Exercise 8.2 - Chapter 8 Vector Algebra–I 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

E.x.8.2

Verify whether the following ratios are direction cosines of some vector or not.
(i) $\frac{1}{5}, \frac{3}{5}, \frac{4}{5}$
(ii) $\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2}$
(iii) $\frac{4}{3}, 0, \frac{3}{4}$
Solution:
(i) Here the ratios are $\frac{1}{5}, \frac{3}{5}, \frac{4}{5}$
$
\text { Now }\left(\frac{1}{5}\right)^2+\left(\frac{3}{5}\right)^2+\left(\frac{4}{5}\right)^2=\frac{1}{25}+\frac{9}{25}+\frac{16}{25}=\frac{26}{25} \neq 1
$
So, the given ratios are not the direction cosines of a vector.
(ii) $\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2}$
Now, $\left(\frac{1}{\sqrt{2}}\right)^2+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2=\frac{1}{2}+\frac{1}{4}+\frac{1}{4}=1$
$\Rightarrow \frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2}$ are the direction cosines of a vector.
(iii) $\frac{4}{3}, 0, \frac{3}{4}$
-. Now, $\left(\frac{4}{3}\right)^2+0+\left(\frac{3}{4}\right)^2=\frac{16}{9}+\frac{9}{16}=\frac{256+81}{144}=\frac{337}{144} \neq 1$
So, the given ratios are not the direction cosines of a vector.
Question 2.
Find the direction cosines of a vectors whose direction ratios are
(i) $1,2,3$
(ii) $3,-1,3$
(iii) $0,0,7$

Solution:
(i) Let the vector be $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}$
So,
$
|\vec{a}|=\sqrt{1^2+2^2+3^2}=\sqrt{1+4+9}=\sqrt{14}
$
Hence and d.c's of $\vec{a}=\left(\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}\right)$
(ii) Let
d.c's of
$
\begin{aligned}
& \vec{a}=3 \hat{i}-\hat{j}+3 \hat{k} \\
&|\vec{a}|=\sqrt{3^2+1^2+3^2}=\sqrt{9+1+9}=\sqrt{19} \\
& \vec{a}=\left(\frac{3}{\sqrt{19}}, \frac{-1}{\sqrt{19}}, \frac{3}{\sqrt{19}}\right)
\end{aligned}
$
(iii) Let
$
\begin{aligned}
\vec{a} & =0 \hat{i}+0 \hat{j}+7 \hat{k} \\
|\vec{a}| & =\sqrt{0+0+7^2}=7 \\
\text { d.c's of } \vec{a} & =\left(\frac{0}{7}, \frac{0}{7}, \frac{7}{7}\right) \quad \text { (i.e.,). }=(0,0,1)
\end{aligned}
$
Question 3.
Find the direction cosines and direction ratios for the following vectors
(i) $3 \hat{i}-4 \hat{j}+8 \hat{k}$
(ii) $3 \hat{i}+\hat{j}+\hat{k}$
(iii) $\hat{j}$
(iv) $5 \hat{i}-3 \hat{j}-48 \hat{k}$
(v) $3 \hat{i}-3 \hat{k}+4 \hat{j}$
(vi) $\hat{i}-\hat{k}$
Solution:
If $\vec{a}=\vec{a} \hat{i}+\vec{b} \hat{j}+\vec{c} \hat{k}$
$
\begin{aligned}
|\vec{a}| & =\sqrt{a^2+b^2+c^2}=r \text { (say) } \\
\text { d.c's of } \vec{a} & =\left(\frac{\vec{a}}{r}, \frac{\vec{b}}{r}, \frac{\vec{c}}{r}\right) \\
\text { d.r's of } \vec{a} & =(\vec{a}, \vec{b}, \vec{c})
\end{aligned}
$

(i) Let
$
\begin{aligned}
\vec{a} & =3 \hat{i}-4 \hat{j}+8 \hat{k} \\
\text { d.r's of } \vec{a} & =(3,-4,8) \\
|\vec{a}| & =\sqrt{9+16+64}=\sqrt{89} \\
\text { d.c's of }|\vec{a}| & =\left(\frac{3}{\sqrt{89}}, \frac{-4}{\sqrt{89}}, \frac{8}{\sqrt{89}}\right)
\end{aligned}
$

(ii) Let
$
\begin{aligned}
\vec{a} & =3 \hat{i}+\hat{j}+\hat{k} \\
|\vec{a}| & =\sqrt{9+1+1}=\sqrt{11} \\
\text { d.c's of } \vec{a} & =\left(\frac{3}{\sqrt{11}}, \frac{1}{\sqrt{11}}, \frac{1}{\sqrt{11}}\right) \\
\text { d.r's of } \vec{a} & =(3,1,1)
\end{aligned}
$

