Exercise 8.2 - Chapter 8 - Vector Algebra – I - 11th Maths Guide Samacheer Kalvi Solutions

You can Download the Exercise 8.2 - 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

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.