Exercise 8.3 - Chapter 8 Vector Algebra–I 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex 8.3
Question 1.
Find $\vec{a} \cdot \vec{b}$ when
(i) $\vec{a}=\hat{i}-2 \hat{j}+\hat{k}$ and $\vec{b}=3 \hat{i}-4 \hat{j}-2 \hat{k}$
(ii) $\vec{a}=2 \hat{i}+2 \hat{j}-\hat{k}$ and $\vec{b}=6 \hat{i}-3 \hat{j}+2 \hat{k}$
Solution:
For
$
\begin{gathered}
\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k} \\
\vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k} \\
\vec{a} \cdot \vec{b}=a_1 b_1+a_2 b_2+a_3 b_3
\end{gathered}
$
(i) Here $\vec{a}=\hat{i}-2 \hat{j}+\hat{k}$ and $\vec{b}=3 \hat{i}-4 \hat{j}-2 \hat{k}$
$
\begin{aligned}
\therefore \quad \vec{a} \cdot \vec{b} & =(1)(3)+(-2)(-4)+(1)(-2) \\
& =3+8-2=9
\end{aligned}
$

(ii)
$
\begin{aligned}
\vec{a} & =2 \hat{i}+2 \hat{j}-\hat{k} \text { and } \vec{b}=6 \hat{i}-3 \hat{j}+2 \hat{k} \\
\vec{a} \cdot \vec{b} & =(2)(6)+(2)(-3)+(-1)(2) \\
& =12-6-2=4 .
\end{aligned}
$
Question 2.
Find the value of $\lambda$ for which the vectors $\vec{a}$ and $\vec{b}$ are perpendicular, where
(i) $\vec{a}=2 \hat{i}+\lambda \hat{j}+\hat{k}$ and $\vec{b}=\hat{i}-2 \hat{j}+3 \hat{k}$
(ii) $\vec{a}=2 \hat{i}+4 \hat{j}-\hat{k}$ and $\vec{b}=3 \hat{i}-2 \hat{j}+\lambda \hat{k}$

When $\vec{a}$ and $\vec{b}$ are $\perp^{\mathrm{I}}$ then $\vec{a} \cdot \vec{b}=0$ $\vec{a} \perp^{\mathrm{r}} \vec{b} \Rightarrow \vec{a} \cdot \vec{b}=0$
(i) (2) (1) $+(\lambda)(-2)+(1)(3)=0 \Rightarrow \lambda=5 / 2$
(ii) $(2)(3)+(4)(-2)+(-1)(\lambda)=0$
$6-8-\lambda=0$
$-\lambda-2=0 \Rightarrow-\lambda=2 \Rightarrow \lambda=-2$
Question 3.
If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}|=10,|\vec{b}|=15$ and $\vec{a} \cdot \vec{b}=75 \sqrt{2}$, find the angle between $\vec{a}$ and $\vec{a}$.
Solution:
Given $\quad|\vec{a}|=10 ;|\vec{b}|=15, \vec{a} \cdot \vec{b}=75 \sqrt{2}$
The angle between $\vec{a}$ and $\vec{b}$ is given by
$
\begin{aligned}
\cos \theta & =\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{75 \sqrt{2}}{10 \times 15}=\frac{75 \sqrt{2}}{150} \\
& =\frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{\sqrt{2} \sqrt{2}}=\frac{1}{\sqrt{2}} \\
\Rightarrow \quad \theta & =\pi / 4 .
\end{aligned}
$
Question 4.
Find the angle between the vectors
(i) $2 \hat{i}+3 \hat{j}-6 \hat{k}$ and $6 \hat{i}-3 \hat{j}+2 \hat{k}$
(ii) $\hat{i}-\hat{j}$ and $\hat{j}-\hat{k}$