(iii) Let
$
\begin{aligned}
\vec{a} & =0 \hat{i}+\hat{j}+0 \hat{k} \\
|\vec{a}| & =\sqrt{1}=1 \\
\text { d.c's of } \vec{a} & =(0,1,0) \\
\text { d.r's of } \vec{a} & =(0,1,0)
\end{aligned}
$

(iv) Let
$
\begin{aligned}
\vec{a} & =5 \hat{i}-3 \hat{j}-48 \hat{k} \\
|\vec{a}| & =\sqrt{25+9+48^2}=\sqrt{25+9+2304} \\
& =\sqrt{2338} \\
\text { d.c's of } \vec{a} & =\left(\frac{5}{\sqrt{2338}}, \frac{-3}{\sqrt{2338}}, \frac{-48}{\sqrt{2338}}\right) \\
\text { d.r's of } \vec{a} & =(5,-3,-48)
\end{aligned}
$

(v) Let
$
\begin{aligned}
& \vec{a}=3 \hat{i}+4 \hat{j}-3 \hat{k} \\
& \vec{a}=3 \hat{i}+4 \hat{j}-3 \hat{k}
\end{aligned}
$
$
|\vec{a}|=\sqrt{9+16+9}=\sqrt{34}
$
d.c's of $\vec{a}=\left(\frac{3}{\sqrt{34}}, \frac{4}{\sqrt{34}}, \frac{-3}{\sqrt{34}}\right)$; d.r's of $\vec{a}=(3,4,-3)$

(vi) Let
$
\begin{aligned}
\vec{a} & =\hat{i}+0 \hat{j}-\hat{k} \\
|\vec{a}| & =\sqrt{1+1}=\sqrt{2} \\
\text { d.c's of } \quad \vec{a} & =\left(\frac{1}{\sqrt{2}}, 0,-\frac{1}{\sqrt{2}}\right) \\
\text { d.r's of } \quad \vec{a} & =(1,0,-1)
\end{aligned}
$

Question 4.
A triangle is formed by joining the points $(1,0,0),(0,1,0)$ and $(0,0,1)$. Find the direction cosines of the medians.
Solution:
$
\overrightarrow{\mathrm{OA}}=\hat{i}, \overrightarrow{\mathrm{OB}}=\hat{j}, \overrightarrow{\mathrm{OC}}=\hat{k}
$

$\mathrm{D}$ is the mid point of $\mathrm{BC}$
$\therefore \quad \overrightarrow{\mathrm{OD}}=\frac{\overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{OC}}}{2}=\frac{\hat{j}+\hat{k}}{2}$
$\mathrm{E}$ is the mid point of $\mathrm{AC}$
$\therefore \quad \overrightarrow{\mathrm{OE}}=\frac{\hat{i}+\hat{k}}{2}$ and
$F$, is the mid point of $\mathrm{AB}$
$\therefore \quad \overrightarrow{\mathrm{OF}}=\frac{\hat{i}+\hat{j}}{2}$
Now the medians are $\overrightarrow{\mathrm{AD}}, \overrightarrow{\mathrm{BE}}$ and $\overrightarrow{\mathrm{CF}}$
(i)
$
\begin{aligned}
\overrightarrow{\mathrm{AD}}=\overrightarrow{\mathrm{OD}}-\overrightarrow{\mathrm{OA}} & =\frac{\hat{j}+\hat{k}}{2}-\hat{i} \\
& =-\hat{i}+\frac{\hat{j}}{2}+\frac{\hat{k}}{2} \\
|\overrightarrow{\mathrm{AD}}| & =\sqrt{1+\frac{1}{4}+\frac{1}{4}}=\sqrt{1+\frac{1}{2}}=\frac{\sqrt{3}}{\sqrt{2}}
\end{aligned}
$

$\begin{aligned}
& \text { and d.c's of } \overrightarrow{\mathrm{AD}}=\left(\frac{-1}{\sqrt{3} / \sqrt{2}}, \frac{1 / 2}{\sqrt{3} / \sqrt{2}}, \frac{1 / 2}{\sqrt{3} / \sqrt{2}}\right) \\
&=\left(-\frac{\sqrt{2}}{\sqrt{3}}, \frac{\sqrt{2}}{2 \sqrt{3}}, \frac{\sqrt{2}}{2 \sqrt{3}}\right) \\
&=\left(-\frac{\sqrt{2}}{\sqrt{3}}, \frac{1}{\sqrt{2} \sqrt{3}}, \frac{1}{\sqrt{2} / \sqrt{3}}\right) \\
&=\left(-\frac{\sqrt{2}}{\sqrt{3}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}\right) \\
& {\left[\text { Now } \frac{\sqrt{2}}{\sqrt{3}}=\frac{\sqrt{2}}{\sqrt{3}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{2}{\sqrt{6}}\right]=\left(-\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}\right) }
\end{aligned}$

