Exercise 10.3 (Revised) - Chapter 10 - Vector Algebra - Ncert Solutions class 12 - Maths

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NCERT Class 12 Maths: Chapter 10 - Vector Algebra Solutions

Ex 10.3 Question 1.

Find the angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitude $\sqrt{3}$ and 2 respectively having $\vec{a} \cdot \vec{b}=\sqrt{6}$.

Answer.

Given: $|\vec{a}|=\sqrt{3},|\vec{b}|=2$ and $\vec{a} \cdot \vec{b}=\sqrt{6}$
Let $\theta$ be the angle between the vector $\vec{a}$ and $\vec{b}$.
We know that $\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot|\vec{b}|}$
$
\begin{aligned}
& \Rightarrow \cos \theta=\frac{\sqrt{6}}{\sqrt{3} \cdot 2} \\
& =\frac{\sqrt{3} \cdot \sqrt{2}}{\sqrt{3} \cdot \sqrt{2} \cdot \sqrt{2}}=\frac{1}{\sqrt{2}} \\
& \Rightarrow \cos \theta=\cos \frac{\pi}{4} \\
& \Rightarrow \theta=\frac{\pi}{4}
\end{aligned}
$

Ex 10.3 Question 2.

Find the angle between the vectors $\hat{i}-2 \hat{j}+3 \hat{k}$ and $3 \hat{i}-2 \hat{j}+\hat{k}$.

Answer.

Given: Let $\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}$ and $\vec{b}=3 \hat{i}-2 \hat{j}+\hat{k}$
$
\Rightarrow|\vec{a}|=\sqrt{1+4+9}=\sqrt{14} \text { and }|\vec{b}|=\sqrt{9+4+1}=\sqrt{14}
$

$
\left[\because x \hat{i}+y \hat{j}+z \hat{k}=\sqrt{x^2+y^2+z^2}\right]
$

Also $\vec{a} \cdot \vec{b}$
$
\begin{aligned}
& =\text { Product of coefficients of } \hat{i}+\text { Product of coefficients of } \hat{j}+\text { Product of coefficients } \hat{k} \\
& =1(3)+(-2)(-2)+3(1)=3+4+3=10
\end{aligned}
$

Let $\theta$ be the angle between the vector $\vec{a}$ and $\vec{b}$.
We know that $\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot|\vec{b}|}$
$
\begin{aligned}
& \Rightarrow \cos \theta=\frac{10}{\sqrt{14} \cdot \sqrt{14}} \\
& =\frac{10}{14}=\frac{5}{7} \\
& \Rightarrow \cos \theta=\frac{5}{7} \\
& \Rightarrow \theta=\cos ^{-1} \frac{5}{7}
\end{aligned}
$

Ex 10.3 Question 3.

Find the projection of the vector $\hat{i}-\hat{j}$ on the vector $\hat{i}+\hat{j}$.

Answer.

Let $\vec{a}=\hat{i}-\hat{j}=\hat{i}-\hat{j}+0 \hat{k}$ and $\vec{b}=\hat{i}+\hat{j}=\hat{i}+\hat{j}+0 \hat{k}$
Projection of vector $\vec{a}$ and $\vec{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$
$
=\frac{(1)(1)+(-1)(1)+0(0)}{\sqrt{(1)^2+(1)^2+(0)^2}}
$

$
=\frac{1-1+0}{\sqrt{2}}=\frac{0}{\sqrt{2}}=0
$

If projection of vector $\vec{a}$ and $\vec{b}$ is zero, then vector $\vec{a}$ is perpendicular to $\vec{b}$.

Ex 10.3 Question 4.

Find the projection of the vector $\hat{i}+3 \hat{j}+7 \hat{k}$ on the vector $7 \hat{i}-\hat{j}+8 \hat{k}$.

Answer.

