Exercise 8.5 - Chapter 8 - Vector Algebra – I - 11th Maths Guide Samacheer Kalvi Solutions

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Ex 8.5
Choose the correct or the most suitable answer from the given four alternatives:
Question 1.
The value of $\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{DA}}+\overrightarrow{\mathrm{CD}}$ is
(a) $\overline{\mathbf{A D}}$
(b) $\overrightarrow{\mathbf{C A}}$
(c) $\overrightarrow{0}$
(d) $-\overrightarrow{\mathrm{AD}}$
Solution:
(c) $\overrightarrow{0}$
(c) $\overrightarrow{0}$
$
\begin{aligned}
\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}} & =\overrightarrow{\mathrm{AC}} \\
\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CD}} & =\overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{CD}}=\overrightarrow{\mathrm{AD}} \\
\therefore \quad \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CD}}+\overrightarrow{\mathrm{DA}} & =\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{DA}} \\
& =\overrightarrow{\mathrm{AD}}-\overrightarrow{\mathrm{AD}}=0
\end{aligned}
$
Question 2.
If $\vec{a}+2 \vec{b}$ and $3 \vec{a}+m \vec{b}$ are parallel, then the value of $m$ is
(a) 3
(b) $\frac{1}{3}$
(c) 6
(d) $\frac{1}{6}$
Solution:
(c) 6
$
\begin{aligned}
\vec{a}+2 \vec{b} & =t(3 \vec{a}+m \vec{b}) \\
1 & =3 t \Rightarrow t=1 / 3 \\
2 & =t_m \Rightarrow 2=\frac{1}{3} m \\
\therefore m & =6
\end{aligned}
$
Question 3.
The unit vector parallel to the resultant of the vectors $\hat{i}+\hat{j}-\hat{k}$ and $\hat{i}-2 \hat{j}+\hat{k}$ is
(a) $\frac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{5}}$
(b) $\frac{2 \hat{i}+\hat{j}}{\sqrt{5}}$
(c) $\frac{2 \hat{i}-\hat{j}+\hat{k}}{\sqrt{5}}$
(d) $\frac{2 \hat{i}-\hat{j}}{\sqrt{5}}$

Solution:
(d) $\frac{2 \hat{i}-\hat{j}}{\sqrt{5}}$
$\vec{a}=\hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}-2 \hat{j}+\hat{k}$
$
\begin{aligned}
\vec{a}+\vec{b} & =\hat{i}+\hat{j}-\hat{k}+\hat{i}-2 \hat{j}+\hat{k}=2 \hat{i}-\hat{j} \\
|\vec{a}+\vec{b}| & =\sqrt{4+1}=\sqrt{5}
\end{aligned}
$
The unit vector in the direction of $\vec{a}+\vec{b}$ is $\frac{2 \hat{i}-\hat{j}}{\sqrt{5}}$.
Question 4.
A vector $\overrightarrow{O P}$ makes $60^{\circ}$ and $45^{\circ}$ with the positive direction of the $\mathrm{x}$ and y axes respectively. Then the angle between $\overrightarrow{O P}$ and the z-axis is
(a) $45^{\circ}$
(b) $60^{\circ}$
(c) $90^{\circ}$
(d) $30^{\circ}$
Solution:
(b) $60^{\circ}$
$\alpha=60^{\circ}, \beta=45^{\circ}$
We know $\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1$
$
\begin{aligned}
& \text { (i.e.,) }\left(\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{2}}\right)^2+\cos ^2 \gamma=1 \\
& \cos ^2 \gamma=1-\frac{1}{4}-\frac{1}{2}=\frac{1}{4} \\
& \cos \gamma=\frac{1}{2} \Rightarrow y=\pi / 3=60^{\circ}
\end{aligned}
$