(ii)
$
\begin{aligned}
\overrightarrow{\mathrm{BE}} & =\overrightarrow{\mathrm{OE}}-\overrightarrow{\mathrm{OB}}=\frac{\hat{i}+\hat{k}}{2}-\hat{j}=\frac{\hat{i}}{2}-\hat{j}+\frac{\hat{k}}{2} \\
|\overrightarrow{\mathrm{BE}}| & =\sqrt{\frac{1}{4}+1+\frac{1}{4}}=\frac{\sqrt{3}}{\sqrt{2}} \\
\text { d.c's of } \overrightarrow{\mathrm{BE}} & =\left(\frac{1 / 2}{\sqrt{3} / \sqrt{2}}, \frac{-1}{\sqrt{3} / \sqrt{2}}, \frac{1 / 2}{\sqrt{3} / \sqrt{2}}\right)=\left(\frac{1}{\sqrt{6}}, \frac{-2}{\sqrt{6}}, \frac{1}{\sqrt{6}}\right)
\end{aligned}
$

(iii)
$
\begin{aligned}
\overrightarrow{\mathrm{CF}}=\overrightarrow{\mathrm{OF}}-\overrightarrow{\mathrm{OC}} & =\frac{\hat{i}+\hat{j}}{2}-\hat{k}=\frac{\hat{i}}{2}+\frac{\hat{j}}{2}-\hat{k} \\
|\overrightarrow{\mathrm{CF}}| & =\sqrt{\frac{1}{4}+\frac{1}{4}+1}=\frac{\sqrt{3}}{\sqrt{2}} \\
\text { d.c's of } \overrightarrow{\mathrm{CF}} & =\left(\frac{1 / 2}{\sqrt{3} / \sqrt{2}}, \frac{1 / 2}{\sqrt{3} / \sqrt{2}}, \frac{-1}{\sqrt{3} / \sqrt{2}}\right) \\
& =\left(\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{-2}{\sqrt{6}}\right)
\end{aligned}
$

Question 5.
If $\frac{1}{2}, \frac{1}{\sqrt{2}}$, a are the direction cosines of some vector, then find a.

Solution:
Let $\cos \alpha=\frac{1}{2}, \cos \beta=\frac{1}{\sqrt{2}}$, and $\cos \gamma=a$
$
\begin{aligned}
& \text { Now, } \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1 \\
& \Rightarrow \quad\left(\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{2}}\right)^2+a^2=1 \\
& \Rightarrow \quad \frac{1}{4}+\frac{1}{2}+a^2=1 \\
& a^2=1-\frac{1}{2}-\frac{1}{4}=\frac{1}{4} \Rightarrow a=\sqrt{\frac{1}{4}}= \pm \frac{1}{2} \\
&
\end{aligned}
$
Question 6.
If $(a, a+b, a+b+c)$ is one set of direction ratios of the line joining $(1,0,0)$ and $(0,1,0)$, then find a set of values of $\mathrm{a}, \mathrm{b}, \mathrm{c}$.
Solution:
Let $\mathrm{A}$ be the point $(1,0,0)$ and $\mathrm{B}$ be the point $(0,1,0)$ (i.e.,) $\overrightarrow{\mathrm{OA}}=\hat{i}$ and $\overrightarrow{\mathrm{OB}}=\hat{j}$
Then $\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=\hat{j}-\hat{i}=-\hat{i}+\hat{j}$
$
\begin{aligned}
& =(-1,1,0) \\
& =(\mathrm{a}, \mathrm{a}+\mathrm{b}, \mathrm{a}+\mathrm{b}+\mathrm{c}) \\
& \Rightarrow \mathrm{a}=-1, \mathrm{a}+\mathrm{b}=1 \text { and } \mathrm{a}+\mathrm{b}+\mathrm{c}=0 \\
& \text { Now } \mathrm{a}=-1 \Rightarrow-1+\mathrm{b}=1 ; \mathrm{a}+\mathrm{b}+\mathrm{c}=0 \\
& \Rightarrow \mathrm{b}=2 ;-1+2+\mathrm{c}=0 \Rightarrow \mathrm{c}+1=0 \\
& \Rightarrow \mathrm{c}=-1 \\
& \therefore \mathrm{a}=-1 ; \mathrm{b}=2 ; \mathrm{c}=-1 .
\end{aligned}
$
Note: If we taken $\overrightarrow{\mathrm{BA}}$ then we get $\mathrm{a}=1, \mathrm{~b}=-2$ and $\mathrm{c}=1$.
Question 7.
Show that the vectors $2 \hat{i}-\hat{j}+\hat{k}, 3 \hat{i}-4 \hat{j}-4 \hat{k}, \hat{i}-3 \hat{j}-5 \hat{k}$ form a right angled triangle.
Sol:

Let the given vectors be $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}, \vec{b}=3 \hat{i}-4 \hat{j}-4 \hat{k}$ and $\vec{c}=\hat{i}-3 \hat{j}-5 \hat{k}$
$
\begin{aligned}
\vec{a}+\vec{c} & =(2 \hat{i}-\hat{j}+\hat{k})+(\hat{i}-3 \hat{j}-5 \hat{k}) \\
& =3 \hat{i}-4 \hat{j}-4 \hat{k}=\vec{b}
\end{aligned}
$
$\Rightarrow$ The given vectors form the sides of a triangle.
Now
$
\begin{aligned}
\therefore \quad|\vec{a}|^2 & =6 \\
|\vec{b}| & =\sqrt{3^2+4^2+4^2}=\sqrt{9+16+16}=\sqrt{41}
\end{aligned}
$
$\mathrm{So}$
$
\begin{aligned}
|\vec{b}|^2 & =41 \\
|\vec{c}| & =\sqrt{1+9+25}=\sqrt{35}
\end{aligned}
$
$\therefore \quad|\vec{c}|^2=35$
Now $|\vec{b}|^2=41=35+6=|\vec{c}|^2+|\vec{a}|^2$.
$\Rightarrow$ The given vectors form the sides of a right angled triangle.
$\Rightarrow$ The given vectors form the sides of a right angled triangle.

Question 8.
Find the value of $\mathrm{k}$ for which the vectors $\vec{a}=3 \hat{i}+2 \hat{j}+9 \hat{k}$ and $\vec{b}=\hat{i}+\lambda \hat{j}+3 \hat{k}$ are parallel.
Solution:
Given $\vec{a}$ and $\vec{b}$ are parallel $\Rightarrow \vec{a}=t \vec{b} \quad$ (where $t$ is a scalar)
(i.e.,) $3 \hat{i}+2 \hat{j}+9 \hat{k}=t(\hat{i}+\lambda \hat{j}+3 \hat{k})$
equating $\hat{i}$ components we get $3=t$
equating $\hat{j}$ components
(i.e.)
$
\begin{aligned}
& 2=t \lambda \\
& 2=3 \lambda \Rightarrow \lambda=2 / 3
\end{aligned}
$
Question 9.
Show that the following vectors are coplanar.
(i) $\hat{i}-2 \hat{j}+3 \hat{k},-2 \hat{i}+3 \hat{j}-4 \hat{k},-\hat{j}+2 \hat{k}$
(ii) $2 \hat{i}+3 \hat{j}+\hat{k}, \hat{i}-\hat{j}, 7 \hat{i}+3 \hat{j}+2 \hat{k}$
Solution:
Let the given three vectors be $\vec{a}, \vec{b}$ and $\vec{c}$. When we are able to write one vector as a linear combination of the other two vectors, then the given vectors are called coplanar vectors.
Let $\vec{a}=m \vec{b}+n \vec{c}$ where
(i.e.) $\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}$

$
\begin{gathered}
\vec{b}=-2 \hat{i}+3 \hat{j}-4 \hat{k} \text { and } \vec{c}=-\hat{j}+2 \hat{k} \\
\Rightarrow \quad \hat{i}-2 \hat{j}+3 \hat{k}=m(-2 \hat{i}+3 \hat{j}-4 \hat{k})+n(-\hat{j}+2 \hat{k})
\end{gathered}
$
Equating the $\hat{i}, \hat{j}$ and $\hat{k}$ components
$
\text { (i.e.) } \begin{aligned}
1 & =-2 m \\
-2 & =3 m-n \\
3 & =-4 m+2 n
\end{aligned}
$
Now we have to solve (1) and (2) and substitute the value in (3).
Solving (1) and (2)
$
\begin{aligned}
(1) \Rightarrow-2 m & =1 \\
\therefore m & =-\frac{1}{2}
\end{aligned}
$
Substituting $m=-\frac{1}{2}$ in (2) we get,
$
\begin{aligned}
3\left(\frac{-1}{2}\right)-n & =-2 \\
-\frac{3}{2}-n & =-2 \\
\Rightarrow \quad \therefore-n & =-2+\frac{3}{2}=\frac{-4+3}{2}=-\frac{1}{2} \\
\Rightarrow \quad n & =\frac{1}{2}
\end{aligned}
$

$
\therefore m=-\frac{1}{2} ; n=\frac{1}{2}
$
Substituting the values of $m$ and $n$ in (3).
$
\begin{aligned}
\text { LHS } & =3 \\
\text { RHS } & =-4 m+2 n=-4\left(\frac{-1}{2}\right)+2\left(\frac{1}{2}\right) \\
& =2+1=3 \\
\Rightarrow \quad \text { LHS } & =\text { RHS }
\end{aligned}
$
$\therefore$ we are able to write one vector as a linear combination of the other two $\Rightarrow$ the given vectors are coplanar.
(ii) Let the given vectors be $\vec{a}, \vec{b}$ and $\vec{c}$.
$
\vec{a}=2 \hat{i}+3 \hat{j}+\hat{k}, \vec{b}=\hat{i}-\hat{j}, \vec{c}=7 \hat{i}+3 \hat{j}+2 \hat{k}
$
If we are able to write $\vec{a}=m \vec{b}+n \vec{c}$ (where $m$ and $n$ are scalars) then we say that the vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar.