Let $\vec{a}=\hat{i}+3 \hat{j}+7 \hat{k}$ and $\vec{b}=7 \hat{i}-\hat{j}+8 \hat{k}$
Projection of vector $\vec{a}$ and $\vec{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$
$
\begin{aligned}
& =\frac{(1)(7)+(3)(-1)+7(8)}{\sqrt{(7)^2+(-1)^2+(8)^2}} \\
& =\frac{7-3+56}{\sqrt{49+1+64}}=\frac{60}{\sqrt{114}}
\end{aligned}
$

Ex 10.3 Question 5.

Show that each of the given three vectors is a unit vector:
$
\frac{1}{7}(2 \hat{i}+3 \hat{j}+6 \hat{k}), \frac{1}{7}(3 \hat{i}-6 \hat{j}+2 \hat{k})=\frac{1}{7}(6 \hat{i}+2 \hat{j}-3 \hat{k})
$

Also show that they are mutually perpendicular to each other.
Answer.

Let $\vec{a}=\frac{1}{7}(2 \hat{i}+3 \hat{j}+6 \hat{k})=\frac{2}{7} \hat{i}+\frac{3}{7} \hat{j}+\frac{6}{7} \hat{k}$
$
\begin{aligned}
& \vec{b}=\frac{1}{7}(3 \hat{i}-6 \hat{j}+2 \hat{k})=\frac{3}{7} \hat{i}-\frac{6}{7} \hat{j}+\frac{2}{7} \hat{k} \\
& \vec{c}=\frac{1}{7}(6 \hat{i}+2 \hat{j}-3 \hat{k})=\frac{6}{7} \hat{i}+\frac{2}{7} \hat{j}-\frac{3}{7} \hat{k}
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow|\vec{a}|=\sqrt{\left(\frac{2}{7}\right)^2+\left(\frac{3}{7}\right)^2+\left(\frac{6}{7}\right)^2}=\sqrt{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}}=\sqrt{\frac{49}{49}}=\sqrt{1}=1 \\
& |\vec{b}|=\sqrt{\left(\frac{3}{7}\right)^2+\left(\frac{-6}{7}\right)^2+\left(\frac{2}{7}\right)^2}=\sqrt{\frac{9}{49}+\frac{36}{49}+\frac{4}{49}}=\sqrt{\frac{49}{49}}=\sqrt{1}=1 \\
& |\vec{c}|=\sqrt{\left(\frac{6}{7}\right)^2+\left(\frac{2}{7}\right)^2+\left(\frac{-3}{7}\right)^2}=\sqrt{\frac{36}{49}+\frac{4}{49}+\frac{9}{49}}=\sqrt{\frac{49}{49}}=\sqrt{1}=1
\end{aligned}
$
$\therefore$ Each of the three given vectors $\vec{a}, \vec{b}, \vec{c}$ is a unit vector.
From eq. (i) and (ii),
$
\begin{aligned}
& \vec{a} \cdot \vec{b}=\left(\frac{2}{7}\right) \cdot\left(\frac{3}{7}\right)+\left(\frac{3}{7}\right) \cdot\left(\frac{-6}{7}\right) \cdot\left(\frac{6}{7}\right) \cdot\left(\frac{2}{7}\right)\left[\because \vec{a} \cdot \vec{b}=a_1 b_1+a_2 b_2+a_3 b_3\right] \\
& =\frac{6}{49}-\frac{18}{49}+\frac{12}{49}=\frac{6-18+12}{49}=\frac{0}{49}=0
\end{aligned}
$
$\therefore \vec{a}$ and $\vec{b}$ are perpendicular to each other.
From eq. (ii) and eq. (iii),
$
\begin{aligned}
& \vec{b} \cdot \vec{c}=\left(\frac{3}{7}\right) \cdot\left(\frac{6}{7}\right)+\left(\frac{-6}{7}\right) \cdot\left(\frac{2}{7}\right) \cdot\left(\frac{2}{7}\right) \cdot\left(\frac{-3}{7}\right)\left[\because \vec{a} \cdot \vec{b}=a_1 b_1+a_2 b_2+a_3 b_3\right] \\
& =\frac{18}{49}-\frac{12}{49}-\frac{6}{49}=\frac{18-12-6}{49}=\frac{0}{49}=0
\end{aligned}
$
$\therefore \vec{a}$ and $\vec{b}$ are perpendicular to each other.

