RD Sharma Solutions Class 12 Mathematics Chapter 22 VSA

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RD Sharma Solutions Class 12 Mathematics Chapter 22 VSA

RD Sharma Solutions Class 12 Mathematics Chapter 22 VSA

Satyajeet KumarUpdated on 24 Jan 2022, 06:21 PM IST

The RD Sharma mathematics collection is said to be the topmost solution for students to prepare themselves for board examinations. The CBSE students are most likely to get recommended by their teachers to get help from the RD Sharma class 12 solution of Algebra of vectors exercise VSA to guide for better understanding. RD Sharma solutions Therefore, without a doubt, any student can opt for the RD Sharma class 12th exercise VSA for a better understanding of maths.

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RD Sharma Class 12 Solutions Chapter22 VSA Algebra of vectors - Other Exercise
Algebra of Vectors Excercise: VSA
RD Sharma Chapter wise Solutions

RD Sharma Class 12 Solutions Chapter22 VSA Algebra of vectors - Other Exercise

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

Answer:

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

Answer:

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

Answer: 2
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

Answer: 13
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 $B=\frac{\text { Position vector of A+Position vector of } \mathrm{C}}{2}$
$\overrightarrow{O B}=\frac{\overline{O A}+\overrightarrow{O C}}{2}$
$\vec{b}=\frac{\vec{a}+\overrightarrow{O C}}{2}$
$2 \vec{b}=\vec{a}+\overrightarrow{O C}$
$\overrightarrow{O C}=2 \vec{b}-\vec{a}$

Algebra of Vectors Exercise very Short Answer type Question 52

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.
The solution also helps in solving the homework as it provides you with solved questions and examples and also because teachers use the solutions to assign homework.
The RD Sharma class 12 chapter 22 exercise VSA is trusted by many students across the country as they have seen that a good practice of the RD Sharma solution has helped them to score big.
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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