Solution:
If $\theta$ is the angle between $\vec{a}$ and $\vec{b}$ then $\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$
(i) Here
$
\begin{aligned}
\vec{a} & =2 \hat{i}+3 \hat{j}-6 \hat{k} \text { and } \\
\vec{b} & =6 \hat{i}-3 \hat{j}+2 \hat{k} \\
\vec{a} \cdot \vec{b} & =(2)(6)+(3)(-3)+(-6)(2) \\
& =12-9-12=-9 \\
|\vec{a}| & =\sqrt{2^2+3^2+6^2}=\sqrt{4+9+36}=\sqrt{49}=7 \\
|\vec{b}| & =\sqrt{6^2-3^2+2^2}=\sqrt{36+9+4}=\sqrt{49}=7
\end{aligned}
$
So,
$
\begin{aligned}
\cos \theta & =\frac{-9}{7 \times 7}=\frac{-9}{49} \\
\theta & =\cos ^{-1}\left(\frac{-9}{49}\right)
\end{aligned}
$
$
\Rightarrow \quad \theta=\cos ^{-1}\left(\frac{-9}{49}\right)
$
(ii) $\vec{a}=\hat{i}-\hat{j} ; \vec{b}=\hat{j}-\hat{k}$
$
\begin{aligned}
\vec{a} \cdot \vec{b} & =(1)(0)+(-1)(1)+(0)(-1)=-1 \\
|\vec{a}| & =\sqrt{1^2+1^2}=\sqrt{2} ; \\
|\vec{b}| & =\sqrt{1^2+1^2}=\sqrt{2} \\
\cos \theta & =\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{-1}{\sqrt{2} \sqrt{2}}=\frac{-1}{2} \\
\theta \quad \theta & =\pi-\pi / 3=2 \pi / 3 .
\end{aligned}
$

Question 5 .
If $\vec{a}, \vec{b}, \overrightarrow{\boldsymbol{c}}$ are three vectors such that $\vec{a}+2 \vec{b}+\vec{c}=\overrightarrow{0}$ and $|\vec{a}|=3,|\vec{b}|=4,|\vec{c}|=7$ find the angle between $\vec{a}$ and $\vec{b}$

Solution:
We are given $\quad \vec{a}+2 \vec{b}+\vec{c}=0 \Rightarrow \vec{a}+2 \vec{b}=-\vec{c}$
$
\begin{aligned}
& (\vec{a}+2 \vec{b})^2=(-\vec{c})^2=c^2 \\
& \text { (i.e.,) } \quad \vec{a}^2+4 \vec{b}^2+4 \vec{a} \cdot \vec{b}=\vec{c}^2 \\
& \text { Here } \quad|\vec{a}|=3,|\vec{b}|=4 \text { and }|\vec{c}|=7 \\
& \Rightarrow \quad 3^2+4\left(4^2\right)+4 \vec{a} \cdot \vec{b}=7^2 \\
& 9+64+4 \vec{a} \cdot \vec{b}=49 \\
& 4 \vec{a} \cdot \vec{b}=49-9-64=-24 \\
& \vec{a} \cdot \vec{b}=\frac{-24}{4}=-6 \\
& \vec{a} \cdot \vec{b}=|\vec{b}||\vec{b}| \cos \theta \\
& \text { (i.e.,) } \\
& \text { (3) (4) } \cos \theta=-6 \\
& \Rightarrow \quad \cos \theta=\frac{-6}{12}=-\frac{1}{2} \\
& \Rightarrow \quad \theta=\pi-\pi / 3=\frac{2 \pi}{3} \\
&
\end{aligned}
$
Question 6.
Show that the vectors $\vec{a}=2 \hat{i}+3 \hat{j}+6 \hat{k}, \vec{b}=6 \hat{i}+2 \hat{j}-3 \hat{k}$ and $\vec{c}=3 \hat{i}-6 \hat{j}+2 \hat{k}$ are mutually orthogonal.

Solution:
We are given
$
\begin{aligned}
\vec{a} & =2 \hat{i}+3 \hat{j}+6 \hat{k} \\
\vec{b} & =6 \hat{i}+2 \hat{j}-3 \hat{k} \text { and } \\
\vec{c} & =3 \hat{i}-6 \hat{j}+2 \hat{k} \\
\vec{a} \cdot \vec{b} & =(2)(6)+(3)(2)+(6)(-3) \\
& =12+6-18=0 \\
\vec{a} & \perp^r \vec{b} \\
\vec{b} \cdot \vec{c} & =(6)(3)+(2)(-6)+(-3)(2) \\
& =18-12-6=0 \\
\vec{b} & \perp^r \vec{c} \\
\vec{a} \cdot \vec{c} & =(2)(3)+(3)(-6)+(6)(2) \\
& =6-18+12=0 \\
\vec{a} & \perp^r \vec{c}
\end{aligned}
$
$
\Rightarrow \quad \vec{a} \perp^r \vec{b}
$
$
\Rightarrow \quad \vec{b} \perp^r \vec{c}
$
$
\Rightarrow \quad \vec{a} \perp^r \vec{c}
$
from (i), (ii), (iii) we see that, $\vec{a}, \vec{b}, \vec{c}$ are mutually orthogonal.