From eq. (i) and (iii),
$
\vec{a} \cdot \vec{c}=\left(\frac{2}{7}\right) \cdot\left(\frac{6}{7}\right)+\left(\frac{3}{7}\right) \cdot\left(\frac{2}{7}\right) \cdot\left(\frac{6}{7}\right) \cdot\left(\frac{-3}{7}\right)\left[\because \vec{a} \cdot \vec{b}=a_1 b_1+a_2 b_2+a_3 b_3\right]
$

$
=\frac{12}{49}+\frac{6}{49}-\frac{18}{49}=\frac{12+6-18}{49}=\frac{0}{49}=0
$
$\therefore \vec{a}$ and $\vec{b}$ are perpendicular to each other.
Hence, $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular vectors.

Ex 10.3 Question 6.

Find $|\vec{a}|$ and $|\vec{b}|$, if $(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=8$ and $|\vec{a}|=8|\vec{b}|$.

Answer.

Given: $(\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=8$ and $|\vec{a}|=8|\vec{b}| \ldots \ldots \ldots$. (i)
$
\begin{aligned}
& \Rightarrow \vec{a} \cdot \vec{a}-\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}-\vec{b} \cdot \vec{b}=8 \\
& \Rightarrow|\vec{a}|^2-\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{b}-|\vec{b}|^2=8 \\
& \Rightarrow|\vec{a}|^2-|\vec{b}|^2=8
\end{aligned}
$

Putting $|\vec{a}|=8|\vec{b}|$ in eq. (ii),
$
\begin{aligned}
& 64|\stackrel{\rightharpoonup}{b}|^2-|\stackrel{\rightharpoonup}{b}|^2=8 \\
& \Rightarrow(64-1)|\vec{b}|^2=8 \\
& \Rightarrow 63|\vec{b}|^2=8 \\
& \Rightarrow|\vec{b}|^2=\frac{8}{63} \\
& \Rightarrow|\vec{b}|=\sqrt{\frac{8}{63}}=\frac{2 \sqrt{2}}{3 \sqrt{7}}
\end{aligned}
$

Putting $|\vec{b}|=\frac{2 \sqrt{2}}{3 \sqrt{7}}$ in eq (i),

$|\vec{a}|=8\left(\frac{2 \sqrt{2}}{3 \sqrt{7}}\right)=\frac{16}{3} \sqrt{\frac{2}{7}}$

Ex 10.3 Question 7.

Evaluate the product $(3 \bar{a}-5 \bar{b}) \cdot(2 \bar{a}+7 \bar{b})$.

Answer.

Given: $(3 \vec{a}-5 \vec{b}) \cdot(2 \vec{a}+7 \vec{b})=(3 \vec{a}) \cdot(2 \vec{a})+(3 \vec{a}) \cdot(7 \vec{b})-(5 \vec{b}) \cdot(2 \vec{a})-(5 \vec{b}) \cdot(7 \vec{b})$
$
\begin{aligned}
& =6 \vec{a} \cdot \vec{a}+21 \vec{a} \cdot \vec{b}-10 \vec{b} \cdot \vec{a}-35 \vec{b} \cdot \vec{b} \\
& =6|\vec{a}|^2+21 \vec{a} \cdot \vec{b}-10 \vec{a} \cdot \vec{b}-35|\vec{b}|^2 \\
& =6|\vec{a}|^2+11 \vec{a} \cdot \vec{b}-35|\vec{b}|^2
\end{aligned}
$

Ex 10.3 Question 8.

Find the magnitude of two vectors $\vec{a}$ and $\vec{b}$, having the same magnitude such that the angle between them is $60^{\circ}$ and their scalar product is $\frac{1}{2}$.

Answer.