Let $\vec{a}=m \vec{b}+n \vec{c}$
$
2 \hat{i}+3 \hat{j}+\hat{k}=m(\hat{i}-\hat{j})+n(7 \hat{i}+3 \hat{j}+2 \hat{k})
$
Equating $\hat{i}, \hat{j}$ and $\hat{k}$ components
$
\begin{aligned}
2 & =m+7 n \\
3 & =-m+3 n \\
1 & =2 n \\
n & =\frac{1}{2}
\end{aligned}
$
Substituting $n=\frac{1}{2}$ in (1) we get,
$
\begin{aligned}
m+7\left(\frac{1}{2}\right) & =2 \Rightarrow m+\frac{7}{2}=2 \Rightarrow m=2-\frac{7}{2} \\
m & =\frac{4-7}{2} \Rightarrow m=\frac{-3}{2} \\
\therefore m & =\frac{-3}{2}, n=\frac{1}{2}
\end{aligned}
$
Substituting $m$ and $n$ values in (1) we get,
$
\begin{aligned}
& \text { LHS }=2 \\
& \text { RHS }=\frac{-3}{2}+7\left(\frac{1}{2}\right)=\frac{-3}{2}+\frac{7}{2}=\frac{4}{2}=2 \\
& \text { LHS }=\text { RHS }
\end{aligned}
$
We are able to write $\vec{a}$ as a linear combination of $\vec{b}$ and $\vec{c}$
$\therefore$ The vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar

Question 10 .
Show that the points whose position vectors $4 \hat{i}+5 \hat{j}+\hat{k},-\hat{j}-\hat{k}, 3 \hat{i}+9 \hat{j}+4 \hat{k}$ and $-4 \hat{i}+4 \hat{j}+4 \hat{k}$ are coplanar
Solution:
Let the given points be A, B, C and D. To prove that the points A, B, C, D are coplanar, we have to prove that the vectors $\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{AC}}$ and $\overrightarrow{\mathrm{AC}}$ are coplanar

Now $\overrightarrow{\mathrm{OA}}=4 \hat{i}+5 \hat{j}+\hat{k} ; \overrightarrow{\mathrm{OB}}=-\hat{j}-\hat{k} ; \overrightarrow{\mathrm{OC}}=3 \hat{i}+9 \hat{j}+4 \hat{k}$ and $\overrightarrow{\mathrm{OD}}=-4 \hat{i}+4 \hat{j}+4 \hat{k}$
$
\begin{aligned}
& \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=(-\hat{j}-\hat{k})-(4 \hat{i}+5 \hat{j}+\hat{k})=-4 \hat{i}-6 \hat{j}-2 \hat{k} \\
& \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}}=(3 \hat{i}+9 \hat{j}+4 \hat{k})-(4 \hat{i}+5 \hat{j}+\hat{k})=-\hat{i}+4 \hat{j}+3 \hat{k} \\
& \overrightarrow{\mathrm{AD}}=\overrightarrow{\mathrm{OD}}-\overrightarrow{\mathrm{OA}}=(-4 \hat{i}+4 \hat{j}+4 \hat{k})-(4 \hat{i}+5 \hat{j}+\hat{k})=-8 \vec{i}-\vec{j}+3 \vec{k}
\end{aligned}
$
Let $\overrightarrow{\mathrm{AB}}=\vec{a}=-4 \hat{i}-6 \hat{j}-2 \hat{k}$
$
\begin{aligned}
& \overrightarrow{\mathrm{AC}}=\vec{b}=-\hat{i}+4 \hat{j}+3 \hat{k} \\
& \overrightarrow{\mathrm{AD}}=\vec{c}=-8 \hat{i}-\hat{j}+3 \hat{k}
\end{aligned}
$

To prove the vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar we have to prove that $\vec{a}=m \vec{b}+n \vec{c}$
Let $-4 \hat{i}-6 \hat{j}-2 \hat{k}=m(-\hat{i}+4 \hat{j}+3 \hat{k})+n(-8 \hat{i}-\hat{j}+3 \hat{k})$
Equation $\hat{i}, \hat{j}$ and $\hat{k}$ components we get
$
\begin{aligned}
& -4=-m-8 n \\
& \Rightarrow \quad m+8 n=4 \\
& -6=4 m-n \\
& \Rightarrow \quad 4 m-n=-6 \\
& -2=3 m+3 n \\
& \Rightarrow \quad 3 m+3 n=-2 \\
&
\end{aligned}
$
Solving $(i)$ and (ii)