Question 7.
Show that the vectors $-\hat{i}-2 \hat{j}-6 \hat{k}, 2 \hat{i}-\hat{j}+\hat{k}$ and $-\hat{i}+3 \hat{j}+5 \hat{k}$ form a right angled triangle.
Solution:
The given vectors are $-\hat{i}-2 \hat{j}-6 \hat{k}, 2 \hat{i}-\hat{j}+\hat{k}$ and $-\hat{i}+3 \hat{j}+5 \hat{k}$
Here sum of the vectors $=\overrightarrow{0}$
$\Rightarrow$ We are given the sides of a triangle (i.e.,)
Now
$
\begin{aligned}
\vec{a} & =-\hat{i}-2 \hat{j}-6 \hat{k} ; \\
\vec{b} & =2 \hat{i}-\hat{j}+\hat{k} \text { and } \vec{c}=-\hat{i}+3 \hat{j}+5 \hat{k} \\
\vec{a} \cdot \vec{b} & =(-1)(2)+(-2)(-1)+(-6)(1) \\
& =-2+2-6=-6 \\
\vec{b} \cdot \vec{c} & =(2)(-1)+(-1)(3)+(1)(5) \\
& =-2-3+5=0 \\
\vec{b} \cdot \vec{c}=0 & \Rightarrow \vec{b} \perp^r \text { to } \vec{c}
\end{aligned}
$
also $\vec{a}, \vec{b}, \vec{c}$ are the sides of a $\Delta$.
So, the given vectors form the sides of a right angled triangle
Question 8.
If $|\vec{a}|=5,|\vec{b}|=6,|\vec{c}|=7$ and $\vec{a}+\vec{b}+\vec{c}=0$, find $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$.

Solution:
$
\begin{aligned}
& \text { Given } \quad \vec{a}+\vec{b}+\vec{c}=0 \\
& \Rightarrow \quad(\vec{a}+\vec{b}+\vec{c})^2=0 \\
& \text { (i.e.) }|\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 \\
& \Rightarrow 5^2+6^2+7^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0 \\
& \Rightarrow \quad 2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=-25-36-49=-110 \\
& \Rightarrow \quad \vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=\frac{-110}{2}=-55 \text {. } \\
&
\end{aligned}
$
Question 9.
Show that the points $(2,-1,3)(4,3,1)$ and $(3,1,2)$ are collinear
Solution:
Let the given points be A, B, C
i.e., and
Now
$
\begin{aligned}
\overrightarrow{\mathrm{AB}} & =\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=(4 \hat{i}+3 \hat{j}+\hat{k})-(2 \hat{i}-\hat{j}+3 \hat{k}) \\
& =2 \hat{i}+4 \hat{j}-2 \hat{k} \\
\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}} & =(3 \hat{i}+\hat{j}+2 \hat{k})-(2 \hat{i}-\hat{j}+3 \hat{k}) \\
& =\hat{i}+2 \hat{j}-\hat{k}
\end{aligned}
$
Here
$
\begin{aligned}
\overrightarrow{\mathrm{AB}} & =2 \hat{i}+4 \hat{j}-2 \hat{k}=2(\hat{i}+2 \hat{j}-\hat{k}) \\
& =2 \overrightarrow{\mathrm{AC}}
\end{aligned}
$
$\Rightarrow \overrightarrow{\mathrm{AB}}$ and $\overrightarrow{\mathrm{AC}}$ are collinear vectors.
$\Rightarrow$ The point $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are collinear.