Given: $|\vec{a}|=|\vec{b}|$, angle $\theta$ (say) between $\vec{a}$ and $\vec{b}$ is $60^{\circ}$ and their scalar (i.e., dot)
$
\begin{aligned}
& \text { product }=\frac{1}{2} \\
& \Rightarrow \vec{a} \cdot \vec{b}=\frac{1}{2} \Rightarrow|\vec{a}| \cdot|\vec{b}| \cos \theta=\frac{1}{2}
\end{aligned}
$

Putting $|\vec{a}|=|\vec{b}|$ and $\theta=60^{\circ}$, we have $|\vec{a}| .|\vec{a}| \cos 60^{\circ}=\frac{1}{2}$
$
\begin{aligned}
& \Rightarrow|\vec{a}|^2 \cdot\left(\frac{1}{2}\right)=\frac{1}{2} \\
& \Rightarrow|\vec{a}|^2=1
\end{aligned}
$

$\begin{aligned}
& \Rightarrow|\vec{a}|=1 \\
& \therefore|\vec{b}|=|\vec{a}|=1 \\
& \therefore|\vec{a}|=1 \text { and }|\vec{b}|=1
\end{aligned}$

Ex 10.3 Question 9.

Find $|\vec{x}|$, if for a unit vector $\vec{a},(\bar{x}-\bar{a}) \cdot(\bar{x}+\bar{a})=12$

Answer.

Given: $\vec{a}$ is a unit vector $\Rightarrow|\vec{a}|=1$
$
\begin{aligned}
& (\vec{x}-\vec{a})(\vec{x}+\vec{a})=12 \\
& \Rightarrow \vec{x} \cdot \vec{x}+\vec{x} \cdot \vec{a}-\vec{a} \cdot \vec{x}-\vec{a} \cdot \vec{a}=12 \\
& \Rightarrow|\vec{x}|^2+\vec{x} \cdot \vec{a}-\vec{a} \cdot \vec{x}-|\vec{a}|^2=12 \\
& \Rightarrow|\vec{x}|^2+\vec{a} \cdot \vec{x}-\vec{a} \cdot \vec{x}-|\vec{a}|^2=12 \\
& \Rightarrow|\vec{x}|^2-|\vec{a}|^2=12
\end{aligned}
$

Putting $|\vec{a}|=1$ from eq. (i), $|\vec{x}|^2-1=12$
$
\Rightarrow|\vec{x}|^2=13 \Rightarrow|\vec{x}|=\sqrt{13}
$

Ex 10.3 Question 10.

If $\vec{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{c}=3 \hat{i}+\hat{j}$ are such that $\vec{a}+\lambda \vec{b}$ is perpendicular to $\vec{c}$, then find the value of $\lambda$.

Answer.

Given: $\vec{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{c}=3 \hat{i}+\hat{j}$
Now $\vec{a}+\lambda \vec{b}=2 \hat{i}+2 \hat{j}+3 \hat{k}+\lambda(-\hat{i}+2 \hat{j}+\hat{k})=$

$
\begin{aligned}
& =2 \hat{i}+2 \hat{j}+3 \hat{k}-\hat{\lambda} \hat{i}+2 \lambda \hat{j}+\lambda \hat{k} \\
& \Rightarrow \vec{a}+\lambda \vec{b}=(2-\lambda) \hat{i}+(2+2 \lambda) \hat{j}+(3+\lambda) \hat{k} \\
& \text { Again, } \vec{c}=3 \hat{i}+\hat{j}=3 \hat{i}+\hat{j}+0 \hat{k}
\end{aligned}
$

Since, $(\vec{a}+\lambda \vec{b})$ is perpendicular to $\vec{c}$, therefore, $(\vec{a}+\lambda \vec{b}) \cdot \vec{c}=0$
$
\begin{aligned}
& \Rightarrow(2-\lambda) 3+(2+2 \lambda) 1+(3+\lambda) 0=0 \\
& \Rightarrow 6-3 \lambda+2+2 \lambda=0 \Rightarrow-\lambda+8=0 \\
& \Rightarrow-\lambda=-8 \Rightarrow \lambda=8
\end{aligned}
$

Ex 10.3 Question 11.