$
m=-\frac{44}{33}=-\frac{4}{3}
$
Substituting $m=-\frac{4}{3}$ in $(i)$ we get,
$
\begin{aligned}
-\frac{4}{3}+8 n & =4 \\
8 n & =4+\frac{4}{3}=\frac{12+4}{3}=\frac{16}{3} \\
n & =\frac{16}{3 \times 8}=\frac{2}{3}
\end{aligned}
$

Substituting $m=-\frac{4}{3}$ and $n=\frac{2}{3}$ in (iii) we get,
$
\begin{aligned}
\text { LHS } & =3\left(\frac{-4}{3}\right)+3\left(\frac{2}{3}\right)=-\frac{12}{3}+\frac{6}{3} \\
& =\frac{-12+6}{3}=\frac{-6}{3}=-2=\text { RHS }
\end{aligned}
$
$\therefore$ we are able to write one vector as a linear combination of the other two vectors $\Rightarrow$ the given vectors $\vec{a}, \vec{b}$, $\vec{c}$ are coplanar.
(i.e.) the given points $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$ are coplanar.

Question 11.
If $\vec{a}=2 \hat{i}+3 \hat{j}-4 \hat{k}, \vec{b}=3 \hat{i}-4 \hat{j}-5 \hat{k}$ and $\vec{c}=-3 \hat{i}+2 \hat{j}+3 \hat{k}$, find the magnitude and direction cosines of
(i) $\vec{a}+\vec{b}+\vec{c}$
(ii) $3 \vec{a}-2 \vec{b}+5 \vec{c}$
Solution:
Given $\vec{a}=2 \hat{i}+3 \hat{j}-4 \hat{k}, \vec{b}=3 \hat{i}-4 \hat{j}-5 \hat{k}$ and $\vec{c}=-3 \hat{i}+2 \hat{j}+3 \hat{k}$
(i)
$
\begin{aligned}
\vec{a}+\vec{b}+\vec{c} & =2 \hat{i}+\hat{j}-6 \hat{k} \\
|\vec{a}+\vec{b}+\vec{c}| & =\sqrt{4+1+36}=\sqrt{41}
\end{aligned}
$
d.c's of $\vec{a}+\vec{b}+\vec{c}=\left(\frac{2}{\sqrt{41}}, \frac{1}{\sqrt{41}}, \frac{-6}{\sqrt{41}}\right)$

(ii)
$
\begin{aligned}
3 \vec{a}-2 \vec{b}+5 \vec{c} & =3(2 \hat{i}+3 \hat{j}-4 \hat{k})-2(3 \hat{i}-4 \hat{j}-5 \hat{k})+5(-3 \hat{i}+2 \hat{j}+3 \hat{k}) \\
& =6 \hat{i}+9 \hat{j}-12 \hat{k}-6 \hat{i}+8 \hat{j}+10 \hat{k}-15 \hat{i}+10 \hat{j}+15 \hat{k} \\
& =-15 \hat{i}+27 \hat{j}+13 \hat{k} \\
& =|3 \vec{a}-2 \vec{b}+5 \vec{c}|=\sqrt{15^2+27^2+13^2} \\
& =\sqrt{225+729+169}=\sqrt{1123} \\
\text { d.c's of } 3 \vec{a}-2 \vec{b}+5 \vec{c} & =\left(\frac{-15}{\sqrt{1123}}, \frac{27}{\sqrt{1123}}, \frac{13}{\sqrt{1123}}\right) .
\end{aligned}
$
Question 12.
The position vectors of the vertices of a triangle are $\hat{i}+2 \hat{j}+3 \hat{k} ; 3 \hat{i}-4 \hat{j}+5 \hat{k}$ and $-2 \hat{i}+3 \hat{j}-7 \hat{k}$ Find the perimeter of the triangle
Solution:

Let A, B, C be the vertices of the triangle $\mathrm{ABC}$,

Now,
$
\begin{aligned}
\overrightarrow{\mathrm{AB}} & =\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=(3 \hat{i}-4 \hat{j}+5 \hat{k})-(\hat{i}+2 \hat{j}+3 \hat{k}) \\
& =3 \hat{i}-4 \hat{j}+5 \hat{k}-\hat{i}-2 \hat{j}-3 \hat{k} \\
& =2 \hat{i}-6 \hat{j}+2 \hat{k} \\
|\overrightarrow{\mathrm{AB}}| & =\sqrt{4+36+4}=\sqrt{44}=\mathrm{AB} \\
\mathrm{BC} & =\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}}=(-2 \hat{i}+3 \hat{j}-7 \hat{k})-(3 \hat{i}-4 \hat{j}+5 \hat{k}) \\
& =-2 \hat{i}+3 \hat{j}-7 \hat{k}-3 \hat{i}+4 \hat{j}-5 \hat{k} \\
& =-5 \hat{i}+7 \hat{j}-12 \hat{k} \\
|\overrightarrow{\mathrm{BC}}| & =\sqrt{25+49+144}=\sqrt{218}=\mathrm{BC} \\
\overrightarrow{\mathrm{AC}} & =\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}}=(-2 \hat{i}+3 \hat{j}-7 \hat{k})-(\hat{i}+2 \hat{j}+3 \hat{k}) \\
& =-2 \hat{i}+3 \hat{j}-7 \hat{k}-\hat{i}-2 \hat{j}-3 \hat{k} \\
& =-3 \hat{i}+\hat{j}-10 \hat{k} \\
|\overrightarrow{\mathrm{AC}}| & =\sqrt{9+1+100}=\sqrt{110}=\mathrm{AC} \\
\text { Perimeter of } \Delta \mathrm{ABC} & =\mathrm{AB}+\mathrm{BC}+\mathrm{AC} \\
& =\sqrt{44}+\sqrt{218}+\sqrt{110} .
\end{aligned}
$
Question 13.
Find the unit vector parallel to $3 \vec{a}-2 \vec{b}+4 \vec{c}$ if $\vec{a}=3 \hat{i}-\hat{j}-4 \hat{k}, \vec{b}=-2 \hat{i}+4 \hat{j}-3 \hat{k}$, and $\vec{c}=\hat{i}+2 \hat{j}-\hat{\boldsymbol{k}}$
Solution

Given $\vec{a}=3 \hat{i}-\hat{j}-4 \hat{k}, \vec{b}=-2 \hat{i}+4 \hat{j}-3 \hat{k}$ and $\vec{c}=\hat{i}+2 \hat{j}-\hat{k}$
Now
$
\begin{aligned}
3 \vec{a}-2 \vec{b}+4 \vec{c} & =3(3 \hat{i}-\hat{j}-4 \hat{k})-2(-2 \hat{i}+4 \hat{j}-3 \hat{k})+4(\hat{i}+2 \hat{j}-\hat{k}) \\
& =9 \hat{i}-3 \hat{j}-12 \hat{k}+4 \hat{i}-8 \hat{j}+6 \hat{k}+4 \hat{i}+8 \hat{j}-4 \hat{k} \\
& =17 \hat{i}-3 \hat{j}+10 \hat{k}
\end{aligned}
$
Let
$
\begin{aligned}
3 \vec{a}-2 \vec{b}+4 \vec{c} \text { be } & =\vec{r} \\
|\vec{r}| & =|3 \vec{a}-2 \vec{b}+4 \vec{c}|=\sqrt{17^2+3^2+10^2}=\sqrt{289+9+100} \\
& =\sqrt{398}
\end{aligned}
$
The unit vector in the direction of $|\vec{r}|$ is $\pm \frac{\vec{r}}{|\vec{r}|}$ i.e., $\pm \frac{17 \hat{i}-3 \hat{j}-10 \hat{k}}{\sqrt{398}}$
Question 14.
The position vector $\vec{a}, \vec{b}, \vec{c}$ three points satisfy the relation $2 \vec{a}-7 \vec{b}+5 \vec{c}=\overrightarrow{0}$. Are these points collinear?

Solution:
Given $\quad \overrightarrow{\mathrm{OA}}=\vec{a} ; \overrightarrow{\mathrm{OB}}=\vec{b} ; \overrightarrow{\mathrm{OC}}=\vec{c}$
Now we are given that $2 \vec{a}-7 \vec{b}+5 \vec{c}=\overrightarrow{0}$
$
\begin{aligned}
\Rightarrow & 2 \vec{a}-2 \vec{b}-5 \vec{b}+5 \vec{c} & =0 \\
\text { (i.e.,) } & 2(\vec{a}-\vec{b})-5(\vec{b}-\vec{c}) & =\overrightarrow{0} \\
\Rightarrow & 2 \overrightarrow{\mathrm{BA}}-5 \overrightarrow{\mathrm{CB}} & =0 \\
\Rightarrow & 2 \overrightarrow{\mathrm{BA}} & =5 \overrightarrow{\mathrm{CB}}
\end{aligned}
$
$\Rightarrow \overrightarrow{\mathrm{BA}}$ and $\overrightarrow{\mathrm{CB}}$ are parallel.
But $\mathrm{B}$ is a common point $\Rightarrow$ The point $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are collinear.
Question 15.
The position vectors of the point $\mathrm{P}, \mathrm{Q}, \mathrm{R}, \mathrm{S}$ are $\hat{i}+\hat{j}+\hat{k}, 2 \hat{i}+5 \hat{j}, 3 \hat{i}+2 \hat{j}-3 \hat{k}$ and $\hat{i}-6 \hat{j}-\hat{k}$ respectively. Prove that the line $\mathrm{PQ}$ and RS are parallel.

