RD Sharma Solutions Class 12 Mathematics Chapter 22 VSA

# RD Sharma Solutions Class 12 Mathematics Chapter 22 VSA

Edited By Satyajeet Kumar | Updated on Jan 24, 2022 06:21 PM IST

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## Algebra of Vectors Excercise: VSA

Algebra of Vectors Exercise Very Short Answer Question 1

Given: Define Zero vector
Explanation:
A vector whose initial and final point coincident or the length of vector is zero known as zero vector or null vector. The null vector is denoted by $\bar{O}$. Also, the magnitude of zero vector is same.

Algebra of Vectors Exercise Very Short Answer Question 2

Given: Define Unit vector
Explanation:
A unit vector is a vector that has a magnitude of 1. The unit vector in the direction of a vector $\vec{a}$is denoted by $\hat{a}$
$\therefore \mid \hat{a}\mid =1$

Algebra of Vectors Exercise Very Short Answer Question 3

Given: Define position vector of a point
Explanation:
$\Rightarrow$The position vector is said to be a straight line having one end fixed to a body like origin and the second end that is attached to a moving point and this is used to describe the position of the point relative to the body.
$\Rightarrow$A point O is fixed as origin in space (or plane) and P is any point, then
$\bar{OP}$ is called a position vector of P w.r.t O

Algebra of Vectors Exercise Very Short Answer Question 4

Hint: You must know the rules of vector functions.
Given: Write $\overrightarrow{P Q}+\overrightarrow{R P}+\overrightarrow{Q R}$ in simplest form
Solution: We have,
$\begin{array}{ll} \overrightarrow{P Q}+\overrightarrow{R P}+\overline{Q R} \\ \end{array}$
$\begin{array}{ll} \Rightarrow \overline{P Q}+\overline{Q R}+\overline{R P} \end{array}$ $\begin{array}{ll} \quad & {[\overline{P R}=\overline{P Q}+\overline{Q R}]} \\ \end{array}$
$\begin{array}{ll} \Rightarrow \overline{P R}+\overline{R P} \end{array}$ $\begin{array}{ll} \quad & {[\overline{P R}=-\overline{R P}]} \end{array}$ because of change in direction

Hence $\Rightarrow \bar{0}$

Algebra of Vectors Exercise Very Short Answer Question 5

Answer:$X=0,Y=0$
Hint: You must know the rules of vector functions.
Given: If $\vec{a}$ and $\vec{b}$are two non-collinear vectors, such that $x\vec{a}+y\vec{b}=\vec{0}$, then write the values of $x$ and $y$
Solution: We have, $\vec{a}$ and $\vec{b}$ are non-collinear vectors and $x\vec{a}+y\vec{b}=\vec{0}$
$\Rightarrow \frac{a}{b}=-\frac{y}{x}$
Given that $\vec{a}$ and $\vec{b}$ are non collinear , so $\vec{a}$ and $\vec{b}$ should not be parallel ,
Therefore, $\frac{y}{x}=0$
$\Rightarrow y=0$
Similarly , the given equation can also be written as $\frac{b}{a}=-\frac{x}{y}$
Given that $\vec{a}$ and $\vec{b}$ are non collinear , so $\vec{a}$ and $\vec{b}$ should not be parallel ,
Therefore, $\frac{x}{y}=0$
$\Rightarrow x=0$
Hence we get , $x=0$ and $y=0$

Algebra of Vectors Exercise Very Short Answer Question 6

Answer: $\vec{a}+\vec{b},\vec{a}-\vec{b}$
Hints: You must know the rules of vector functions.
Given: If $\vec{a}$ and $\vec{b}$represents two adjacent sides of a parallelogram, then write vectors representing its diagonals.
Solution:

Let $\vec{a}$ and $\vec{b}$ represents two adjacent sides of a parallelogram ABCD.
∴ AB = DC and AD = BC [Because diagonals of parallelogram is equal]
\begin{aligned} &\Rightarrow \overrightarrow{D C}=\overline{A B}=\vec{a} \\ &\overrightarrow{A D}=\overrightarrow{B C}=\vec{b} \end{aligned}
In ABC
\begin{aligned} &\overrightarrow{A B}+\overrightarrow{B C}=\overrightarrow{A C} \\ &\vec{a}+\vec{b}=\overrightarrow{A C} \end{aligned}
Now, In ABD
\begin{aligned} & \overrightarrow{A D}+\overline{D B}=\overline{A B} \\ & \vec{b}+\overrightarrow{D B}=\vec{a} \\ \Rightarrow \overrightarrow{D B} &=\vec{a}-\vec{b} \end{aligned}
Vectors representing its diagonals are $\left ( \vec{a}+\vec{b} \right ),\left ( \vec{a}-\vec{b} \right )$

Algebra of Vectors Exercise Very Short Answer Question 7

Answer: $\vec{0}$
Hint: You must know the rules of vector functions.
Given: If $\vec{a},\vec{b},\vec{c}$ represents the sides of a triangle taken in order, then write the value of $\vec{a}+\vec{b}+\vec{c}$
Solution: Let ABC be a triangle, such that $\overrightarrow{B C}=\vec{a}, \overline{C A}=\vec{b} \& \overrightarrow{A B}=\vec{c}$
Then,
\begin{aligned} \vec{a}+\vec{b}+\vec{c}=\overrightarrow{B C}+\overline{C A}+\overline{A B} \\ \end{aligned}
\begin{aligned} &=\overrightarrow{B A}+\overrightarrow{A B} \\ \end{aligned} \begin{aligned} \left [ we\; can \; write \vec{BC}+\vec{CA}=\vec{BA} \right ] \end{aligned}
\begin{aligned} \text { But } \overrightarrow{A B}=-\overrightarrow{B A} \\ \end{aligned} $[Because \; of \; opposite\; direction]$
\begin{aligned} \therefore \overrightarrow{B A}+\overrightarrow{A B} \Rightarrow \overrightarrow{0} \end{aligned}

Algebra of Vectors Exercise Very Short Answer Question 8

Answer: $\vec{0}$
Hint: You must know the rules of vector functions.
Given: If $\vec{a},\vec{b},\vec{c}$ are position vectors of vertices A, B and C respectively of a triangle ABC, write the value of $\vec{AB},\vec{BC},\vec{CA}$
Solution: Given $\vec{a},\vec{b},\vec{c}$ are position vectors of vertices A, B and C respectively
Then,
\begin{aligned} &\overrightarrow{A B}=\vec{b}-\vec{a} \\ &\overrightarrow{B C}=\vec{c}-\vec{b} \\ &\overrightarrow{C A}=\vec{a}-\vec{c} \end{aligned}

Consider,
$\vec{AB}+\vec{BC}+\vec{CA}=\vec{b}-\vec{a}+\vec{c}-\vec{b}+\vec{a}-\vec{c}$
$\Rightarrow \vec{0}$

Algebra of Vectors Exercise Very Short Answer Question 9

Answer: $2\left ( \vec{c}-\vec{a} \right )$
Hint: You must know the rules of vector functions.
Given: If $\vec{a},\vec{b},\vec{c}$ are position vectors of points A, B and C respectively, write the value of $\vec{AB},\vec{BC},\vec{AC}$
Solution: $\vec{a},\vec{b},\vec{c}$are position vectors of points A, B and C respectively.
Then,
\begin{aligned} &\overrightarrow{A B}=\vec{b}-\vec{a} \\ &\overrightarrow{B C}=\vec{c}-\vec{b} \quad \text { because }[\overline{C A}=\vec{a}-\vec{c}=-\overrightarrow{A C}=\vec{c}-\vec{a}] \text { Opposite direction } \\ &\overline{C A}=\vec{c}-\vec{a} \end{aligned}
Therefore,
\begin{aligned} \overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{A C} &=\vec{b}-\vec{a}+\vec{c}-\vec{b}+\vec{c}-\vec{a} \\ &=2 \vec{c}-2 \vec{a} \\ & \Rightarrow 2(\vec{c}-\vec{a}) \end{aligned}