Question 10.
If $\vec{a}, \vec{b}$ are unit vectors and $\theta$ is the angle between them, show that
(i) $\sin \frac{\theta}{2}=\frac{1}{2}|\vec{a}-\vec{b}|$
(ii) $\cos \frac{\theta}{2}=\frac{1}{2}|\vec{a}+\vec{b}|$
(iii) $\tan \frac{\theta}{2}=\frac{|\vec{a}-\vec{b}|}{|\vec{a}+\vec{b}|}$

Solution:

(i)
Given $|\vec{a}|=|\vec{b}|=1$ and $\theta$ is the angle between the vector $\vec{a}$ and $\vec{b}$.
$
\begin{aligned}
&(\vec{a}-\vec{b})^2=|\vec{a}|^2+|\vec{b}|^2-2 \vec{a} \cdot \vec{b} \\
&=1+1-2|\vec{a}||\vec{b}| \cos \theta=2-2 \cos \theta \\
&=2(1-\cos \theta)=2\left(2 \sin ^2 \frac{\theta}{2}\right)=4 \sin ^2 \frac{\theta}{2} \\
& \therefore|\vec{a}-\vec{b}|=\sqrt{4 \sin ^2 \frac{\theta}{2}}=2 \sin \frac{\theta}{2} \\
& \Rightarrow \frac{1}{2}|\vec{a}-\vec{b}|=\sin \frac{\theta}{2}
\end{aligned}
$

(ii)
$
\begin{aligned}
(\vec{a}+\vec{b})^2 & =|\vec{a}|^2+|\vec{b}|^2+2 \vec{a} \cdot \vec{b}=1+1+2|\vec{a}||\vec{b}| \cos \theta \\
& =2+2 \cos \theta=2(1+\cos \theta) \\
& =2\left(2 \cos ^2 \frac{\theta}{2}\right)=4 \cos ^2 \frac{\theta}{2} \\
\therefore|\vec{a}+\vec{b}| & =\sqrt{4 \cos ^2 \frac{\theta}{2}}=2 \cos \frac{\theta}{2} \\
\Rightarrow \frac{1}{2}|\vec{a}+\vec{b}| & =\cos \frac{\theta}{2}
\end{aligned}
$
(iii) From (1) $\sin \frac{\theta}{2}=\frac{1}{2}|\vec{a}-\vec{b}|$; From (2) $\cos \frac{\theta}{2}=\frac{1}{2}|\vec{a}+\vec{b}|$
$(1) /(2) \Rightarrow \tan \frac{\theta}{2}=\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}=\frac{\frac{1}{2}|\vec{a}-\vec{b}|}{\frac{1}{2}|\vec{a}+\vec{b}|}=\frac{|\vec{a}-\vec{b}|}{|\vec{a}+\vec{b}|}$
Question 11.
Let $\vec{a}, \vec{b}, \vec{c}$ be the three vectors such that $|\vec{a}|=3,|\vec{b}|=4,|\vec{c}|=5$ and each one of them being perpendicular to the sum of the other two, find $|\vec{a}+\vec{b}+\vec{c}|$

Solution:
Given $|\vec{a}|=3 ;|\vec{b}|=4 ;|\vec{c}|=5$
Now,
$
\begin{aligned}
(\vec{a}+\vec{b}+\vec{c})^2 & =\vec{a}^2+\vec{b}^2+\vec{c}^2+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{a} \cdot \vec{c}) \\
& =3^2+4^2+5^2+2 \vec{a} \cdot \vec{b}+2 \vec{b} \cdot \vec{c}+2 \vec{a} \cdot \vec{c} \\
& =9+16+25+\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{b} \cdot \vec{c}+\vec{a} \cdot \vec{c}+\vec{a} \cdot \vec{c} \\
& =50+\vec{a} \cdot(\vec{b}+\vec{c})+\vec{b} \cdot(\vec{c}+\vec{a})+\vec{c} \cdot(\vec{a}+\vec{b}) \\
& =50+0+0+0=50
\end{aligned}
$
( $\because$ one vector is $\perp^r$ to the sum of other two vectors)
$
|\vec{a}+\vec{b}+\vec{c}|=\sqrt{50}=\sqrt{25 \times 2}=5 \sqrt{2}
$
Question 12.
Find the projection of the vector $\hat{i}+3 \hat{j}+7 \hat{k}$ on the vector $2 \hat{i}+6 \hat{j}+3 \hat{k}$

Solution:
. Projection of $\vec{a}$ on $\vec{b}$ is given by $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$
Here $\vec{a}=\hat{i}+3 \hat{j}+7 \hat{k}$ and $\vec{b}=2 \hat{i}+6 \hat{j}+3 \hat{k}$
Now
$
\begin{aligned}
\vec{a} \cdot \vec{b} & =(1)(2)+(3)(6)+(7)(3) \\
& =2+18+21=41 \\
|\vec{b}| & =\sqrt{2^2+6^2+3^2}=\sqrt{4+36+9}=\sqrt{49}=7
\end{aligned}
$
Projection of $\vec{a}$ on $\vec{b}$ is $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}=\frac{41}{7}$.