Question 5.
If $\overrightarrow{B A}=3 \hat{i}+2 \hat{j}+\hat{k}$ and the position vector of $\mathrm{B}$ is $\hat{i}+3 \hat{j}-\hat{k}$, then the position vector $\mathrm{A}$ is
(a) $4 \hat{i}+2 \hat{j}+\hat{k}$
(b) $4 \hat{i}+5 \hat{j}$
(c) $4 \hat{i}$
$(d)-4 \hat{i}$
Solution:
(b) $4 \hat{i}+5 \hat{j}$
$
\begin{aligned}
\overrightarrow{\mathrm{BA}} & =3 \hat{i}+2 \hat{j}+\hat{k} \\
\overrightarrow{\mathrm{OB}} & =\hat{i}+3 \hat{j}-\hat{k} \\
\text { Now } \overrightarrow{\mathrm{OA}} & =\overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{BA}}=\hat{i}+3 \hat{j}-\hat{k}+3 \hat{i}+2 \hat{j}+\hat{k} \\
& =4 \hat{i}+5 \hat{j}
\end{aligned}
$

Question 6.
A vector makes equal angle with the positive direction of the coordinate axes. Then each angle is equal to
(a) $\cos ^{-1}\left(\frac{1}{3}\right)$
(b) $\cos ^{-1}\left(\frac{2}{3}\right)$
(c) $\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
(d) $\cos ^{-1}\left(\frac{2}{\sqrt{3}}\right)$
Solution:
(c) $\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
Given $\alpha=\beta=\gamma$
We know $\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1$
$
\begin{aligned}
\Rightarrow \quad 3 \cos ^2 \alpha & =1 \Rightarrow \cos ^2 \alpha=\frac{1}{3} \\
\cos \alpha & =\frac{1}{\sqrt{3}} \\
\Rightarrow \quad \alpha & =\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)
\end{aligned}
$

Question 7.
The vectors $\vec{a}-\vec{b}, \vec{b}-\vec{c}, \vec{c}-\vec{a}$ are
(a) parallel to each other
(b) unit vectors
(c) mutually perpendicular vectors
(d) coplanar vectors
Solution:
(d) coplanar vectors
Question 8.
If $\mathrm{ABCD}$ is a parallelogram, then $\overrightarrow{A B}+\overrightarrow{A D}+\overrightarrow{C B}+\overrightarrow{C D}$ is equal to
(a) $\mathbf{2}(\overrightarrow{\mathbf{A B}}+\overrightarrow{\mathbf{A D}})$
(b) $4 \overrightarrow{\mathrm{AC}}$
(c) $4 \overrightarrow{\mathrm{BD}}$
(d) $\overrightarrow{0}$
Solution:
(d) $\overrightarrow{0}$

$
\begin{aligned}
\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{CB}}+\overrightarrow{\mathrm{CD}} & =(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{CD}})+(\overrightarrow{\mathrm{AD}}+\overrightarrow{\mathrm{CB}}) \\
& =\overrightarrow{0}+\overrightarrow{0}=\overrightarrow{0}
\end{aligned}
$

Question 9.
One of the diagonals of parallelogram $\mathrm{ABCD}$ with $\vec{a}$ and $\vec{b}$ as adjacent sides is $\vec{a}+\vec{b}$. The other diagonal $\overrightarrow{\mathrm{BD}}$ is
(a) $\vec{a}-\vec{b}$
(b) $\vec{b}-\vec{a}$
(c) $\vec{a}+\vec{b}$
(d) $\frac{\vec{a}+\vec{b}}{2}$

Solution:
(b) $\vec{b}-\vec{a}$
Given $\overrightarrow{\mathrm{AC}}=\vec{a}+\vec{b}$

$
\begin{aligned}
\overrightarrow{\mathrm{BD}} & =\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CD}} \\
& =\vec{b}-\vec{a}
\end{aligned}
$