Algebra of Vectors Exercise Very Short Answer Question 10

Answer: $\frac{\vec{a}+\vec{b}+\vec{c}}{3}$
Hint: You must know the rules of vector functions
Given: If $\vec{a}+\vec{b}+\vec{c}$are position vectors of vertices of a triangle, then write the position vector of centroid.
Solution: Let ABC be a triangle and D,E and F are the mid-points of sides BC, CA and AB respectively.
Also, Let $\vec{a}+\vec{b}+\vec{c}$are position vectors of A, B, C respectively.
Then position vectors of D, E and F are
$\left(\frac{\vec{b}+\vec{c}}{2}\right) \cdot\left(\frac{\vec{c}+\vec{a}}{2}\right),\left(\frac{\vec{a}+\vec{b}}{2}\right)$ respectively.
The position vector of a point divides AD in the ratio of 2 is

$\frac{1 \cdot \vec{a}+2\left(\frac{\vec{b}+\vec{c}}{2}\right)}{3}=\frac{\vec{a}+\vec{b}+\vec{c}}{3}$

Similarly, position vectors of the points divides BE, CF in the ratio of 2:1 are equal to

$\frac{\vec{a}+\vec{b}+\vec{c}}{3}$

Thus, the points dividing AD in ratio 2:1 also divides BE, CF in the ratio.

Hence, medians of triangle are concurrent and the position of centroid is $\frac{\vec{a}+\vec{b}+\vec{c}}{3}$

Algebra of Vectors Exercise Very Short Answer Type Question 11

Answer: $\vec{0}$
Hint: You must know the rules of vector functions
Given: If G is denotes the centroid of ?ABC , then write the value of $\vec{GA}+\vec{GB}+\vec{GC}$
Solution: Let $\vec{a},\vec{b},\vec{c}$be the position vectors of the vertices A, B and C respectively.
Then, the position of centroid G is $\frac{\vec{a}+\vec{b}+\vec{c}}{3}$
Thus,
\begin{aligned} \overrightarrow{G A}+\overrightarrow{G B}+\overrightarrow{G C} &=\vec{a}-\left(\frac{\vec{a}+\vec{b}+\vec{c}}{3}\right)+\vec{b}-\left(\frac{\vec{a}+\vec{b}+\vec{c}}{3}\right)+\vec{c}-\left(\frac{\vec{a}+\vec{b}+\vec{c}}{3}\right) \\ & \Rightarrow(\vec{a}+\vec{b}+\vec{c})-3\left(\frac{\vec{a}+\vec{b}+\vec{c}}{3}\right) \\ & \Rightarrow(\vec{a}+\vec{b}+\vec{c})-(\vec{a}+\vec{b}+\vec{c}) \\ & \Rightarrow \overrightarrow{0} \end{aligned}

Algebra of Vectors Exercise Very Short Answer Type Question 12

Answer: $\frac{1}{3}\left ( 2\vec{b}+\vec{a} \right )$
Hint: You must know the rules of vector functions
Given: If $\vec{a}$ & $\vec{b}$denote the position vectors of points A and B respectively and C is appoint on AB, such that 3AC = 2AB, then write the position vector of C
Solution: Let $\vec{c}$is the position vector of c
Now,
$\vec{AB}=\vec{b}-\vec{a}$
$\vec{AC}=\vec{c}-\vec{a}$

Consider,

\begin{aligned} &3 \overrightarrow{A C}=2 \overrightarrow{A B} \\ &3(\vec{c}-\vec{a})=2(\vec{b}-\vec{a}) \\ &3 \vec{c}-3 \vec{a}=2 \vec{b}-2 \vec{a} \\ &3 \vec{c}=2 \vec{b}-2 \vec{a}+3 \vec{a} \\ &3 \vec{c}=2 \vec{b}+\vec{a} \\ &\vec{c}=\frac{1}{3}(2 \vec{b}+\vec{a}) \end{aligned}

Algebra of Vectors Exercise Very Short Answer Type Question 13

Answer: $\lambda =2$
Hint: You must know the rules of vector functions
Given: If D is the mid-point of sides BC of a triangle ABC such $\vec{AB}+\vec{AC}=\lambda \vec{AD}$ find $\lambda$
Solution: D is mid-point of side BC of a ABC, $\vec{AB}+\vec{AC}=\lambda \vec{AD}$
Let $\vec{a},\vec{b},\vec{c}$is a position vectors of AB, BC, CA
Now, the position vector of D is $\frac{\vec{b}+\vec{c}}{2}$
Then,
$\vec{AB}=\vec{b}-\vec{a}$
$\vec{AC}=\vec{c}-\vec{a}$
$\begin{gathered} \overrightarrow{A D}=\frac{\vec{b}+\vec{c}}{2}-\vec{a} \\ \therefore \overrightarrow{A B}+\overline{A C}=\lambda \overrightarrow{A D} \\ (\vec{b}-\vec{a})+(\vec{c}-\vec{a})=\lambda\left(\frac{\vec{b}+\vec{c}-2 \vec{a}}{2}\right) \\ \vec{b}-\vec{a}+\vec{c}-\vec{a}=\lambda\left(\frac{\vec{b}+\vec{c}-2 \vec{a}}{2}\right) \\ (\vec{b}+\vec{c}-2 \vec{a})\left(\frac{2}{\vec{b}+\vec{c}-2 \vec{a}}\right)=\lambda \\ \therefore \lambda=2 \end{gathered}$

Algebra of Vectors Exercise Very Short Answer Type Question 14

Answer: $\vec{0}$
Hint: You must know the rules of vector functions
Given: If D,E, F are the mid-points of sides BC,CA and AB respectively of $\Delta ABC$
Write the value of $\vec{AD}+\vec{BE}+\vec{CF}$
Solution: D, E, F are the mid-points of sides BC, CA, AB respectively
Then, the position vectors of the mid-points
D, E, F are given by $\frac{\vec{b}+\vec{c}}{2},\frac{\vec{c}+\vec{a}}{2},\frac{\vec{a}+\vec{b}}{2}$
Now,
\begin{aligned} &\overrightarrow{A D}+\overrightarrow{B E}+\overline{C F} \\ &\left(\frac{\vec{b}+\vec{c}}{2}\right)-\vec{a}+\left(\frac{\vec{c}+\vec{a}}{2}\right)-\vec{b}+\left(\frac{\vec{a}+\vec{b}}{2}\right)-\vec{c} \\ &2\left(\frac{\vec{a}+\vec{b}+\vec{c}}{2}\right)-(\vec{a}+\vec{b}+\vec{c}) \\ &\Rightarrow(\vec{a}+\vec{b}+\vec{c})-(\vec{a}+\vec{b}+\vec{c}) \\ &\Rightarrow \overrightarrow{0} \end{aligned}

Algebra of Vectors Exercise Very Short Answer Type Question 15

Answer: $m=\pm \frac{1}{a}$
Hint: You must know the rules of vector functions
Given: If $\vec{a}$ is a non-zero vector of modulus a and m is a non-zero scalar such that $m\vec{a}$ is a unit vector, find m.
Solution: $\vec{a}$is non-zero vector with modulus a and m
Also $m\vec{a}$ is unit vector
Therefore,
\begin{aligned} &\qquad\mid m \vec{a} \mid=1 \\ \end{aligned}
\begin{aligned} &\Rightarrow|m||\vec{a}|=1 \\ \end{aligned}
\begin{aligned} \Rightarrow|m| a=1 \\ \end{aligned}
\begin{aligned} |m|=\frac{1}{a} \\ \end{aligned}
\begin{aligned} &\Rightarrow m=\pm \frac{1}{a} \end{aligned}

