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

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Additional Questions

Question 1.
Find $\lambda$ so that the vectors $2 \hat{i}+\lambda \hat{j}+\hat{\boldsymbol{k}}$ and $\hat{i}-\mathbf{2} \hat{j}+\hat{\boldsymbol{k}}$ are perpendicular to each other.

Solution:
Let $\vec{a}=2 \hat{i}+\lambda \hat{j}+\hat{k}$ and $\vec{b}=\hat{i}-2 \hat{j}+\hat{k}$
Given $\vec{a} \perp^r$ to $\vec{b} \Rightarrow \vec{a} \cdot \vec{b}=0$
Here
i.e.,
$
\begin{aligned}
\vec{a} \cdot \vec{b} & =(2 \hat{i}+\lambda \hat{j}+\hat{k}) \cdot(\hat{i}-2 \hat{j}+\hat{k}) \\
& =(2)(1)+(\lambda)(-2)+(1)(1)=2-2 \lambda+1=3-2 \lambda \\
\vec{a} \cdot \vec{b} & =0 \Rightarrow 3-2 \lambda=0 \\
2 \lambda & =3 ; \quad \therefore \quad \lambda=\frac{3}{2}
\end{aligned}
$
Question 2.
If $|\vec{a}+\vec{b}|=60,|\vec{a}-\vec{b}|=40$ and $|\vec{b}|=46$, find $|\vec{a}|$.
Solution:
$
\begin{aligned}
|\vec{a}+\vec{b}| & =60 \\
(\vec{a}+\vec{b})^2 & =60^2=3600 \\
\text { i.e., } \quad \vec{a}^2+\vec{b}^2+2 \vec{a} \cdot \vec{b} & =3600 \\
|\vec{a}-\vec{b}| & =40 ; \quad \therefore \quad(\vec{a}-\vec{b})^2=40^2=1600 \\
\text { i.e. } \quad \vec{a}^2+\vec{b}^2-2 \vec{a} \cdot \vec{b} & =1600 \\
(1)+(2) \Rightarrow 2\left(\vec{a}^2+\vec{b}^2\right) & =5200
\end{aligned}
$

$
\begin{aligned}
& \therefore \quad \vec{a}^2+\vec{b}^2=2600 \\
& |\vec{b}|=46 ; \quad \therefore \quad \vec{b}^2=|\vec{b}|^2=46^2=2116 \\
&
\end{aligned}
$
Substituting $b^2$ value in (3) we get,
$
\begin{array}{rlrl} 
& & |\vec{a}|^2+2116 & =2600 \\
\therefore & |\vec{a}|^2 & =2600-2116=484 \\
\therefore & |\vec{a}| & =\sqrt{484}=22
\end{array}
$
Question 3.
If the sum of two unit vectors is a unit vector prove that the magnitude of their difference is $\sqrt{3}$

Solution:
Let $\vec{a}$ and $\vec{b}$ be the two unit vectors i.e., $|\vec{a}|=|\vec{b}|=1$.
Also $\vec{a}+\vec{b}$ is a unit vector.
$
\begin{aligned}
\Rightarrow \quad|\vec{a}+\vec{b}| & =1 ; \quad \therefore \quad(\vec{a}+\vec{b})^2=1^2=1 \\
\text { i.e., } \quad|\vec{a}|^2+|\vec{b}|^2+2 \vec{a} \cdot \vec{b} & =1 \\
\text { i.e., } \quad(1)+(1)+2|\vec{a}| \cdot|\vec{b}| \cos \theta & =1 \\
1+1+2 \cos \theta & =1 \\
2 \cos \theta & =1-2=-1 \\
\cos \theta & =\frac{-1}{2} \Rightarrow \theta=\frac{2 \pi}{3} \\
\therefore \quad \therefore \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 \frac{2 \pi}{3} \quad\left[\because \text { angle between } \vec{a} \text { and } \vec{b} \text { is } \frac{2 \pi}{3}\right] \\
& =1+1-2(1)(1)\left(\frac{-1}{2}\right)=2+1=3 \\
\therefore|\vec{a}-\vec{b}| & =\sqrt{3}
\end{aligned}
$
Question 4 .
Show that the vectors $3 \hat{i}-2 \hat{j}+\hat{k}, \hat{i}-3 \hat{j}+5 \hat{k}$ and $2 \hat{i}+\hat{j}-4 \hat{k}$ form a right angled triangle.
Solution:
Let $\vec{a}=3 \hat{i}-2 \hat{j}+\hat{k} ; \vec{b}=\hat{i}-3 \hat{j}+5 \hat{k} ; \vec{c}=2 \hat{i}+\hat{j}-4 \hat{k}$
$
\begin{array}{ll}
|\vec{a}|^2=9+4+1=14 ; & |\vec{b}|^2=1+9+25=35 \\
|\vec{c}|^2=4+1+16=21 & \therefore \vec{b}^2=\vec{a}^2+\vec{c}^2
\end{array}
$
$\Rightarrow$ The given vectors form the sides of a right of a right angled triangle.
Question 5.
Find the projection of
(i) $\hat{i}-\hat{j}$ on $\mathbf{Z}$-axis
(ii) $\hat{i}+2 \hat{j}-2 \hat{k}$ on $2 \hat{i}-\hat{j}+5 \hat{k}$
(iii) $3 \hat{i}+\hat{j}-\hat{k}$ on $4 \hat{i}-\hat{j}+2 \hat{k}$
Solution:

Projection of $\vec{a}$ on $\vec{b}$ is $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$
(i) Here $\vec{a}=\hat{i}-\hat{j} ; \vec{b}=\hat{k} ; \vec{a} \cdot \vec{b}=(\hat{i}-\hat{j}) \cdot \hat{k}=0$
So projection of $\hat{i}-\hat{j}$ on Z-axis $=0$.
(ii) $\vec{a}=\hat{i}+2 \hat{j}-2 \hat{k} ; \vec{b}=2 \hat{i}-\hat{j}+5 \hat{k}$
$\vec{a} \cdot \vec{b}=2-2-10=-10 ; \quad|\vec{b}|=\sqrt{4+1+25}=\sqrt{30}$
$\therefore \quad$ Projection of $\vec{a}$ on $\vec{b}=\frac{-10}{\sqrt{30}}$ units.
(iii) $\vec{a}=3 \hat{i}+\hat{j}-\hat{k} ; \vec{b}=4 \hat{i}-\hat{j}+2 \hat{k}$
$\vec{a} \cdot \vec{b}=12-1-2=9 ; \quad|\vec{b}|=\sqrt{16+1+4}=\sqrt{21}$
$\therefore \quad$ Projection of $\vec{a}$ on $\vec{b}=\frac{9}{\sqrt{21}}$ units.
Question 6.
Show that the vector $\hat{i}+\hat{j}+\hat{k}$ is equally inclined with the coordinate axes.

Solution:
Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}$
$
\therefore \quad|\vec{a}|=\sqrt{1+1+1}=\sqrt{3}
$
The unit vectors along the $\mathrm{X}, \mathrm{Y}, \mathrm{Z}$ axes are respectively $\hat{i}, \hat{j}$ and $\hat{k}$. Let $\theta_1$ be the angle between $\vec{a}$ and X-axis.
$
\begin{aligned}
\therefore \quad \cos \theta_1 & =\frac{\vec{a} \cdot \hat{i}}{|\vec{a}||\hat{i}|}=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot \hat{i}}{(\sqrt{3})(1)}=\frac{(1)(1)}{\sqrt{3}}=\frac{1}{\sqrt{3}} \\
\therefore \quad \theta_1 & =\cos ^{-1} \frac{1}{\sqrt{3}}
\end{aligned}
$
Let $\theta_2$ be the angle between $\vec{a}$ and Y-axis.
$
\begin{aligned}
\therefore \quad \cos \theta_2 & =\frac{\vec{a} \cdot \hat{j}}{|\vec{a}||\hat{j}|}=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot \hat{j}}{(\sqrt{3})(1)} \\
& =\frac{(1)(1)}{\sqrt{3}}=\frac{1}{\sqrt{3}} \Rightarrow \theta_2=\cos ^{-1} \frac{1}{\sqrt{3}}
\end{aligned}
$
Let $\theta_3$ be the angle between $\vec{a}$ and $\mathrm{Z}$-axis.
$
\begin{aligned}
\therefore \quad \cos \theta_3 & =\frac{\vec{a} \cdot \hat{k}}{|\vec{a}||\hat{k}|}=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot \hat{k}}{(\sqrt{3})(1)} \\
& =\frac{(1)(1)}{\sqrt{3}}=\frac{1}{\sqrt{3}} \theta_3=\cos ^{-1} \frac{1}{\sqrt{3}}
\end{aligned}
$
From (1), (2) and (3) we see that, $\theta_1=\theta_2=\theta_3 \Rightarrow \vec{a}($ i.e., ) $\hat{i}+\hat{j}+\hat{k}$ is equally inclined with the coordinate axis.
Question 7.
If $\vec{a}, \vec{b}, \vec{c}$ are three mutually perpendicular unit vectors, then prove that $|\vec{a}+\vec{b}+\vec{c}|=\sqrt{3}$
Solution:
Given $\quad|\vec{a}|=|\vec{b}|=|\vec{c}|=1$
and $\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{a}=0$
$
\begin{aligned}
& \text { To prove } \\
& \qquad \begin{aligned}
|\vec{a}+\vec{b}+\vec{c}| & =\sqrt{3} \\
(\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{c} \cdot \vec{a}) \\
& =1+1+1+2(0+0+0)=3 \\
|\vec{a}+\vec{b}+\vec{c}| & =\sqrt{3}
\end{aligned}
\end{aligned}
$

