Are vectors important for boards?

Yes, each and every chapter given in the RD Sharma textbook is definitely important for board exams.

Can I skip the algebra of vectors in board exams?

It is always said to consider each and every chapter of RD Sharma to be important for board exams, therefore skipping any chapter will not be a smart move

Is RD Sharma solution class 12 helpful for homework?

Yes, it is a benefit for students to use the RD Sharma solutions when solving homework to get help from the solved questions.

Are the questions in this exercise of the latest version?

Yes, the solutions that are provided in this exercise are thoroughly revised and updated for better knowledge.

Is this material available online?

Yes, it is available on the Careers360 website and also all other RD Sharma solutions can be found on this website.

RD Sharma Class 12 Exercise 22.9 Algebra of Vectors Solutions Maths - Download PDF Free Online

Schools
RD Sharma Solutions
RD Sharma Class 12 Exercise 22.9 Algebra of Vectors Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 22.9 Algebra of Vectors Solutions Maths - Download PDF Free Online

Updated on 24 Jan 2022, 06:20 PM IST

The RD Sharma class 12 solution of Algebra of Vectors exercise 22.9 is a tough chapter to be solved but not a tricky one. The RD Sharma class 12th exercise 22.9 explains some of the important factors of this chapter and helps you to go further by making previous concepts clear. The Class 12 RD Sharma chapter 22 exercise 22.9 solution consists of a total of 16 level 1 questions to acknowledge the concepts of the chapter.

This Story also Contains

RD Sharma Class 12 Solutions Chapter22 FBQ Algebra of vectors - Other Exercise
Algebra of Vectors Excercise: 22.9
RD Sharma Chapter wise Solutions

Also read - RD Sharma Solutions For Class 9 to 12 Maths

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

Algebra of Vectors Excercise: 22.9

Algebra of Vectors Exercise 22.9 Question 1

Answer: Yes
Given: Can a vector have direction angles $45^{o},60^{o},120^{o}$
Hint: Verify using $l^{2}+m^{2}+n^{2}=1$
Explanation: Given angles are $45^{o},60^{o},120^{o}$
Now the cosines of the direction angles are $\cos 45^{o},\cos 60^{o},\cos 120^{o}$
i.e. $\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{-1}{2}\left[\because \cos 45^{\circ}=\frac{1}{\sqrt{2}}, \cos 60^{\circ}=\frac{1}{2}, \cos 120^{\circ}=-\frac{1}{2}\right]$
We know if l ,m ,n be the direction cosines of any line then $l^{2}+m^{2}+n^{2}=1$
$\therefore$ L.H.S= $l^{2}+m^{2}+n^{2}$
$\begin{aligned} &=\left(\frac{1}{\sqrt{2}}\right)^{2}+\left(\frac{1}{2}\right)^{2}+\left(-\frac{1}{2}\right)^{2} \\ &=\frac{1}{2}+\frac{1}{4}+\frac{1}{4} \\ &=1 \end{aligned}$
=R.H.S
Hence $\frac{1}{\sqrt{2}},\frac{1}{2},\frac{-1}{2}$ are the direction cosines of the vector having direction angles $\cos 45^{o},\cos 60^{o},\cos 120^{o}$
So, the given angles can be the direction angles of a vector.

Algebra of Vectors Exercise 22.9 Question 2

Answer: 1, 1, 1 cannot be the direction cosines of a straight line.
Given: Prove that 1, 1, 1 cannot be the direction cosines of a straight line.
Hint: Check by using $l^{2}+m^{2}+n^{2}=1$
Explanation: We know if l, m, n are the direction cosines of any line then $l^{2}+m^{2}+n^{2}=1$
But here $1^{2}+1^{2}+1^{2}=3\neq 1$
So 1, 1, 1 can’t be the direction cosines for any line.

Algebra of Vectors Exercise 22.9 Question 3

Answer: $\frac{\pi }{2}$
Given: A vector makes an angle of $\frac{\pi }{4}$ with each of x-axis and y-axis.
Hint: Using $l^{2}+m^{2}+n^{2}=1$
Explanation: Let $\alpha ,\beta ,\gamma$ be the angle made with x- axis , y- axis and z- axis respectively.
According to given: $\alpha=\frac{\pi }{4};\beta=\frac{\pi }{4} ;\gamma =\gamma$
l, m, n be the direction cosines $\Rightarrow l^{2}+m^{2}+n^{2}=1$----- (A)
Here $I=\cos \alpha=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}} \Rightarrow I=\frac{1}{\sqrt{2}}$
$\begin{aligned} &m=\cos \beta=\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}} \Rightarrow m=\frac{1}{\sqrt{2}} \\ &n=\cos \gamma=\cos \gamma \Rightarrow n=\gamma \\ &\therefore \operatorname{By}(A) \\ &\left(\frac{1}{\sqrt{2}}\right)^{2}+\left(\frac{1}{2}\right)^{2}+\cos ^{2} \gamma=1 \\ &\Rightarrow \frac{1}{2}+\frac{1}{2}+\cos ^{2} \gamma=1 \\ &\Rightarrow 1+\cos ^{2} \gamma=1 \\ &\Rightarrow \cos ^{2} \gamma=0 \\ &\Rightarrow \cos \gamma=0 \\ &\Rightarrow \gamma=\cos ^{-1} 0=\cos ^{-1}\left(\cos \frac{\pi}{2}\right)=\frac{\pi}{2} \end{aligned}$
$\therefore$ The angle made by the vector with the z-axis is $\frac{\pi }{2}$ .