Show that $|\bar{a}| \bar{b}+|\vec{b}| \bar{a}$ is perpendicular to $|\bar{a}| \vec{b}-|\vec{b}| \bar{a}$, for any two non-zero vectors $\vec{a}$ and $\vec{b}$.

Answer.

Let $\vec{c}=|\vec{a}| \vec{b}+|\vec{b}| \vec{a}=l \vec{b}+m \vec{a}$, where $l=|\vec{a}|$ and $m=|\vec{b}|$
Let $\vec{d}=|\vec{a}| \vec{b}-|\vec{b}| \vec{a}=l \vec{b}-m \vec{a}$
Now $|\vec{c}| \cdot|\vec{d}|=(l \vec{b}+m \vec{a})(l \vec{b}-m \vec{a})$
$
\begin{aligned}
& =l^2 \vec{b} \cdot \vec{b}-l m \vec{b} \cdot \vec{a}+\operatorname{lm} \vec{a} \cdot \vec{b}-m^2 \vec{a} \cdot \vec{a} \\
& =l^2|\vec{b}|^2-l m \vec{a} \cdot \vec{b}+\operatorname{lm} \vec{a} \cdot \vec{b}-m^2|\vec{a}|^2 \\
& =l^2|\vec{b}|^2-m^2|\vec{a}|^2
\end{aligned}
$

Putting, $l=|\vec{a}|$ and $m=|\vec{b}|$
$
=|\vec{a}|^2|\vec{b}|^2-|\vec{b}|^2|\vec{a}|^2=0
$

$
\Rightarrow|\vec{c}| \cdot|\vec{d}|=0
$

Therefore, vectors $\vec{c}$ and $\vec{d}$ are perpendicular ot each other.

Ex 10.3 Question 12.

If $\vec{a} \cdot \vec{a}=0$ and $\vec{a} \cdot \vec{b}=0$, then what can be concluded about the vector $\vec{b}$ ?

Answer.

Given: $\vec{a} \cdot \vec{a}=0 \Rightarrow|\vec{a}|^2=0$
$
\begin{aligned}
& \Rightarrow|\vec{a}|=0 \\
& \text { Again } \vec{a} \cdot \vec{b}=0 \Rightarrow|\vec{a}| \cdot|\vec{b}| \cos \theta=0 \\
& \Rightarrow 0 \cdot|\vec{b}| \cos \theta=0[\because|\vec{a}|=0] \\
& \Rightarrow 0=0 \text { for all (any vector } \vec{b}_{.} \text {) }
\end{aligned}
$

Therefore, $\vec{b}$ can be any vector.

Ex 10.3 Question 13.

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$, find the value of $\vec{a} \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$

Answer.

Since, $\vec{a}, \vec{b}, \vec{c}$ are unit vectors.
Therefore, $|\vec{a}|=1,|\vec{b}|=1$ and $|\vec{c}|=1$
Also given $\vec{a}+\vec{b}+\vec{c}=0$
$
\begin{aligned}
& \Rightarrow(\vec{a}+\vec{b}+\vec{c})^2=0 \\
& \Rightarrow(\vec{a})^2+(\vec{b})^2+(\vec{c})^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0 \\
& {\left[\because(a+b+c)^2=a^2+b^2+c^2+2 a b+2 b c+2 c a\right]}
\end{aligned}
$

$
\Rightarrow|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0
$

Putting the values from eq. (i), we get
$
\begin{aligned}
& \Rightarrow 1+1+1+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0 \\
& \Rightarrow 2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=-3 \\
& \Rightarrow(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=\frac{-3}{2}
\end{aligned}
$

Ex 10.3 Question Ex 10.3 Question I

f either vector $\vec{a}=\overrightarrow{0}$ or $\vec{b}=\overrightarrow{0}$, then $\vec{a} \cdot \vec{b}=0$. But the converse need not be true. Justify your answer with an example.

Answer.