Question 8.
Show that the points whose positions vectors
$4 \hat{i}-3 \hat{j}+\hat{k}, 2 \hat{i}-4 \hat{j}+5 \hat{k}, \hat{i}-\hat{j}$ from a right angled
triangle.
Solution:
Let the given points be $\mathrm{A}, \mathrm{B}, \mathrm{C}$.
$\overrightarrow{\mathrm{OA}}=4 \hat{i}-3 \hat{j}+\hat{k}, \overrightarrow{\mathrm{OB}}=2 \hat{i}-4 \hat{j}+5 \hat{k}$ and $\overrightarrow{\mathrm{OC}}=\hat{i}-\hat{j}$
Now

$
\begin{aligned}
\overrightarrow{\mathrm{AB}} & =\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}} \\
& =(2 \hat{i}-4 \hat{j}+5 \hat{k})-(4 \hat{i}-3 \hat{j}+\hat{k}) \\
& =-2 \hat{i}-\hat{j}+4 \hat{k} \\
|\overrightarrow{\mathrm{AB}}|^2 & =4+1+16=21=c^2 \\
\overrightarrow{\mathrm{BC}} & =\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}}=(\vec{i}-\vec{j})-(2 \vec{i}-4 \vec{j}+5 \vec{k}) \\
& =-\hat{i}+3 \hat{j}-5 \hat{k} \\
|\overrightarrow{\mathrm{BC}}|^2 & =1+9+25=35=a^2 \\
\overrightarrow{\mathrm{AC}} & =\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}} \\
& =(\hat{i}-\hat{j})-(4 \hat{i}-3 \hat{j}+\hat{k})=-3 \hat{i}+2 \hat{j}-\hat{k} \\
|\overrightarrow{\mathrm{AC}}|^2 & =9+4+1=14=b^2 ; \quad \text { Here } c^2+b^2=a^2
\end{aligned}
$
$\Rightarrow$ The given points form a right angled triangle.

Question 9.
Let $\vec{u}, \vec{v}$ and $\vec{w}$ be vectors such that $\vec{u}+\vec{v}+\vec{w}=\overrightarrow{0}$. If $|\vec{u}|=3,|\vec{v}|=4$ and $|\vec{w}|=5$ then find $\vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u}$
Solution:
$
\begin{aligned}
& \vec{u}+\vec{v}+\vec{w}=\overrightarrow{0} ; \quad \therefore \quad(\vec{u}+\vec{v}+\vec{w})^2=0^2=0 \\
& \text { i.e., } \quad|\vec{u}|^2+|\vec{v}|^2+|\vec{w}|^2+2(\vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u})=0 \\
& \text { Given }|\vec{u}|=3,|\vec{v}|=4 \text { and }|\vec{w}|=5 \\
& \Rightarrow \quad 9+16+25+2(\vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u})=0 \\
& \Rightarrow \quad 2[\vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u}]=-50 \\
& \therefore \quad \vec{u} \cdot \vec{v}+\vec{v} \cdot \vec{w}+\vec{w} \cdot \vec{u}=\frac{-50}{2}=-25
\end{aligned}
$