Algebra of Vectors Exercise Very Short Answer Type Question 16

Answer: $\vec{0}$
Hint: You must know the rules of vector functions
Given: If $\vec{a},\vec{b},\vec{c}$ are the position vectors of the vertices of equilateral triangle. Write value of $\vec{a}+\vec{b}+\vec{c}$
Solution: Let ABC be a given equilateral triangle and vertices are $A\vec{\left ( a \right )},B\vec{\left ( b \right )},C\vec{\left ( c \right )}$. Also $O\vec{\left ( o \right )},$ be your orthocentre.
We know the centroid and orthocenter of equilateral triangles coincide at a point.
Orthocentre of $\Delta ABC=0$
Centroid of $\Delta ABC=\vec{0}$
$\Rightarrow \frac{\vec{a}+\vec{b}+\vec{c}}{3}=\vec{0}$
$\vec{a}+\vec{b}+\vec{c}=\vec{0}$

Algebra of Vectors Exercise Very Short Answer Type Question 17

A nswer: $\frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}+\frac{1}{\sqrt{3}} \hat{k}$
Hint: You must know the rules of vector functions
Given: Write a unit vector making equal acute angles with the coordinate axes
Solution: Suppose $\vec{r}$ makes an angle $\alpha$ with each of the axes OX, OY and OZ
Then its direction cosines are $l=\cos \alpha, m=\cos \alpha, n=\cos \alpha$
Now,
\begin{aligned} &l^{2}+m^{2}+n^{2}=1 \\ &\Rightarrow l^{2}+l^{2}+l^{2}=1 \; \; \; \; \; \; \; \; \; \; \; \; \quad[l=m=n] \\ &\Rightarrow 3 l^{2}=1 \\ &\Rightarrow l^{2}=\frac{1}{3} \\ &\Rightarrow l=\pm \frac{1}{\sqrt{3}} \end{aligned}
Since, we know angle is acute, Hence we only take positive values
∴ Unit vector is $\frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}+\frac{1}{\sqrt{3}} \hat{k}$

Algebra of Vectors Exercise Very Short Answer Type Question 18

Hint: You must know the rules of vector functions
Given: If a vector makes angles with $\alpha ,\beta$ &$\gamma$ with OX, OY and OZ respectively, Write $\sin ^{2}\alpha +\sin ^{2}\beta +\sin ^{2}\gamma$
Solution: Suppose, a vector $\vec{OP}$makes angles $\alpha ,\beta$ &$\gamma$ with OX, OY and OZ respectively
Then direction cosines of vectors are given by $l=\cos \alpha, m=\cos \beta, n=\cos \gamma$
Consider,
$\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma=\left(1-\cos ^{2} \alpha\right)+\left(1-\cos ^{2} \beta\right)+\left(1-\cos ^{2} \gamma\right)$
\begin{aligned} &=3-\left(\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma\right)\\ &=3-\left(l^{2}+m^{2}+n^{2}\right)\\ &=3-1 \quad\left[l^{2}+m^{2}+n^{2}=1\right]\\ &=2 \end{aligned}

Algebra of Vectors Exercise Very Short Answer Type Question 19

Answer: $6(\sqrt{2} \hat{i}+\hat{j}-\hat{k})$
Hint: You must know the rules of vector functions
Given: Write a vector of magnitude 12 units which makes $45^{0}$ angles with x-axis, $60^{0}$angle with y-axis, obtuse angle with z- axis.
Solution: Suppose a vector $\vec{r}$makes an angle $45^{0}$ with OX, $60^{0}$ with OY and having magnitude 12 units.
$l=\cos 45^{\circ}=\frac{1}{\sqrt{2}} \text { and } m=\cos 60^{\circ}=\frac{1}{2}$
Now,
\begin{aligned} &l^{2}+m^{2}+n^{2}=1 \\ &\frac{1}{2}+\frac{1}{4}+n^{2}=1 \\ &n^{2}=1-\frac{1}{2}-\frac{1}{4} \end{aligned}
\begin{aligned} &n^{2}=\frac{1}{4} \end{aligned}
\begin{aligned} &n^{2}=\pm \frac{1}{4} \end{aligned}
But angle obtuse angle along z-axis, so we use negative value.
\begin{aligned} &n^{2}= \frac{-1}{4} \end{aligned}
Therefore,
\begin{aligned} &\vec{r}=|\vec{r}|(\hat{i}+m \hat{j}+n \hat{k}) \\ &=12\left(\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{2} \hat{j}-\frac{1}{2} \hat{k}\right) \\ &=6(\sqrt{2} \hat{i}+\hat{j}-\hat{k}) \end{aligned}

Algebra of Vectors Exercise Very Short Answer Type Question 20

Hint: You must know the rules of vector functions
Given: Whose projections on coordinate axis are 12, 3, 4 units. Write length of vector
Solution:Projections on coordinate axis are 12, 3, 4 units
Therefore, length of vector
\begin{aligned} &=\sqrt{(12)^{2}+(3)^{2}+(4)^{2}} \\ &=\sqrt{144+9+16} \\ &=\sqrt{169} \\ &=13 \end{aligned}

Algebra of Vectors Exercise Very Short Answer Type Question 21

Answer: $3 \hat{i}+\frac{11}{3} \hat{j}+5 \hat{k}$
Hint: You must know the rules of vector functions
Given: Write the position vector of a point dividing the line segment joining points A and B with position vectors $\vec{a}$ &$\vec{b}$ externally in the ratio 1:4 , where,
\begin{aligned} &\vec{a}=2 \hat{i}+3 \hat{j}+4 \hat{k} \\ &b=-\hat{i}+\hat{j}+\hat{k} \end{aligned}
Solution: The position vectors of A and B are
\begin{aligned} &\vec{a}=2 \hat{i}+3 \hat{j}+4 \hat{k} \\ &b=-\hat{i}+\hat{j}+\hat{k} \end{aligned}
Let C divides AB in the ratio such that AB:CB=1:4
Position vector of C=
\begin{aligned} &\frac{1(-\hat{i}+\hat{j}+\hat{k})-4(2 \hat{i}+3 \hat{j}+4 \hat{k})}{1-4} \\ &\Rightarrow \frac{-\hat{i}+\hat{j}+\hat{k}-8 \hat{i}-12 \hat{j}-16 \hat{k}}{-3} \\ &\Rightarrow \frac{-9 \hat{i}-11 \hat{j}-15 \hat{k}}{-3} \Rightarrow 3 \hat{i}+\frac{11 \hat{j}}{3}+5 \hat{k} \end{aligned}

Algebra of Vectors Exercise Very Short Answer Type Question 22

Answer: $\frac{6}{7},\frac{-2}{7},\frac{3}{7}$
Hint: You must know the rules of vector functions
Solution: $\vec{r}=6\hat{i}-2\hat{j}+3\hat{k}$
The direction cosines are
\begin{aligned} &\frac{6}{\sqrt{(6)^{2}+(-2)^{2}+(3)^{2}}}, \frac{-2}{\sqrt{(6)^{2}+(-2)^{2}+(3)^{2}}}, \frac{3}{\sqrt{(6)^{2}+(-2)^{2}+(3)^{2}}} \\ &\begin{array}{l} \end{array} \end{aligned}
$\Rightarrow \frac{6}{\sqrt{49}}, \frac{-2}{\sqrt{49}}, \frac{3}{\sqrt{49}} \\$
$\Rightarrow \frac{6}{7}, \frac{-2}{7}, \frac{3}{7}$