Question 10.
If $\vec{a}, \vec{b}$ are the position vectors $A$ and $B$, then which one of the following points whose position vector lies on $\mathrm{AB}$, is ...........
(a) $\vec{a}+\vec{b}$
(b) $\frac{2 \vec{a}-\vec{b}}{2}$
(c) $\frac{2 \vec{a}+\vec{b}}{3}$
(d) $\frac{\vec{a}-\vec{b}}{3}$
Solution:
(c) $\frac{2 \vec{a}+\vec{b}}{3}$
Question 11.
If $\vec{a}, \vec{b}, \vec{c}$ are the position vectors of three collinear points, then which of the following is true?
(a) $\vec{a}=\vec{b}+\vec{c}$
(b) $2 \vec{a}=\vec{b}+\vec{c}$
(c) $\vec{b}=\vec{c}+\vec{a}$
(d) $4 \vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$
Solution:
(b) $2 \vec{a}=\vec{b}+\vec{c}$
Question 12.
If $\vec{r}=\frac{9 \vec{a}+7 \vec{b}}{16}$, then the point $\mathrm{P}$ whose position vector $\vec{r}$ divides the line joining the points with position vectors $\vec{a}$ and $\vec{b}$ in the ratio
(a) $7: 9$ internally
(b) $9: 7$ internally
(c) 9:7 externally
(d) $7: 9$ externally
Solution:

(a) $7: 9$ internally
$\overrightarrow{\mathrm{OP}}=\vec{r}=\frac{9 \vec{a}+7 \vec{b}}{9+7} \Rightarrow \vec{r}$ divides
$\mathrm{AB}$ is the ratio $7: 9$ internally.
Question 13.
If $\lambda \hat{i}+2 \lambda \hat{j}+2 \lambda \hat{k}$ is a unit vector, then the value of $\lambda$ is
(a) $\frac{1}{3}$
(b) $\frac{1}{4}$
(c) $\frac{1}{9}$

(d) $\frac{1}{2}$
Solution:
(a) $\frac{1}{3}$
Given $\lambda \hat{i}+2 \lambda \hat{j}+2 \lambda \hat{k}$ is a unit vector.
$
\begin{aligned}
& \Rightarrow|\lambda \hat{i}+2 \lambda \hat{j}+2 \lambda \hat{k}|=1 \Rightarrow\left|\lambda^2+4 \lambda^2+4 \lambda^2\right|=1 \\
& \text { (i.e., } 9 \lambda^2=1 \Rightarrow \lambda^2=1 / 9 \Rightarrow \lambda=\frac{1}{3}
\end{aligned}
$
Question 14.
Two vertices of a triangle have position vectors $3 \hat{i}+4 \hat{j}-4 \hat{k}$ and $2 \hat{i}+3 \hat{j}+4 \hat{k}$.If the position vector of the centroid is $\hat{i}+2 \hat{j}+3 \hat{k}$, then the position vector of the third vertex is
(a) $-2 \hat{i}-\hat{j}+9 \hat{k}$
(b) $-2 \hat{i}-\hat{j}-6 \hat{k}$
(c) $2 \hat{i}-\hat{j}+6 \hat{k}$
(d) $-2 \hat{i}+\hat{j}+6 \hat{k}$
Solution:
(a) $-2 \hat{i}-\hat{j}+9 \hat{k}$
Let $\overrightarrow{\mathrm{OA}}=3 \hat{i}+4 \hat{j}-4 \hat{k}$ and $\overrightarrow{\mathrm{OB}}=2 \hat{i}+3 \hat{j}+4 \hat{k}$
$
\begin{aligned}
& \text { Centroid of } \triangle \mathrm{ABC}=\frac{\overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{OC}}}{3}=\hat{i}+2 \hat{j}+3 \hat{k} \\
& \Rightarrow \quad \overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{OC}}=3(\hat{i}+2 \hat{j}+3 \hat{k})=3 \hat{i}+6 \hat{j}+9 \hat{k} \\
& \text { (i.e., } 3 \hat{i}+4 \hat{j}-4 \hat{k}+2 \hat{i}+3 \hat{j}+4 \hat{k}+\overrightarrow{\mathrm{OC}}=3 \hat{i}+6 \hat{j}+9 \hat{k} \\
& \Rightarrow \quad 5 \hat{i}+7 j+\overrightarrow{\mathrm{OC}}=3 \hat{i}+6 \hat{j}+9 \hat{k} \\
& \quad \overrightarrow{\mathrm{OC}}=-2 \hat{i}-\hat{j}+9 \hat{k} \\
& \text { (i.e.,) Third vertex }=-2 \hat{i}-\hat{j}+9 \hat{k}
\end{aligned}
$
Question 15 .
If $|\vec{a}+\vec{b}|=60,|\vec{a}-\vec{b}|=40$ and $|\vec{b}|=46$, then $|\vec{a}|$ is
(a) 42
(b) 12
(c) 22
(d) 32
Solution:

(c) 22
$
|\vec{a}+\vec{b}|=60 ;|\vec{a}-\vec{b}|=40 ;|\vec{b}|=46
$
We know $(\vec{a}+\vec{b})^2+(\vec{a}-\vec{b})^2=2\left[|\vec{a}|^2+|\vec{b}|^2\right]$
$
\begin{aligned}
& \text { (i.e.,) } 60^2+40^2=2\left[|\vec{a}|^2+46^2\right] \\
& \Rightarrow \quad 3600+1600=2\left[|\vec{a}|^2+2116\right] \\
& \Rightarrow \quad 2\left(|\vec{a}|^2+2116\right)=5200 \\
& \Rightarrow \quad|\vec{a}|^2+2116=\frac{520 G}{2}=2600 \\
& |\vec{a}|^2=2600-2116=484 \\
& \text { (i.e.,) } \\
& |\vec{a}|^2=22^2 \\
& \Rightarrow \quad|\vec{a}|=22 \\
&
\end{aligned}
$
Question 16.
If $\vec{a}$ and $\vec{b}$ having same magnitude and angle between them is $60^{\circ}$ and their scalar product is $\frac{1}{2}$ then $|\vec{a}|$ is
(a) 2
(b) 3
(c) 7
(d) 1
Solution:
(d) 1
Given
$
|\vec{a}|=|\vec{b}|
$
$
\begin{aligned}
\vec{a} \cdot \vec{b} & =|\vec{a}||\vec{b}| \cos 60^{\circ}=\frac{1}{2} \\
\Rightarrow \quad \vec{a}^2\left(\frac{1}{2}\right) & =\frac{1}{2} \\
\vec{a}^2 & =\frac{1}{2} \times \frac{2}{1}=1 \\
\vec{a}^2 & =1 \Rightarrow|\vec{a}|=1
\end{aligned}
$

Question 17.
The value of $\theta \in\left(0, \frac{\pi}{2}\right)$ for which the vectors $\vec{a}=(\sin \theta) \hat{i}+(\cos \theta) \hat{j}$ and $\vec{b}=\hat{i}-\sqrt{3} \hat{j}+2 \hat{k}$ are perpendicular, is equal to
(a) $\frac{\pi}{3}$
(b) $\frac{\pi}{6}$
(c) $\frac{\pi}{4}$
(d) $\frac{\pi}{2}$
Solution:
(a) $\frac{\pi}{3}$
$
\vec{a}=(\sin \theta) \hat{i}+(\cos \theta) \hat{j}
$
$
\begin{aligned}
\vec{b} & =\hat{i}-\sqrt{3} \hat{j}+2 \hat{k} \\
\vec{a} \perp^r \text { to } \vec{b} & \Rightarrow \vec{a} \cdot \vec{b}=0
\end{aligned}
$
$
\text { (i.e., } \begin{aligned}
\sin \theta-\sqrt{3} \cos \theta & =0 \\
\sin \theta & =\sqrt{3} \cos \theta \\
\Rightarrow \quad \frac{\sin \theta}{\cos \theta}=\sqrt{3} \text { (i. e.,) } \tan \theta & =\sqrt{3} \Rightarrow \theta=\frac{\pi}{3}
\end{aligned}
$