Question 13.
Find $\lambda$, when the projection of $\vec{a}=\lambda \hat{i}+\hat{j}+4 \hat{k}$ on $\vec{b}=2 \hat{i}+6 \hat{j}+3 \hat{k}$ is 5 units.

Solution:
$\vec{a}=\lambda \hat{i}+\hat{j}+4 \hat{k}$ and $\vec{b}=2 \hat{i}+6 \hat{j}+3 \hat{k}$ projection of $\vec{a}$ on $\vec{b}$ is $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}=4$ (given) $$ \begin{aligned} \text { Now } & \vec{a} \cdot \vec{b}=(\lambda)(2)+(1)(6)+(4)(3)\end{aligned} $$
Here
$
\begin{array}{rlrl}
\Rightarrow & 2 \lambda+18 & =4 \times 7=28 \\
\cdot & 2 \lambda & =28-18=10 \\
\lambda & =10 / 2=5
\end{array}
$

Question 14 ,
Three vectors $\vec{a}, \vec{b}$ and $\vec{c}$ are such that $|\vec{a}|=2,|\vec{b}|=3,|\vec{c}|=4$ and $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$. Find $4 \vec{a} \cdot \vec{b}+3 \vec{b} \cdot \vec{c}+3 \vec{c} \cdot \vec{a}$
Solution:
Given $\vec{a}+\vec{b}+\vec{c}=0$
$
\begin{aligned}
& \Rightarrow \quad \vec{a}+\vec{b}=-\vec{c} \\
& \text { So } \quad(\vec{a}+\vec{b})^2=\vec{c}^2 \\
& \text { (i.e.) } \quad a^2+b^2+2 \vec{a} \cdot \vec{b}=\vec{c}^2 \\
& \Rightarrow \quad 4+9+2 \vec{a} \cdot \vec{b}=16 \\
& \Rightarrow \quad 2 \vec{a} \cdot \vec{b}=16-4-9=3 \\
& \vec{a} \cdot \vec{b}=3 / 2 \\
& \text { Again } \quad \vec{a}+\vec{b}+\vec{c}=0 \\
& \Rightarrow \quad \vec{a}+\vec{c}=-\vec{b} \\
& (\bar{a}+\bar{c})^2=\vec{b}^2 \\
& \vec{a}^2+\vec{c}^2+2 \vec{a} \cdot \vec{c}=\vec{b}^2 \\
& 4+16+2 \vec{a} \cdot \vec{c}=9 \\
& 2 \vec{a} \cdot \vec{c}=9-4-16=-11 \\
&
\end{aligned}
$

$
\vec{a} \cdot \vec{c}=\frac{-11}{2}(\text { i.e., }) \vec{c} \cdot \vec{a}=\frac{-11}{2} \quad(\because \vec{a} \cdot \vec{c}=\vec{c} \cdot \vec{a})
$
Also $\quad \begin{aligned} \vec{a}+\vec{b}+\vec{c} & =0 \\ \vec{b}+\vec{c} & =-\vec{a} \\ (\vec{b}+\vec{c})^2 & =\vec{a}^2 \\ 9+16+2 \vec{b} \cdot \vec{c} & =4 \\ 2 \vec{b} \cdot \vec{c} & =4-9-16=-21 \\ 9 & \vec{b} \cdot \vec{c}=\frac{-21}{2}\end{aligned}$
Here, $\quad \vec{a} \cdot \vec{b}=3 / 2 ; \vec{b} \cdot \vec{c}=-\frac{21}{2}$ and $\vec{c} \cdot \vec{a}=\frac{-11}{2}$
$
\text { So, } \begin{aligned}
4(\vec{a} \cdot \vec{b})+3(\vec{b} \cdot \vec{c})+3(\vec{c} \cdot \vec{a}) & =4\left(\frac{3}{2}\right)+3\left(\frac{-21}{2}\right)+3\left(\frac{-11}{2}\right) \\
& =6-\frac{63}{2}-\frac{33}{2}=6-\frac{96}{2}=6-48=-42 .
\end{aligned}
$