Algebra of Vectors Exercise Very Short Answer Type Question 23

Answer:$\frac{-\hat{i}+2\hat{j}-\hat{k}}{\sqrt{6}}$
Hint: You must know the rules of vector functions
Given: $\vec{a}=\hat{i}+\hat{j}, \vec{b}=\hat{j}+\hat{k} \& \vec{c}=\hat{k}+\hat{i}$ find unit vector parallel to $\vec{a}+\vec{b}-2\vec{c}$
Solution: $\vec{a}=\hat{i}+\hat{j}, \vec{b}=\hat{j}+\hat{k} \& \vec{c}=\hat{k}+\hat{i}$
Now,
\begin{aligned} &\vec{a}+\vec{b}-2 \vec{c}=\hat{i}+\hat{j}+\hat{j}+\hat{k}-2 \hat{k}-2 \hat{i} \\ &=-\hat{i}+2 \hat{j}-\hat{k} \end{aligned}
Unit vector parallel to,
\begin{aligned} &\vec{a}+\vec{b}-2 \vec{c}=\frac{-\hat{i}+2 \hat{j}-\hat{k}}{\sqrt{(-1)^{2}+(2)^{2}+(1)^{2}}} \\ &=\frac{-\hat{i}+2 \hat{j}-\hat{k}}{\sqrt{6}} \end{aligned}

Algebra of Vectors Exercise Very Short Answer Type Question 24

Answer: $\frac{1}{\sqrt{41}}\left ( 3\hat{i}+4\hat{j}-4\hat{k} \right )$
Hint: You must know the rules of vector functions
Given: $\vec{a}=\hat{i}+2\hat{j},\vec{b}=\hat{j}+2\hat{k}$, find unit vector $3\vec{a}-2\vec{b}$
Solution: $\vec{a}=\hat{i}+2\hat{j},\vec{b}=\hat{j}+2\hat{k}$
Hence,
\begin{aligned} &3 \vec{a}-2 \vec{b}=3(\hat{i}+2 \hat{j})-2(\hat{j}+2 \hat{k}) \\ &=3 \hat{i}+6 \hat{j}-2 \hat{j}-4 \hat{k} \\ &=3 \hat{i}+4 \hat{j}-4 \hat{k} \end{aligned}

Hence, unit vector along,

\begin{aligned} &3 \vec{a}-2 \vec{b}=\frac{3 \hat{i}+4 \hat{j}-4 \hat{k}}{\sqrt{(3)^{2}+(4)^{2}+(-4)^{2}}} \\ &=\frac{3 \hat{i}+4 \hat{j}-4 \hat{k}}{\sqrt{9+16+16}} \\ &=\frac{1}{\sqrt{41}}(3 \hat{i}+4 \hat{j}-4 \hat{k}) \end{aligned}

Algebra of Vectors Exercise Very Short Answer Type Question 25

Answer: $-\hat{i}+5\hat{j}-12\hat{k}$
Hint: You must know the rules of vector functions
Given: Find position vector of point dividing line segment
$\hat{i}+\hat{j}-2\hat{k},2\hat{i}-\hat{j}+3\hat{k}$ Externally in 2:3
Solution: Let A and B be the points with vectors
\begin{aligned} &\vec{a}=\hat{i}+\hat{j}-2 \hat{k} \\ &\vec{b}=2 \hat{i}-\hat{j}+3 \hat{k} \end{aligned}respectively
Let C divide AB externally with ratio 2:3 such AC: CB=2:3
Position vector of C=
\begin{aligned} &=\frac{2(2 \hat{i}-\hat{j}+3 \hat{k})-3(\hat{i}+\hat{j}-2 \hat{k})}{2-3} \\ &=\frac{(4 \hat{i}-2 \hat{j}+6 \hat{k})-3 \hat{i}-3 \hat{j}+6 \hat{k}}{-1} \\ &=\frac{1}{-1}(\hat{i}-5 \hat{j}+12 \hat{k}) \\ &=-\hat{i}+5 \hat{j}-12 \hat{k} \end{aligned}

Algebra of Vectors Exercise Very Short Answer Type Question 26

Answer: $\frac{1}{\sqrt{3}}\left ( \hat{i}+\hat{j}+\hat{k} \right )$
Hint: You must know the rules of vector functions
Given:$\vec{a}=\hat{i}+\hat{j}, \vec{b}=\hat{j}+\hat{k}, \vec{c}=\hat{k}+\hat{i}$ find unit vector in the direction of $\vec{a}+\vec{b}+\vec{c}$
Solution:$\vec{a}=\hat{i}+\hat{j}, \vec{b}=\hat{j}+\hat{k}, \vec{c}=\hat{k}+\hat{i}$
Then,
\begin{aligned} &\vec{a}+\vec{b}+\vec{c}=\hat{i}+\hat{j}+\hat{j}+\hat{k}+\hat{k}+\hat{i} \\ &=2(\hat{i}+\hat{j}+\hat{k}) \\ &\therefore|\vec{a}+\vec{b}+\vec{c}|=\sqrt{2^{2}+2^{2}+2^{2}} \\ &=\sqrt{4+4+4} \\ &=\sqrt{12} \\ &=2 \sqrt{3} \end{aligned}
Therefore, unit vector in the direction of

$\vec{a}+\vec{b}+\vec{c}=\frac{2(\hat{i}+\hat{j}+\hat{k})}{2 \sqrt{3}}=\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$

Algebra of Vectors Exercise Very Short Answer Type Question 27

Answer: $\sqrt{398}$
Hint: You must know the rules of vector functions
Given: If
\begin{aligned} &\vec{a}=3 \hat{i}-\hat{j}-4 \hat{k} \\ &\vec{b}=-2 \hat{i}+4 \hat{j}-3 \hat{k}, \text { find }|3 \vec{a}-2 \vec{b}+4 \vec{c}| \\ &\vec{c}=\hat{i}+2 \hat{j}-\hat{k} \end{aligned}
Solution:
\begin{aligned} &\vec{a}=3 \hat{i}-\hat{j}-4 \hat{k} \\ &\vec{b}=-2 \hat{i}+4 \hat{j}-3 \hat{k}, \\ &\vec{c}=\hat{i}+2 \hat{j}-\hat{k} \end{aligned}
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}
Hence, \begin{aligned} &\therefore|3 \vec{a}-2 \vec{b}+4 \vec{c}|=\sqrt{(17)^{2}-3^{2}-(10)^{2}} \\ \end{aligned}
\begin{aligned} &=\sqrt{289+9+100} \\ &=\sqrt{398} \end{aligned}

Algebra of Vectors Exercise Very Short Answer Type Question 28

Answer: $\theta =30^{o}$
Hint: You must know the rules of vector functions
Given: A unit vector $\vec{r}$make angles $\frac{\pi }{3},\frac{\pi }{2}$ with $\hat{j}$ &$\hat{k}$ and an acute angle $\theta$ with $\hat{i}$.find $\theta$
Solution: Angle $\frac{\pi }{3},\frac{\pi }{2}$ with $\hat{j}$ &$\hat{k}$
Let l , m , n be the direction cosines
\begin{aligned} &l=\cos \theta, m=\cos \frac{\pi}{3}, n=\cos \frac{\pi}{2} \\ &l=\cos \theta, m=\frac{1}{2}, n=0 \\ \end{aligned}
Now,\begin{aligned} &l^{2}+m^{2}+n^{2}=1 \\ \end{aligned}
\begin{aligned} &l^{2}+\frac{1}{4}+0=1 \\ &l^{2}=\frac{3}{4} \\ &l=\pm \sqrt{\frac{3}{4}} \Rightarrow \pm \frac{\sqrt{3}}{2} \\ \end{aligned}

But we know angle made with $\hat{i}$is an acute angle so, we use the positive value.

