Exercise 8.3 - Chapter 8 - Vector Algebra – I - 11th Maths Guide Samacheer Kalvi Solutions

You can Download the Exercise 8.3 - Chapter 8 - Vector Algebra – I - 11th Maths Guide Samacheer Kalvi Solutions with expert answers for all chapters. Perfect for Tamil & English Medium students to revise the syllabus and score more marks in board exams. Download and share it with your friends

Share this to Friend on WhatsApp

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}
$