Question 18.
If $|\vec{a}|=13,|\vec{b}|=5$ and $\vec{a} \cdot \vec{b}=60^{\circ}$ then $|\vec{a} \times \vec{b}|$ is
(a) 15
(b) 35
(c) 45
(d) 25
Solution:
(d) 25
We know $(\vec{a} \cdot \vec{b})^2+(\vec{a} \times \vec{b})^2=|\vec{a}|^2|\vec{b}|^2$
$
\begin{aligned}
60^2+(\vec{a} \times \vec{b})^2 & =(13)^2(5)^2=4225 \\
& =(65)^2 \\
(\vec{a} \times \vec{b})^2 & =4225-3600 \\
& =625 \\
|\vec{a} \times \vec{b}| & =\sqrt{625}=25
\end{aligned}
$
Question 19.
Vectors $\vec{a}$ and $\vec{b}$ are inclined at angle $\theta=120^{\circ}$
If $|\vec{a}|=1,|b|=2$, then $[(\vec{a}+3 \vec{b}) \times(3 \vec{a}-\vec{b})]^2$ is equal to
(a) 225
(b) 275
(c) 325
(d) 300
Solution:
(d) 300

$
\begin{aligned}
& \text { Now }(\vec{a}+3 \vec{b}) \times(3 \vec{a}-\vec{b})=3(\vec{a} \times a)-(\vec{a} \times \vec{b})+9(\vec{b} \times \vec{a})-3(\vec{b} \times \vec{b}) \\
& =-(\vec{a} \times \vec{b})-9(\vec{a} \times \vec{b}) \quad[\because \vec{a} \times \vec{a}=0 \& \vec{b} \times \vec{b}=0] \\
& =-10(\vec{a} \times \vec{b}) \\
& \therefore \quad[(\vec{a}+3 \vec{b}) \times(3 \vec{a}-\vec{b})]^2=100(\vec{a} \times \vec{b})^2 \\
& \text { Now }|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \frac{\pi}{3}=\text { (1) (2) } \frac{\sqrt{3}}{2}=\sqrt{3} \\
& \therefore \quad 100(\vec{a} \times \vec{b})^2=100(\sqrt{3})^2=300 \\
&
\end{aligned}
$
Question 20 .
If $\vec{a}$ and $\vec{b}$ are two vectors of magnitude 2 and inclined at an angle $60^{\circ}$, then the angle between $\vec{a}$ and $\vec{a}+\vec{b}$ is ..............
(a) $30^{\circ}$
(b) $60^{\circ}$
(c) $45^{\circ}$
(d) $90^{\circ}$
Solution:
(a) $30^{\circ}$
The angle between $\vec{a}$ and $\vec{a}+\vec{b}$ is given by $\cos \theta=\frac{\vec{a} \cdot(\vec{a}+\vec{b})}{|\vec{a}||\vec{a}+\vec{a}|}$
$
\text { Now } \begin{aligned}
\vec{a} \cdot(\vec{a}+\vec{b}) & =\vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b} \\
\vec{a} \cdot \vec{a} & =|\vec{a}|^2=2^2=4 \\
\vec{a} \cdot \vec{b} & =|\vec{a}||\vec{b}| \cos 60^{\circ} \\
& =(2)(2) \frac{1}{2}=2 \\
|\vec{a}+\vec{b}|^2 & =\vec{a}^2+\vec{b}^2+2 \vec{a} \cdot \vec{b} \\
& =2^2+2^2+2(2)=4+4+4=12 \\
\cos \theta & =\frac{4+2}{(2) \sqrt{12}}=\frac{6}{2 \times 2 \sqrt{3}}=\frac{6}{4 \sqrt{3}} \\
\Rightarrow \quad \theta \quad & =\frac{3}{2 \sqrt{3}}=\frac{\sqrt{3}}{2} \\
\theta & =30^{\circ}
\end{aligned}
$
Now