$\therefore \theta =30^{o}$ $\left [ \therefore \cos ^{-1}\left ( \frac{\sqrt{3}}{2} \right )=30^{0} \right ]$

Algebra of Vectors Exercise Very Short Answer Type Question 29

Answer: $\frac{3}{7}\hat{i}-\frac{2}{7}\hat{j}+\frac{6}{7}\hat{k}$
Hint: You must know the rules of vector functions
Given: Find unit vector in direction of $\hat{a}=3\hat{i}-2\hat{j}+6\hat{k}$
Solution: We have,
$\hat{a}=3\hat{i}-2\hat{j}+6\hat{k}$
\begin{aligned} &|\vec{a}|=\sqrt{(3)^{2}+(-2)^{2}+(6)^{2}} \Rightarrow \sqrt{49}=7 \\ \end{aligned}
Unit vector in direction of $\hat{a}$ is
\begin{aligned} &\hat{a}=\frac{\vec{a}}{|\vec{a}|}=\frac{1}{7}(3 \hat{i}-2 \hat{j}+6 \hat{k}) \\ &\Rightarrow \frac{3}{7} \hat{i}-\frac{2}{7} \hat{j}+\frac{6}{7} \hat{k} \end{aligned}

Algebra of Vectors Exercise Very Short Answer Type Question 30

Answer: $\frac{1}{3}\left ( \hat{i}+2\hat{j} +2\hat{k}\right )$
Hint: You must know the rules of vector functions
Given: $\vec{a}=\hat{i}+2\hat{j}-3\hat{k}$and $\vec{b}=2\hat{i}+4\hat{j}+9\hat{k}$find vector parallel to $\vec{a}+\vec{b}$
Solution:
$\vec{a}=\hat{i}+2\hat{j}-3\hat{k}$
$\vec{b}=2\hat{i}+4\hat{j}+9\hat{k}$
Now,\begin{aligned} &\vec{a}+\vec{b}=\hat{i}+2 \hat{j}-3 \hat{k}+2 \hat{i}+4 \hat{j}+9 \hat{k} \\ \end{aligned}
\begin{aligned} &\vec{a}+\vec{b}=3 \hat{i}+6 \hat{j}+6 \hat{k} \\ \end{aligned}
$=3(\hat{i}+2 \hat{j}+2 \hat{k}) \\$
$|\vec{a}+\vec{b}|=\sqrt{(3)^{2}+(6)^{2}+(6)^{2}} \\$
$=\sqrt{9+36+36} \\$
$=\sqrt{81} \\$
$=9 \\$

Vector parallel to $\vec{a}+\vec{b}$

$\frac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|}=\frac{3(\hat{i}+2 \hat{j}+2 \hat{k})}{9}=\frac{1}{3}(\hat{i}+2 \hat{j}+2 \hat{k})$

Algebra of Vectors Exercise Very Short Answer Type Question 31

Answer: $\frac{2}{3}\hat{i}+\frac{1}{3}\hat{j}+\frac{2}{3}\hat{k}$
Hint: You must know the rules of vector functions
Given: Find unit vector in direction of $\vec{b}=2\hat{i}+\hat{j}+2\vec{k}$
Solution: $\vec{b}=2\hat{i}+\hat{j}+2\vec{k}$
\begin{aligned} &|\vec{b}|=\sqrt{(2)^{2}+(1)^{2}+(2)^{2}} \\ &=\sqrt{4+1+4} \\ &=\sqrt{9} \\ &=3 \\ \end{aligned}
Unit vector:
\begin{aligned} &\hat{b}=\frac{\vec{b}}{|\vec{b}|}=\frac{1(2 \hat{i}+\hat{j}+2 \hat{k})}{3}=\frac{2}{3} \hat{i}+\frac{1}{3} \hat{j}+\frac{2}{3} \hat{k} \end{aligned}

Algebra of Vectors Exercise Very Short Answer Type Question 32

Answer:$\left ( 2,3,1 \right )$
Hint: You must know the rules of vector functions
Given: Find the position vector of mid-point of line AB where $A\left ( 3,4,-2 \right ),B\left ( 1,2,4 \right )$
Solution: $A\left ( 3,4,-2 \right ),B\left ( 1,2,4 \right )$
Let C is the mid-point
∴ Position vector of $C=\frac{\left ( 3\hat{i}+4\hat{j}-2\hat{k} \right )+\left ( \hat{i}+2\hat{j}+4\hat{k} \right )}{2}$
$=\frac{\left ( 4\hat{i}+6\hat{j}+2\hat{k} \right )}{2}$
$\left ( 2\hat{i}+3\hat{j}+\hat{k} \right )$
Hence position vector$=\left ( 2,3,1 \right )$

Algebra of Vectors Exercise Very Short Answer Type Question 33

Answer: $4\hat{i}-2\hat{j}+4\hat{k}$
Hint: You must know the rules of vector functions
Given: Find a vector in direction of $\vec{a}=2\hat{i}-\hat{j}+2\hat{k}$, which has magnitude 6 units
Solution:$\vec{a}=2\hat{i}-\hat{j}+2\hat{k}$
\begin{aligned} & \begin{aligned} |\vec{a}|=\sqrt{(2)^{2}+(-1)^{2}+(2)^{2}}=\sqrt{4+1+4} \\ \end{aligned} \end{aligned}
\begin{aligned} & \begin{aligned} &=\sqrt{9} \\ &=3 \\ \end{aligned} \end{aligned}
Vector in direction of
\begin{aligned} & \begin{aligned} \hat{a}=6 \times \frac{\vec{a}}{|\vec{a}|}=\frac{6 \times(2 \hat{i}-\hat{j}+2 \hat{k})}{3} \\ \end{aligned} \end{aligned}
\begin{aligned} & \begin{aligned} =& 2(2 \hat{i}-\hat{j}+2 \hat{k}) \\ =& 4 \hat{i}-2 \hat{j}+4 \hat{k} \end{aligned} \end{aligned}

Algebra of Vectors Exercise Very Short Answer Type Question 34

Answer: $\frac{1}{2}$
Hint: You must know the rules of vector functions
Given: What is the cosine of angle with vector $\sqrt{2\hat{i}}+\hat{j}+\hat{k}$ makes with y – axis
Solution: $\sqrt{2\hat{i}}+\hat{j}+\hat{k}$
Direction cosines are
\begin{aligned} &\frac{\sqrt{2}}{\sqrt{(\sqrt{2})^{2}+(1)^{2}+(1)^{2}}}, \frac{1}{\sqrt{(\sqrt{2})^{2}+(1)^{2}+(1)^{2}}}=\frac{1}{\sqrt{(\sqrt{2})^{2}+(1)^{2}+(1)^{2}}} \\ \end{aligned}
\begin{aligned} &\text { OR } \frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2} \end{aligned}

Cosine angle along y-axis is $\frac{1}{2}$

Algebra of Vectors Exercise Very Short Answer Type Question 35

Answer: Both have the same magnitude.
Hint: You must know the rules of solving vectors
Given: Write two different vectors having same magnitude
Solution:
Let \begin{aligned} &\vec{a}=2 \hat{i}-\hat{j}+2 \hat{k} \\ &\vec{b}=-2 \hat{i}+\hat{j}-2 \hat{k} \end{aligned}
\begin{aligned} &|\vec{a}|=\sqrt{(2)^{2}+(-1)^{2}+(2)^{2}}=\sqrt{4+1+4}=\sqrt{9}=3 \\ &|\vec{b}|=\sqrt{(-2)^{2}+(1)^{2}+(-2)^{2}}=\sqrt{4+1+4}=\sqrt{9}=3 \end{aligned}

Hence, both vectors are having the same magnitude.