Case I: Vector $\vec{a}=\overrightarrow{0}$. Therefore by definition of zero vector, $|\vec{a}|=0$
$
\begin{aligned}
& \therefore \vec{a} \cdot \vec{b}=|\vec{a}| \cdot|\vec{b}| \cos \theta \\
& =0 \cdot|\vec{b}| \cos \theta \text { [From eq. (i)] } \\
& \Rightarrow \vec{a} \cdot \vec{b}=0
\end{aligned}
$

Case II: Vector $\vec{b}=\overrightarrow{0}$. Therefore by definition of zero vector, $|\vec{b}|=0$ $\qquad$
$
\begin{aligned}
& \therefore \vec{a} \cdot \vec{b}=|\vec{a}| \cdot|\vec{b}| \cos \theta \\
& =\vec{a} \cdot 0 \cdot \cos \theta \quad \text { [From eq. (ii)] } \\
& \Rightarrow \vec{a} \cdot \vec{b}=0
\end{aligned}
$

But the converse is not true.
Justification: Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}$

Therefore, $|\vec{a}|=\sqrt{(1)^2+(1)^2+(1)^2}=\sqrt{3} \neq 0$
Therefore, $\vec{a} \neq \overrightarrow{0}$
Again let $\vec{b}=\hat{i}+\hat{j}-2 \hat{k}$
$
\therefore|\vec{b}|=\sqrt{(1)^2+(1)^2+(-2)^2}=\sqrt{6} \neq 0
$

Therefore, $\vec{b} \neq \overrightarrow{0}$
But $\vec{a} \cdot \vec{b}=1(1)+1(1)+1(-2)=1+1+-2=0$
Hence, here $\vec{a} \cdot \vec{b}=0$, but $\vec{a} \neq \overrightarrow{0}$ and $\vec{b} \neq \overrightarrow{0}$.

Ex 10.3 Question 15.

If the vertices $\mathbf{A}, \mathbf{B}, \mathbf{C}$ of a triangle $\mathbf{A B C}$ are $(1,2,3),(-1,0,0)$ and $(\mathbf{0}, \mathbf{1}, \mathbf{2})$ respectively, then find $\angle \mathrm{ABC}$.

Answer.

Vertices A, B, C of a triangle are A $(1,2,3), B(-1,0,0)$ and C $(0,1,2)$ respectively.
$\therefore$ Position vector of point $\mathrm{A}=\overrightarrow{\mathrm{OA}}=(1,2,3)=\hat{i}+2 \hat{j}+3 \hat{k}$
Position vector of point $\mathrm{B}=\overrightarrow{\mathrm{OB}}=(-1,0,0)=-\hat{i}+0 \hat{j}+0 \hat{k}$
Position vector of point $\mathrm{C}=\overrightarrow{\mathrm{OC}}=(0,1,2)=0 \hat{i}+1 \hat{j}+2 \hat{k}$
Now $\overrightarrow{\mathrm{BA}}=$ Position vector of point $\mathrm{A}-$ Position vector of point $\mathrm{B}$
$
\begin{aligned}
& =\hat{i}+2 \hat{j}+3 \hat{k}-(-\hat{i}+0 \hat{j}+0 \hat{k}) \\
& =\hat{i}+2 \hat{j}+3 \hat{k}+\hat{i}-0 \hat{j}-0 \hat{k}=2 \hat{i}+2 \hat{j}+3 \hat{k}
\end{aligned}
$

And $\overrightarrow{\mathrm{BC}}=$ Position vector of point $\mathrm{C}-$ Position vector of point $\mathrm{B}$
$
=0 \hat{i}+\hat{j}+2 \hat{k}-(-\hat{i}+0 \hat{j}+0 \hat{k})
$

$
=0 \hat{i}+\hat{j}+2 \hat{k}+\hat{i}-0 \hat{j}-0 \hat{k}=\hat{i}+\hat{j}+2 \hat{k}
$