Question 21.
If the projection of $5 \hat{i}-\hat{j}-3 \hat{k}$ on the vector $\hat{i}+3 \hat{j}+\lambda \hat{k}$ is same as the projection of $\hat{i}+3 \hat{j}+\lambda \hat{k}$ on $5 \hat{i}-\hat{j}-3 \hat{k}$ then $\lambda$ is equal to
(a) \pm 4
(b) \pm 3
(c) \pm 5
(d) \pm 1
Solution:
(c) \pm 5
Let $\vec{a}=5 \hat{i}-\hat{j}-3 \hat{k}$ and $\vec{b}=\hat{i}+3 \hat{j}+\lambda \hat{k}$ projection of $\vec{a}$ on $\vec{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$
$
\begin{aligned}
& =\frac{5-3-3 \lambda}{\sqrt{1+9+\lambda^2}}=\frac{2-3 \lambda}{\sqrt{10+\lambda^2}} \\
& \text { Projection of } \vec{b} \text { on } \vec{a}=\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|} \\
& =\frac{2-3 \lambda}{\sqrt{25+1+9}}=\frac{2-3 \lambda}{\sqrt{35}} \\
&
\end{aligned}
$

Given $(i)=($ ii $) \Rightarrow \frac{2-3}{\sqrt{10+\lambda^2}}=\frac{2-3 \lambda}{\sqrt{35}}$
Equating Dr, on either sides we get
$
\begin{aligned}
\Rightarrow \sqrt{10+\lambda^2} & =\sqrt{35} \\
10+\lambda^2 & =35 \\
\lambda^2 & =35-10=25 \\
\Rightarrow \quad \lambda & = \pm 5
\end{aligned}
$
Question 22.
If $(1,2,4)$ and $(2,-3 \lambda,-3)$ are the initial and terminal points of the vector $\hat{i}+5 \hat{j}-7 \hat{k}$, then the value of $\lambda$ is equal to
(a) $\frac{7}{3}$
(b) $-\frac{7}{3}$
(c) $-\frac{5}{3}$
(d) $\frac{5}{3}$

Solution:
(b) $-\frac{7}{3}$
Given $\overrightarrow{\mathrm{OA}}=\hat{i}+2 \hat{j}+4 \hat{k} ; \overrightarrow{\mathrm{OB}}=2 \hat{i}-3 \lambda \hat{j}-3 \hat{k}$ and $\overrightarrow{\mathrm{AB}}=\hat{i}+5 \hat{j}-7 \hat{k}$
$
\begin{array}{lc}
\Rightarrow & \overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=\hat{i}+5 \hat{j}-7 \hat{k} \\
(\text { i.e.,) }(2 \hat{i}-3 \lambda \hat{j}-3 \hat{k})-(\hat{i}+2 \hat{j}+4 \hat{k})=\hat{i}+5 \hat{j}-7 \hat{k}
\end{array}
$
equating the ' $j$ components we get'
$
\begin{aligned}
-3 \lambda-2 & =5 \\
-3 \lambda & =5+2=7 \\
\lambda & =-\frac{7}{3}
\end{aligned}
$
Question 23.
If the points whose position vector $10 \hat{i}+3 \hat{j}, 12 \hat{i}-5 \hat{j}$ and $\vec{a} \hat{i}+11 \hat{j}$ are collinear then a is equal to
(a) 6
(b) 2
(c) 5
(d) 8
Solution:
$\stackrel{\text { (d) } 8}{\mathrm{OA}}=10 \hat{i}+3 \hat{j}, \overrightarrow{\mathrm{OB}}=12 \hat{i}-5 \hat{j}, \overrightarrow{\mathrm{OC}}=a \hat{i}+11 \hat{j}$
Given that the point A, B, C are collinear
$
\begin{aligned}
& \Rightarrow \quad \overrightarrow{\mathrm{AB}}=t \overrightarrow{\mathrm{AC}} \\
& \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}} \\
& =(12 \hat{i}-5 \hat{j})-(10 \hat{i}+3 \hat{j}) \\
& =2 \hat{i}-8 \hat{j} \\
& \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{OC}}-\overrightarrow{\mathrm{OA}}=(\vec{a} \hat{i}+11 \hat{j})-(10 \hat{i}+3 \hat{j}) \\
& =(a-10) \hat{i}+8 \hat{j} \\
& \text { Given } \\
& \overrightarrow{\mathrm{AB}}=t \overrightarrow{\mathrm{AC}} \\
& \Rightarrow \quad 2 \hat{i}-8 \hat{j}=t\{(a-10) \hat{i}+8 \hat{j}\} \\
& \text { equating } \hat{j} \text { components } \\
& \Rightarrow-8=8 \mathrm{t} \Rightarrow \mathrm{t}=-1 \\
& \text { equation } \hat{i} \text { components } \\
& t(a-10)=2 \\
&
\end{aligned}
$