Algebra of Vectors Exercise Very Short Answer Type Question 36

Answer: $\text { Thus, } \vec{a} \text { is parallel to } \vec{b} \text { and hence in the same direction. }$
Hint: You must know the rules of vector functions
Given: Write two different factors having same direction
Solution:
Let $\vec{a}=2 \hat{i}+\hat{j}+2 \hat{k}, \vec{b}=4 \hat{i}+2 \hat{j}+4 \hat{k}$
The direction cosines of $\hat{a}$ is
\begin{aligned} &l=\frac{2}{\sqrt{(2)^{2}+(1)^{2}+(2)^{2}}}=\frac{2}{\sqrt{9}}=\frac{2}{3} \\ &m=\frac{1}{\sqrt{(2)^{2}+(1)^{2}+(2)^{2}}}=\frac{1}{\sqrt{9}}=\frac{1}{3} \\ &n=\frac{2}{\sqrt{(2)^{2}+(1)^{2}+(2)^{2}}}=\frac{2}{\sqrt{9}}=\frac{2}{3} \\ \end{aligned}

And direction cosines of $\vec{b}$ is
\begin{aligned} &l=\frac{4}{\sqrt{(4)^{2}+(2)^{2}+(4)^{2}}}=\frac{4}{\sqrt{16+4+16}}=\frac{4}{\sqrt{36}}=\frac{4}{6}=\frac{2}{3} \\ &m=\frac{2}{\sqrt{(4)^{2}+(2)^{2}+(4)^{2}}}=\frac{2}{\sqrt{16+4+16}}=\frac{2}{\sqrt{36}}=\frac{2}{6}=\frac{1}{3} \\ &n=\frac{4}{\sqrt{(4)^{2}+(2)^{2}+(4)^{2}}}=\frac{4}{\sqrt{16+4+16}}=\frac{4}{\sqrt{36}}=\frac{4}{6}=\frac{2}{3} \end{aligned}

The direction cosines of $\vec{a}$ and $\vec{b}$ are same.

$\text { Thus, } \vec{a} \text { is parallel to } \vec{b} \text { and hence in the same direction. }$

Algebra of Vectors Exercise Very Short Answer Type Question 37

Answer: $\frac{8}{\sqrt{30}}\left ( 5\hat{i}-\hat{j}+2\hat{k} \right )$
Hint: You must know the rules of vector functions
Given: Write a vector in direction of $5\hat{i}-\hat{j}+2\hat{k}$ having magnitude 8 units
Solution: Let $\vec{a}= 5\hat{i}-\hat{j}+2\hat{k}$
\begin{aligned} &|\vec{a}|=\sqrt{(5)^{2}+(-1)^{2}+(2)^{2}}\\ &=\sqrt{25+1+4}\\ &=\sqrt{30} \end{aligned}
Position vector in the direction of vector is,
\begin{aligned} &=8 \times \frac{\vec{a}}{|\vec{a}|} \\ &=\frac{8}{\sqrt{30}}(5 \hat{i}-\hat{j}+2 \hat{k}) \end{aligned}

Algebra of Vectors Exercise Very Short Answer Type Question 38

Answer: $\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{14}$
Hint: You must know the rules of vector functions
Given: $\hat{i}+2\hat{j}+3\hat{k}$, find direction cosines
Solution: Let $\hat{i}+2\hat{j}+3\hat{k}$
Hence direction cosines are,
\begin{aligned} &\frac{1}{\sqrt{(1)^{2}+(2)^{2}+(3)^{2}}}, \frac{2}{\sqrt{(1)^{2}+(2)^{2}+(3)^{2}}}, \frac{3}{\sqrt{(1)^{2}+(2)^{2}+(3)^{2}}} \\ &=\frac{1}{\sqrt{1+4+9}}, \frac{2}{\sqrt{1+4+9}}, \frac{3}{\sqrt{1+4+9}} \\ &=\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}} \end{aligned}

Algebra of Vectors Exercise Very Short Answer Type Question 39

Answer: $\frac{2}{7}\hat{i}-\frac{3}{7}\hat{j}+\frac{6}{7}\hat{k}$
Hint: You must know the rules of vector functions
Given: Find unit vector in direction of $\vec{a}=2\hat{i}-3\hat{j}+6\hat{k}$
Solution:$\vec{a}=2\hat{i}-3\hat{j}+6\hat{k}$
\begin{aligned} &|\vec{a}|=\sqrt{(2)^{2}+(-3)^{2}+(6)^{2}} \\ &=\sqrt{4+9+36} \\ &=\sqrt{49} \\ &=7 \\ \end{aligned}
Unit vector,
\begin{aligned} &\frac{\vec{a}}{|\vec{a}|}=\frac{2 \hat{i}-3 \hat{j}+6 \hat{k}}{7} \\ &=\frac{2}{7} \hat{i}-\frac{3}{7} \hat{j}+\frac{6}{7} \hat{k} \end{aligned}

Algebra of Vectors Exercise Very Short Answer Type Question 40

Answer: $-4$
Hint: You must know the rules of vector functions
Given: For what value of ‘a’ the vectors $2\hat{i}-3\hat{j}+4\hat{k}$&$a\hat{i}+6\hat{j}-8\hat{k}$are collinear
Solution: Two vectors are
\begin{aligned} &\vec{p}=2 \hat{i}-3 \hat{j}+4 \hat{k}, q=a \hat{i}+6 \hat{j}-8 \hat{k} \\ \end{aligned}

Vectors are collinear,
\begin{aligned} &\vec{p}=\lambda \vec{q} \\ \end{aligned}
\begin{aligned} &2 \hat{i}-3 \hat{j}+4 \hat{k}=\lambda(\hat{a i}+6 \hat{j}-8 \hat{k}) \\ \end{aligned}
\begin{aligned} &2 \hat{i}-3 \hat{j}+4 \hat{k}=\lambda a \hat{i}+6 \lambda \hat{j}-8 \lambda \hat{k} \\ \end{aligned}

By comparing
\begin{aligned} &-8 \lambda=-3 \Rightarrow \lambda=\frac{-1}{2} \\ \end{aligned}
\begin{aligned} &-6 \lambda=-3 \Rightarrow \lambda=\frac{-1}{2} \\ \end{aligned}
\begin{aligned} &\lambda a=2 \Rightarrow \frac{-1}{2} \times a=2 \Rightarrow a=-4 \end{aligned}
Hence $a=-4$

Algebra of Vectors Exercise very Short Answer type Question 41

Answer: $\frac{-2}{\sqrt{30}},\frac{1}{\sqrt{30}},\frac{-5}{\sqrt{30}}$
Hint: You must know the rules of vector functions
Given: Find direction cosines of $-2\hat{i}+\hat{j}-5\hat{k}$
Solution: Let $\vec{a}=-2\hat{i}+\hat{j}-5\hat{k}$
Then its cosines are,
\begin{aligned} &\frac{-2}{\sqrt{(-2)^{2}+(1)^{2}+(-5)^{2}}}, \frac{1}{\sqrt{(-2)^{2}+(1)^{2}+(-5)^{2}}}, \frac{-5}{\sqrt{(-2)^{2}+(1)^{2}+(-5)^{2}}} \\ &\frac{-2}{\sqrt{4+1+25}}, \frac{1}{\sqrt{4+1+25}}, \frac{-5}{\sqrt{4+1+25}} \\ &\frac{-2}{\sqrt{30}}, \frac{1}{\sqrt{30}}, \frac{-5}{\sqrt{30}} \end{aligned}