Let $\theta$ be the angle between the vectors $\overrightarrow{\mathrm{BA}}$ and $\overrightarrow{\mathrm{BC}}$.
$
\begin{aligned}
& \therefore \cos \theta=\frac{\overrightarrow{\mathrm{BA}} \cdot \overrightarrow{\mathrm{BC}}}{|\overrightarrow{\mathrm{BA}}| \cdot|\overrightarrow{\mathrm{BC}}|} \\
& =\frac{2(1)+2(1)+3(2)}{\sqrt{4+4+9} \sqrt{1+1+4}} \text { [Using eq. (i) and (ii)] } \\
& \Rightarrow \cos \theta=\frac{10}{\sqrt{17} \sqrt{6}}=\frac{10}{\sqrt{102}} \\
& \Rightarrow \theta=\cos ^{-1} \frac{10}{\sqrt{102}}
\end{aligned}
$

Ex 10.3 Question 16.

Show that the points $A(1,2,7), B(2,6,3)$ and $C(3,10,-1)$ are collinear.

Answer.

Vertices A, B, C of a triangle are A $(1,2,7), B(2,6,3)$ and C $(3,10,-1)$ respectively.
$\therefore$ Position vector of point $\mathrm{A}=\overrightarrow{\mathrm{OA}}=(1,2,7)=\hat{i}+2 \hat{j}+7 \hat{k}$
Position vector of point $\mathrm{B}=\overrightarrow{\mathrm{OB}}=(2,6,3)=2 \hat{i}+6 \hat{j}+3 \hat{k}$
Position vector of point $\mathrm{C}=\overrightarrow{\mathrm{OC}}=(3,10,-1)=3 \hat{i}+10 \hat{j}-\hat{k}$
Now $\overrightarrow{\mathrm{AB}}=$ Position vector of point $\mathrm{B}-$ Position vector of point $\mathrm{A}$
$
\begin{aligned}
& =2 \hat{i}+6 \hat{j}+3 \hat{k}-(\hat{i}+2 \hat{j}+7 \hat{k}) \\
& =2 \hat{i}+6 \hat{j}+3 \hat{k}-\hat{i}-2 \hat{j}-7 \hat{k}=\hat{i}+4 \hat{j}-4 \hat{k}
\end{aligned}
$

And $\overrightarrow{\mathrm{AC}}=$ Position vector of point $\mathrm{C}-$ Position vector of point $\mathrm{A}$

$
\begin{aligned}
& =3 \hat{i}+10 \hat{j}-\hat{k}-(\hat{i}+2 \hat{j}+7 \hat{k}) \\
& =3 \hat{i}+10 \hat{j}-\hat{k}-\hat{i}-2 \hat{j}-7 \hat{k}=2 \hat{i}+8 \hat{j}-8 \hat{k}=2(\hat{i}+4 \hat{j}-4 \hat{k}) . \\
& \Rightarrow \overrightarrow{\mathrm{AC}}=2 \cdot \overrightarrow{\mathrm{AB}} \quad \text { [Using eq. (i)] }
\end{aligned}
$
$\Rightarrow$ Vectors $\overrightarrow{\mathrm{AB}}$ and $\overrightarrow{\mathrm{AC}}$ are collinear and parallel. $[\because \vec{a}=m \vec{b}]$
Thus, points $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are collinear.
And also vectors $\overrightarrow{\mathrm{AB}}$ and $\overrightarrow{\mathrm{AC}}$ have a common point $\mathrm{A}$ and hence can't be parallel.

Ex 10.3 Question 17.

Show that the vectors $2 \hat{i}-\hat{j}+\hat{k}, \hat{i}-3 \hat{j}-5 \hat{k}$ and $3 \hat{i}-4 \hat{j}-4 \hat{k}$ form the vertices of a right angled triangle.

Answer.