$
\begin{aligned}
& \text { (i.e.) }(-1)(a-10)=2 \\
& a-10=-2 \\
& a=-2+10=-8
\end{aligned}
$
Question 24.
If $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \vec{i}+\vec{x} \hat{j}+\hat{k}, \vec{c}=\hat{i}-\hat{j}+4 \hat{k}$ and $\vec{a} \cdot(\vec{b} \times \vec{c})=70$, then $\mathrm{x}$ is equal to
(a) 5
(b) 7
(c) 26
(d) 10
Solution:
(c) 26
$
\begin{aligned}
\vec{b} \times \vec{c} & =\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & x & 1 \\
1 & -1 & 4
\end{array}\right| \\
& =\hat{i}(4 x+1)-\hat{j}(8-1)+\hat{k}(-2-x) \\
\text { Given } \vec{a} \cdot(\vec{b} \times \vec{c}) & =70 \\
\Rightarrow \quad(\hat{i}+\hat{j}+\hat{k}) \cdot[(4 x+1) \hat{i}-7 \hat{j} & -\hat{k}(2+x)]=70 \\
\Rightarrow \quad(4 x+1)(1)-7(1)-(2+x) & (1)=70 \\
4 x+1-7-2-x & =70 \\
3 x-8 & =70 \\
3 x & =70+8=78 \\
x & =78 / 3=26
\end{aligned}
$

Question 25 .
If $\vec{a}=\hat{i}+2 \hat{j}+2 \hat{k},|\vec{b}|=5$ and the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{6}$, then the area of the triangle formed by these two vectors as two sides, is .............
(a) $\frac{7}{4}$
(b) $\frac{15}{4}$
(c) $\frac{3}{4}$
(d) $\frac{17}{4}$
Solution:
(b) $\frac{15}{4}$
Area of $\Delta$ formed by $\vec{a}$ and $\vec{b}$ is $\frac{1}{2}|\vec{a} \times \vec{b}|$.
Now,
$
\begin{aligned}
&|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \frac{\pi}{6} \\
&|\vec{a}|=\sqrt{1+4+4}=\sqrt{9}=3 ;|\vec{b}|=5 \\
&
\end{aligned}
$
$
\therefore \quad \frac{1}{2}|\vec{a} \times \vec{b}|=\left(\frac{1}{2}\right)(3)(5)=\frac{15}{2}
$
$\therefore \quad$ Area of $\Delta=\frac{1}{2}\left(\frac{15}{2}\right)=\frac{15}{4}$ sq. units