cdot \vec{b}=0
\end{aligned}
$

But $\vec{a} \neq 0, \vec{b} \neq 0$ (given)

Therefore, vectors $\vec{a}$ and $\vec{b}$ are perpendicular to each other.
Miscellaneous Exercise Question 16.

Choose the correct answer:

If $\theta$ is the angle between two vectors $\vec{a}$ and $\vec{b}$, then $\vec{a} \cdot \vec{b} \geq 0$ only when:
(A) $0<\theta<\frac{\pi}{2}$
(B) $0 \leq \theta \leq \frac{\pi}{2}$
(C) $0<\theta<\pi$
(D) $0 \leq \theta \leq \pi$

Answer.

Given: $\vec{a} \cdot \vec{b} \geq 0$
$
\begin{aligned}
& \Rightarrow|\vec{a}| \cdot|\vec{b}| \cos \theta \geq 0 \\
& \Rightarrow \cos \theta \geq 0
\end{aligned}
$
$[\because|\vec{a}|$ and $|\vec{b}|$ being lengths of vectors are always $\geq 0]$
Therefore, option (B) is correct.

Miscellaneous Exercise Question 17.

Choose the correct answer:

Let $\vec{a}$ and $\vec{b}$ be two unit vectors and $\theta$ is the angle between them. Then $\vec{a}+\vec{b}$ is a unit vector if:
(A) $\theta=\frac{\pi}{4}$
(B) $\theta=\frac{\pi}{3}$
(C) $\theta=\frac{\pi}{2}$
(D) $\theta=\frac{2 \pi}{3}$

Answer.

Given: $\vec{a}, \vec{b}$ and $\vec{a}+\vec{b}$ are unit vectors.
$
\Rightarrow|\vec{a}|=1,|\vec{b}|=1 \text { and }|\vec{a}+\vec{b}|=1
$

Now squaring both sides of $|\vec{a}+\vec{b}|=1$, we have,
$
\begin{aligned}
& |\bar{a}+\bar{b}|^2=1 \\
& \Rightarrow(\bar{a}+\bar{b})^2=1 \\
& \Rightarrow(\vec{a})^2+(\vec{b})^2+2 \vec{a} \vec{b}=1 \\
& \Rightarrow|\vec{a}|^2+|\vec{b}|^2+2|\vec{a}| \cdot|\vec{b}| \cos \theta=1 \text {, where } \theta \text { is the given angle between vectors } \vec{a} \text { and } \vec{b}
\end{aligned}
$

Putting $|\vec{a}|=1,|\vec{b}|=1$, we have, $1+1+2 \cos \theta=1$
$
\begin{aligned}
& \Rightarrow 2 \cos \theta=-1 \\
& \Rightarrow \cos \theta=\frac{-1}{2}=\cos 120^{\circ} \\
& \Rightarrow \theta=120^{\circ} \\
& =120^{\circ} \times \frac{\pi}{180^{\circ}}=\frac{2 \pi}{3}
\end{aligned}
$

Therefore, option (D) is correct.

Miscellaneous Exercise Question 18.

Choose the correct answer:

The value of $\hat{i} \cdot(\hat{j} \times \hat{k})+\hat{j} \cdot(\hat{i} \times \hat{k})+\hat{k} \cdot(\hat{i} \times \hat{j})$ is:
(A) 0
(B) -1

(C) 1
(D) 3

Answer.

$\hat{i} \cdot(\hat{j} \times \hat{k})+\hat{j} \cdot(\hat{i} \times \hat{k})+\hat{k} \cdot(\hat{i} \times \hat{j})=\hat{i} \cdot \hat{i}+\hat{j} \cdot(-\hat{j})+\hat{k} \cdot \hat{k}$
Also $\hat{i} \times \hat{k}=-\hat{k} \times \hat{i}=-\hat{j}=1-1+1=1$
Therefore, option (C) is correct.

Miscellaneous Exercise Question 19.

If $\theta$ be the angle between any two vectors $\vec{a}$ and $\vec{b}$, then $|\vec{a} \cdot \vec{b}|=|\vec{a} \times \vec{b}|$, when $\theta$ is equal to:
(A) 0
(B) $\frac{\pi}{4}$
(C) $\frac{\pi}{2}$
(D) $\pi$

Answer.

Given: $|\vec{a} \cdot \vec{b}|=|\vec{a} \times \vec{b}|$
$
\Rightarrow|\vec{a}| \cdot|\vec{b}| \cos \theta=|\vec{a}| \cdot|\vec{b}| \sin \theta
$

And this equation is true only for option (B) namely $\theta=\frac{\pi}{4}$, since $\cos \frac{\pi}{4}=\sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}$
Therefore, option (B) is correct.