Let the given position vectors be $\mathrm{A}, \mathrm{B}, \mathrm{C}$.
$\therefore$ Position vector of point $\mathrm{A}$ is $2 \hat{i}-\hat{j}+\hat{k}$, Position vector of point $\mathrm{B}$ is $\hat{i}-3 \hat{j}-5 \hat{k}$ and Position vector of point $\mathrm{C}$ is $3 \hat{i}-4 \hat{j}-4 \hat{k}$.
$\therefore \overrightarrow{\mathrm{AB}}=$ Position vector of $\mathrm{B}-$ Position vector of $\mathrm{A}$
$
\begin{aligned}
& =\hat{i}-3 \hat{j}-5 \hat{k}-(2 \hat{i}-\hat{j}+\hat{k}) \\
& =\hat{i}-3 \hat{j}-5 \hat{k}-2 \hat{i}+\hat{j}-\hat{k}=-\hat{i}-2 \hat{j}-6 \hat{k}
\end{aligned}
$
$\overrightarrow{\mathrm{BC}}=$ Position vector of $\mathrm{C}-$ Position vector of $\mathrm{B}$
$
\begin{aligned}
& =3 \hat{i}-4 \hat{j}-4 \hat{k}-(\hat{i}-3 \hat{j}-5 \hat{k}) \\
& =3 \hat{i}-4 \hat{j}-4 \hat{k}-\hat{i}+3 \hat{j}+5 \hat{k}=2 \hat{i}-\hat{j}+\hat{k}
\end{aligned}
$
$\overrightarrow{\mathrm{AC}}=$ Position vector of $\mathrm{C}-$ Position vector of $\mathrm{A}$

$
\begin{aligned}
& =3 \hat{i}-4 \hat{j}-4 \hat{k}-(2 \hat{i}-\hat{j}+\hat{k}) \\
& =3 \hat{i}-4 \hat{j}-4 \hat{k}-2 \hat{i}+\hat{j}-\hat{k}=\hat{i}-3 \hat{j}-5 \hat{k}
\end{aligned}
$

Adding eq. (i) and (ii),
$
\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=-\hat{i}-2 \hat{j}-6 \hat{k}+2 \hat{i}-\hat{j}+\hat{k}=\hat{i}-3 \hat{j}-5 \hat{k}=\overrightarrow{\mathrm{AC}} \text { [Using eq. (iii)] }
$

Therefore, by Triangle law of addition of vectors, points A, B, C are the vertices of a triangle $\mathrm{ABC}$.

Now from eq. (i) and (ii),
$
\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{BC}}=(-1)(2)+(-2)(-1)+(-6)(1)=-2+2-6=-6 \neq 0
$

Again from eq. (ii) and (iii),
$
\overrightarrow{\mathrm{BC}} \cdot \overrightarrow{\mathrm{AC}}=(2)(1)+(-1)(-3)+(1)(-5)=2+3-5=0
$
$\Rightarrow \overrightarrow{\mathrm{BC}}$ is perpendicular to $\overrightarrow{\mathrm{AC}}$.
$\Rightarrow$ Angle $\mathrm{C}$ is $90^{\circ}$. Therefore $\triangle \mathrm{ABC}$ is a right angled at $\mathrm{C}$.
Thus, A, B, C are the vertices of a right angled triangle.

Ex 10.3 Question 18.

If $\vec{a}$ is a non-zero vector of magnitude ' $a$ ' and $\lambda$ is a non-zero scalar, then $\hat{\lambda} \vec{a}$ is a unit vector if:
(A) $\lambda=1$
(B) $\lambda=-1$
(C) $a=|\lambda|$
(D) $a=\frac{1}{|\lambda|}$

Answer.

Given: $\vec{a}$ is a non-zero vector of magnitude $a$
$
\Rightarrow|\vec{a}|=1
$

Also given $\lambda \neq 0$ and $\lambda \vec{a}$ is a unit vector.
$
\begin{aligned}
& \Rightarrow|\lambda \vec{a}|=1 \\
& \Rightarrow|\lambda||\vec{a}|=1 \\
& \Rightarrow|\lambda| a=1 \\
& \Rightarrow a=\frac{1}{|\lambda|}
\end{aligned}
$

Therefore, option (D) is correct.