Solution:
Given
$
\begin{aligned}
\overrightarrow{\mathrm{OP}} & =\hat{i}+\hat{j}+\hat{k} \\
\overrightarrow{\mathrm{OQ}} & =2 \hat{i}+5 \hat{j} \\
\overrightarrow{\mathrm{OR}} & =3 \hat{i}+2 \hat{j}-3 \hat{k}
\end{aligned}
$
and
$
\overrightarrow{\mathrm{OS}}=\hat{i}-6 \hat{j}-\hat{k}
$
Now $\begin{aligned} \overrightarrow{\mathrm{PQ}}=\overrightarrow{\mathrm{OQ}}-\overrightarrow{\mathrm{OP}} & =(2 \hat{i}+5 \hat{j})-(\hat{i}+\hat{j}+\hat{k}) \\ & =2 \hat{i}+5 \hat{j}-\hat{i}-\hat{j}-\hat{k}=\hat{i}+4 \hat{j}-\hat{k} \\ \overrightarrow{\mathrm{RS}} & =\overrightarrow{\mathrm{OS}}-\overrightarrow{\mathrm{OR}}=(\hat{i}-6 \hat{j}-\hat{k})-(3 \hat{i}+2 \hat{j}-3 \hat{k}) \\ & =\hat{i}-6 \hat{j}-\hat{k}-3 \hat{i}-2 \hat{j}+3 \hat{k} \\ & =-2 \hat{i}-8 \hat{j}+2 \hat{k} \\ & =-2(\hat{i}+4 \hat{j}-\hat{k}) \\ \overrightarrow{\mathrm{RS}} & =-2 \overrightarrow{\mathrm{PQ}} \\ \Rightarrow \quad \overrightarrow{\mathrm{PQ}} \text { and } \overrightarrow{\mathrm{RS}} \text { are parallel. } & \end{aligned}$
Question 16.
Find the value or values of $m$ for which $m(\hat{i}+\hat{j}+\hat{k})$ is a unit vector Solution:
Given $m(\hat{i}+\hat{j}+\hat{k})$ is a unit vector
$
\begin{array}{llll}
\Rightarrow & & |m(\hat{i}+\hat{j}+\hat{k})| & =1 \\
\Rightarrow & \pm m(\sqrt{1+1+1}) & =1 \Rightarrow \pm m \sqrt{3}=1 \\
\Rightarrow & m & = \pm \frac{1}{\sqrt{3}}
\end{array}
$
Question 17.
Show that the points $\mathrm{A}(1,1,1), \mathrm{B}(1,2,3)$ and $\mathrm{C}(2,-1,1)$ are vertices of an isosceles triangle.

Solution:
Given $\overrightarrow{\mathrm{OA}}=\hat{i}+\hat{j}+\hat{k}, \overrightarrow{\mathrm{OB}}=\hat{i}+2 \hat{j}+3 \hat{k}$ and $\overrightarrow{\mathrm{OC}}=2 \hat{i}-\hat{j}+\hat{k}$
Now
$
\begin{aligned}
\overrightarrow{\mathrm{AB}} & =\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}} \\
& =(\hat{i}+2 \hat{j}+3 \hat{k})-(\hat{i}+\hat{j}+\hat{k}) \\
& =\hat{i}+2 \hat{j}+3 \hat{k}-\hat{i}-\hat{j}-\hat{k}=\hat{j}+2 \hat{k} \\
|\overrightarrow{\mathrm{AB}}| & =\sqrt{1+4}=\sqrt{5} \\
\overrightarrow{\mathrm{BC}} & =\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OB}}=(2 \hat{i}-\hat{j}+\hat{k})-(\hat{i}+2 \hat{j}+3 \hat{k}) \\
& =2 \hat{i}-\hat{j}+\hat{k}-\hat{i}-2 \hat{j}-3 \hat{k} \\
& =\hat{i}-3 \hat{j}-2 \hat{k} \\
|\overrightarrow{\mathrm{BC}}| & =\sqrt{1+9+4}=\sqrt{14} \\
|\overrightarrow{\mathrm{AC}}| & =\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}}=(2 \hat{i}-\hat{j}+\hat{k})-(\hat{i}+\hat{j}+\hat{k}) \\
& =\hat{i}-2 \hat{j} \\
|\overrightarrow{\mathrm{AC}}| & =\sqrt{1+4}=\sqrt{5} \\
|\overrightarrow{\mathrm{AB}}| & =|\overrightarrow{\mathrm{AC}}|=\sqrt{5}
\end{aligned}
$
$
\text { Here } \quad|\overrightarrow{\mathrm{AB}}|=|\overrightarrow{\mathrm{AC}}|=\sqrt{5}
$
$\Rightarrow \mathrm{ABC}$ is an isosceles triangle.