Algebra of Vectors Exercise Very Short Answer Type Question 42

Answer: $5\hat{i}-5\hat{j}+3\hat{k}$
Hint: You must know the rules of vector functions
Given: Find sum of $\vec{a}=\hat{i}-2\hat{j},\vec{b}=2\hat{i}-3\hat{j},\vec{c}=2\hat{i}+3\hat{k}$
So, sum of three vectors
\begin{aligned} &\vec{a}+\vec{b}+\vec{c}=\hat{i}-2 \hat{j}+2 \hat{i}-3 \hat{j}+2 \hat{i}+3 \hat{k} \\ &=5 \hat{i}-5 \hat{j}+3 \hat{k} \end{aligned}

Algebra of Vectors Exercise very Short Answer type Question 43

Answer: $\frac{3}{7}\hat{i}-\frac{2}{7}\hat{j}+\frac{6}{7}\hat{k}$
Hint: You must know the rules of vector functions
Given: Find unit vector in direction of $\vec{a}=3\hat{i}-2\hat{j}+6\hat{k}$
Solution: $\vec{a}=3\hat{i}-2\hat{j}+6\hat{k}$
Unit vector \begin{aligned} &\frac{\vec{a}}{\mid \vec{a}\mid }=\frac{3 \hat{i}-2 \hat{j}+6 \hat{k}}{\sqrt{(3)^{2}+(-2)^{2}+(6)^{2}}} \\ \end{aligned}
\begin{aligned} &=\frac{3 \hat{\imath}-2 \hat{\jmath}+6 \hat{k}}{\sqrt{9+4+36}} \\ &=\frac{3 \hat{\imath}-2 \hat{\jmath}+6 \hat{k}}{\sqrt{49}} \\ &=\frac{3 \hat{\imath}-2 \hat{\jmath}+6 \hat{k}}{7} \\ &=\frac{3}{7} \hat{\imath}-\frac{2}{7} \hat{\jmath}+\frac{6}{7} \hat{k} \end{aligned}

Algebra of Vectors Exercise very Short Answer type Question 44

Answer: $x+y++z=0$
Hint: You must know the rules of vector functions
Given: If $\vec{a}=x \hat{i}+2 \hat{j}-z \hat{k}, \vec{b}=3 \hat{i}-\hat{y} \hat{j}+\hat{k}$are two equal vectors find $x+y++z$
Solution: $\vec{a}=x \hat{i}+2 \hat{j}-z \hat{k}, \vec{b}=3 \hat{i}-\hat{y} \hat{j}+\hat{k}$
They are equal vectors, So, $\vec{a}=\vec{b}$
$x \hat{i}+2 \hat{j}-z \hat{k}=3 \hat{i}-\hat{y} \hat{j}+\hat{k}$
By comparing
$x=3,y=-2,z=-1$
$\therefore x+y+z=3-2-1=0$

Algebra of Vectors Exercise very Short Answer type Question 45

Answer: $\frac{4}{13}\hat{i}+\frac{3}{13}\hat{j}-\frac{12}{13}\hat{k}$
Hint: You must know the rules of vector functions
Given: Find unit vector in direction of sum of vectors
\begin{aligned} &\vec{a}=2 \hat{i}+2 \hat{j}-5 \hat{k}, \\ &\vec{b}=2 \hat{i}+\hat{j}-7 \hat{k} \end{aligned}
Solution: We have,
\begin{aligned} &\vec{a}=2 \hat{i}+2 \hat{j}-5 \hat{k}, \\ &\vec{b}=2 \hat{i}+\hat{j}-7 \hat{k} \end{aligned}
Sum, \begin{aligned} &\vec{p}=\vec{a}+\vec{b}=2 \hat{i}+2 \hat{j}-5 \hat{k}+2 \hat{i}+\hat{j}-7 \hat{k} \\ \end{aligned}
\begin{aligned} &=4 \hat{i}+3 \hat{j}-12 \hat{k} \\ \end{aligned}
∴Required unit vector
\begin{aligned} &\frac{\vec{p}}{|\vec{p}|}=\frac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|}=\frac{4 \hat{i}+3 \hat{j}-12 \hat{k}}{\sqrt{(4)^{2}+(3)^{2}+(-12)^{2}}} \\ &=\frac{4 \hat{i}+3 \hat{j}-12 \hat{k}}{\sqrt{16+9+144}} \\ &=\frac{4 \hat{i}+3 \hat{j}-12 \hat{k}}{\sqrt{169}} \\ &=\frac{4 \hat{i}+3 \hat{j}-12 \hat{k}}{13} \\ &\therefore \frac{4}{13} \hat{i}+\frac{3}{13} \hat{j}-\frac{12}{13} \hat{k} \end{aligned}

Algebra of Vectors Exercise very Short Answer type Question 46

Answer:$p=\frac{-1}{3}$
Hint: You must know the rules of vector functions
Given: Find value of ‘p’
$3\hat{i}+2\hat{j}+9\hat{k},\hat{i}-2\hat{pj}+3\hat{k}$ are parallel
Solution:
Let $\vec{a}=3\hat{i}+2\hat{j}+9\hat{k}$
$\vec{b}=\hat{i}-2\hat{pj}+3\hat{k}$
If $\vec{a}$ &$\vec{b}$ are parallel
$\vec{b}=\lambda\vec{a}$
\begin{aligned} &\hat{i}-2 \hat{p j}+3 \hat{k}=\lambda(3 \hat{i}+2 \hat{j}+9 \hat{k})\\ &\hat{i}-2 \hat{p j}+3 \hat{k}=3 \lambda \hat{i}+2 \lambda \hat{j}+9 \lambda \hat{k}\\ \end{aligned}
$\therefore$ On comparing,
\begin{aligned} &3 \lambda=1 \Rightarrow \lambda=\frac{1}{3}\\ &3=9 \lambda \Rightarrow \lambda=\frac{1}{3}\\ \end{aligned}
\begin{aligned} &-2 p=2 \lambda \Rightarrow-2 p=2 \times \frac{1}{3} \Rightarrow p=\frac{-1}{3} \end{aligned}

Algebra of Vectors Exercise very Short Answer type Question 47

Answer: $\vec{a}=5\left ( \hat{i}+0\hat{j}+\hat{k} \right )$
Hint: You must know the rules of vector functions
Given: Find a vector $\vec{a}$ of magnitude $-5\sqrt{2}$ ,making angle$\frac{\pi }{4}$ with x- axis,$\frac{\pi }{2}$ with y-axis and $\theta$with z-axis
Solution: $\frac{\pi }{4}$ with x- axis, $\frac{\pi }{2}$ with y-axis and $\theta$ with z-axis
\begin{aligned} &l=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}} \\ &m=\cos \frac{\pi}{2}=0 \\ &n=\cos \theta \\ &\therefore l^{2}+m^{2}+n^{2}=1 \\ &\frac{1}{2}+0+\cos ^{2} \theta=1 \\ &\frac{1}{2}=1-\cos ^{2} \theta \\ &\cos ^{2} \theta=1-\frac{1}{2} \\ &\cos ^{2} \theta=\frac{1}{2} \\ &\cos \theta=\frac{1}{\sqrt{2}} \\ \end{aligned}Since $\theta$ is an acute angle
Now,
\begin{aligned} &\vec{a}=|\vec{a}|(l \hat{i}+m \hat{j}+n \hat{k}) \\ &\vec{a}=5 \sqrt{2}\left(\frac{1}{\sqrt{2}} \hat{i}+0 \hat{j}+\frac{1}{\sqrt{2}} \hat{k}\right) \\ &\vec{a}=5(\hat{i}+0 \hat{j}+\hat{k}) \end{aligned}

Algebra of Vectors Exercise very Short Answer type Question 48

Answer: $\frac{1}{7}\left ( 3\hat{i}+2\hat{j}+6\hat{k} \right )$
Hint: You must know the rules of vector functions
Given: Find a unit vector in direction of $\vec{PQ}$ when $P\left ( 1,3,0 \right )$ &$Q\left ( 4,5,6 \right )$
Solution: $P\left ( 1,3,0 \right )$ & $Q\left ( 4,5,6 \right )$
\begin{aligned} &\therefore \overrightarrow{P Q}=(4 \hat{i}+5 \hat{j}+6 \hat{k})-(\hat{i}+3 \hat{j}+0 \hat{k}) \\ &=4 \hat{i}+5 \hat{j}+6 \hat{k}-\hat{i}-3 \hat{j}-0 \hat{k} \\ &\overrightarrow{P Q}=3 \hat{i}+2 \hat{j}+6 \hat{k} \\ &|\overrightarrow{P Q}|=\sqrt{(3)^{2}+(2)^{2}+(6)^{2}} \\ &=\sqrt{9+4+36} \\ &=\sqrt{49} \\ &=7 \\ \end{aligned}
Unit Vector:
\begin{aligned} &\frac{\overline{P Q}}{|\overline{P Q}|}=\frac{1}{7}(3 \hat{i}+2 \hat{j}+6 \hat{k}) \end{aligned}

Algebra of Vectors Exercise very Short Answer type Question 49

Answer: $6\hat{i}-9\hat{j}+18\hat{k}$
Hint: You must know the rules of vector functions
Given: Find vector in direction of $2\hat{i}-3\hat{j}+6\hat{k}$having magnitude 21 units
Solution: Let $\vec{a}=2\hat{i}-3\hat{j}+6\hat{k}$
Unit vector
\begin{aligned} &\frac{\vec{a}}{|\vec{a}|}=\frac{(2 \hat{i}-3 \hat{j}+6 \hat{k})}{\sqrt{2^{2}+(-3)^{2}+(6)^{2}}}\\ \end{aligned}
\begin{aligned} &=\frac{(2 \hat{i}-3 \hat{j}+6 \hat{k})}{\sqrt{49}}\\ \end{aligned}
\begin{aligned} &=\frac{(2 \hat{i}-3 \hat{j}+6 \hat{k})}{7}\\ \end{aligned}

We know, magnitude is 21 units
\begin{aligned} &|\vec{a}|=21\\ \end{aligned}
\begin{aligned} &\frac{\vec{a}}{21}=\frac{1}{7}(2 \hat{i}-3 \hat{j}+6 \hat{k})\\ \end{aligned}
\begin{aligned} &\vec{a}=3(2 \hat{i}-3 \hat{j}+6 \hat{k})\\ \end{aligned}
\begin{aligned} &=6 \hat{i}-9 \hat{j}+18 \hat{k} \end{aligned}

Algebra of Vectors Exercise very Short Answer type Question 50

Answer:$\left [ -12,8 \right ]$
Hint: You must know the rules of vector functions
Given: If $\mid \vec{a}\mid =4$ and $-3\leq \lambda \leq 2$, Writer range of $\mid \lambda \vec{a}\mid$
Solution: $-3\leq \lambda \leq 2$
$\begin{gathered} \Rightarrow-3 \times|\vec{a}| \leq \lambda|\vec{a}| \leq 2 \times|\vec{a}| \\ \end{gathered}$
Value of $\begin{gathered} |\vec{a}|=4 \\ \end{gathered}$
$\begin{gathered} \therefore \Rightarrow-3 \times 4 \leq \lambda|\vec{a}| \leq 2 \times 4 \\ \end{gathered}$
$\begin{gathered} \Rightarrow-12 \leq \lambda|\vec{a}| \leq 8 \\ \end{gathered}$
Range of $\begin{gathered} \lambda|\vec{a}|=>[-12,8] \end{gathered}$

Algebra of Vectors Exercise very Short Answer type Question 51

Answer:$2\vec{b}-\vec{a}$
Hint: You must know the rules of vector functions
Given: $\Delta DAC$ , if B is mid-point of AC and $\vec{OA}=\vec{a}$,$\vec{OB}=\vec{b}$find $\vec{OC}$
Solution:

$\triangle D A C \overrightarrow{O A}=\vec{a}, \overrightarrow{O B}=\vec{b}$
$\mathrm{B} \: \: is \: \: mid-point$
$\therefore Position vector of B=\frac{\text { Position vector of A+Position vector of } \mathrm{C}}{2}$
$\overrightarrow{O B}=\frac{\overline{O A}+\overrightarrow{O C}}{2}$
$\vec{b}=\frac{\vec{a}+\overrightarrow{O C}}{2}$
$2 \vec{b}=\vec{a}+\overrightarrow{O C}$
$\overrightarrow{O C}=2 \vec{b}-\vec{a}$

Answer:$\frac{7}{3}\vec{a}+\frac{4}{3}\vec{b}$
Hint: You must know the rules of vector functions
Given: Write the position vector of the point which divides the join of points with position vectors $3\vec{a}-2\vec{b}$ &$2\vec{a}+3\vec{b}$ in ratio 2:1
Solution: Let R be the point which divides the line joining point with vectors.
$3\vec{a}-2\vec{b}$&$2\vec{a}+3\vec{b}$ in ratio 2:1
And
$\overrightarrow{O A}=3 \vec{a}-2 \vec{b}$
$\overrightarrow{O B}=2 \vec{a}+3 \vec{b}$
Here $m: n=2: 1$
Position vector of $\overline{O R}$is as follows
$\overline{O R}=\frac{m \overline{O B}+n \overline{O A}}{m+n}$
$=\frac{2(2 \vec{a}+3 \vec{b})+1(3 \vec{a}-2 \vec{b})}{2+1}$
$=\frac{4 \vec{a}+6 \vec{b}+3 \vec{a}-2 \vec{b}}{3}$
$=\frac{7 \vec{a}+4 \vec{b}}{3}$
$=\frac{7}{3} \vec{a}+\frac{4}{3} \vec{b}$

Class 12 RD Sharma chapter 22 exercise VSA solution deals with the chapter of Algebra of vectors, which might be a complex chapter for some students to solve. So RD Sharma class 12th exercise VSA provides you with very short answer type questions to take a self-test and evaluate your performance accordingly. The RD Sharma class 12 solutions chapter 22 exercise VSA consists of a total of 52 questions that covers all the essential concepts of the chapter mentioned below-

• Zero vector

• Unit vector

• Position vector

• Vectors in simplified form

• Centroid

• Direction Cosines

• Vectors with same magnitude and direction

Listed below are a few reasons why the RD Sharma class 12th exercise VSA is helpful in the preparation of exams:-

• The questions have a good proportion chance to be asked in the board exams because the exercises designed in the RD Sharma solution cover up each and every concept of every chapter

• The RD Sharma class 12th exercise VSA consists of questions that are frequently asked in board exams so students have to go through each and every exercise thoroughly.

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• The best part about RD Sharma class 12th exercise VSA is students can download the solutions free of cost from the Career360 website.

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## RD Sharma Chapter wise Solutions

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Yes, this material covers the entire syllabus

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