Algebra of Vectors Exercise 22.9 Question 4

Answer:$2\sqrt{3}\left ( \hat{i}+\hat{j}+\hat{k} \right )$
Given: A vector $\vec{\mu }$ is inclined at equal acute angles to x-axis, y-axis and z-axis . If $\mid \vec{\mu }\mid =6$ units, find $\mid \vec{\mu }\mid$
Hint: Find magnitude of $\vec{\mu }$
Explanation: Let any vector $\vec{\mu}=(\cos \alpha \hat{i}+\cos \beta \hat{j}+\cos \gamma \hat{k})$
For equal angles
$\begin{aligned} &\vec{\mu}=k(\hat{i}+\hat{j}+\hat{k}) \\ &|\vec{u}|=\sqrt{k^{2}+k^{2}+k^{2}}=6[\because \text { of given] } \\ &\Rightarrow 3 k^{2}=36 \\ &\Rightarrow k^{2}=12 \\ &\Rightarrow k=2 \sqrt{3} \\ &\therefore \vec{\mu}=2 \sqrt{3}(\hat{i}+\hat{j}+\hat{k}) \end{aligned}$

Algebra of Vectors Exercise 22.9 Question 5

Answer:$4\left ( \sqrt{2\hat{i}}+\hat{j}+\hat{k} \right )$
Given: A vector $\vec{\mu }$ is inclined at equal acute angles to x-axis at 45⁰and y-axis at 60⁰.If $\mid \vec{\mu }\mid =8$ units,
find $\vec{\mu }$
Hint: Use $\mu=|\vec{\mu}|(l \hat{i}+m \hat{j}+n \hat{k})$
Explanation: Here $\vec{\mu }$ makes an angle 45⁰with OX and 60⁰with OY.
So, $I=\cos 45^{\circ}=\frac{1}{\sqrt{2}} \text { and } m=\cos 60^{\circ}=\frac{1}{2}$
Now we know if l , m , n be the direction cosines then $l^{2}+m^{2}+n^{2}=1$
$\begin{aligned} &\Rightarrow\left(\frac{1}{\sqrt{2}}\right)^{2}+\left(\frac{1}{2}\right)^{2}+n^{2}=1 \\ &\Rightarrow \frac{1}{2}+\frac{1}{4}+n^{2}=1 \\ &\Rightarrow n^{2}=1-\frac{1}{2}-\frac{1}{4} \\ &\Rightarrow n^{2}=\frac{4-2-1}{4}=\frac{1}{4} \\ &\Rightarrow n=\pm \frac{1}{2} \end{aligned}$
Therefore $\begin{aligned} &\vec{\mu}=|\vec{u}|(l \hat{i}+m \hat{j}+n \hat{k}) \\ \end{aligned}$
$\begin{aligned} &=8\left(\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{2} \hat{j} \pm \frac{1}{2} \hat{k}\right) \\ &=4(\sqrt{2} \hat{i}+\hat{j} \pm \hat{k})[\because|\vec{\mu}|=8] \\ &\vec{\mu}=8\left(\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{2} \hat{j} \pm \frac{1}{2} \hat{k}\right) \end{aligned}$

Algebra of Vectors Exercise 22.9 Question 6 (i)

Answer:$\frac{2}{3},\frac{2}{3},\frac{-1}{3}$
Given: $2\hat{i}+2\hat{j}-\hat{k}$
Hint: Find $\cos \alpha ,\cos \beta ,\cos \gamma$
Explanation: Let $\vec{\mu }=2\hat{i}+2\hat{j}-\hat{k}$ be the given vector.
Then magnitude of vector $'\mu '$ is $|\vec{u}|=\sqrt{2^{2}+2^{2}+(-1)^{2}}=\sqrt{9}=3$
Let the direction cosines of vector ‘u’ be $\cos \alpha ,\cos \beta ,\cos \gamma$
We have $\begin{aligned} &\cos \alpha=\frac{\mu i}{|\vec{\mu}|}=\frac{2}{3} \\ \end{aligned}$
We have $\begin{aligned} &\cos \beta=\frac{\mu \cdot j}{|\vec{\mu}|}=\frac{2}{3} \\ \end{aligned}$
We have $\begin{aligned} &\cos \gamma=\frac{\mu \cdot k}{|\vec{\mu}|}=\frac{-1}{3} \end{aligned}$
$\therefore$ The direction cosines of given vector are$\frac{2}{3},\frac{2}{3},\frac{-1}{3}$

Algebra of Vectors Exercise 22.9 Question 6 (ii)

Answer:$\frac{6}{7},\frac{-2}{7},\frac{-3}{7}$
Given: $6\hat{i}-2\hat{j}-3\hat{k}$
Hint: Find $\cos \alpha ,\cos \beta ,\cos \gamma$
Explanation: Let $\mu =6\hat{i}-2\hat{j}-3\hat{k}$ be the given vector.
Then magnitude of vector ‘$\mu$ ’ is $|\vec{\mu}|=\sqrt{6^{2}+(-2)^{2}+(-3)^{2}}=\sqrt{49}=7$
Let the direction cosines of vector are $\cos \alpha ,\cos \beta ,\cos \gamma$
We have $\begin{aligned} &\cos \alpha=\frac{\mu i}{|\vec{\mu}|}=\frac{6}{7} \\ \end{aligned}$
We have $\begin{aligned} &\cos \beta=\frac{\mu \cdot j}{|\vec{\mu}|}=\frac{-2}{7} \\ \end{aligned}$
We have $\begin{aligned} &\cos \gamma=\frac{\mu-k}{|\vec{\mu}|}=\frac{-3}{7} \end{aligned}$
$\therefore$ The direction cosines of given vector are $\frac{6}{7},\frac{-2}{7},\frac{-3}{7}$

Algebra of Vectors Exercise 22.9 Question 6 (iii)

Answer:$\frac{3}{5},0,\frac{-4}{5}$
Given: $3\hat{i}-4\hat{k}$
Hint: Find $\cos \alpha ,\cos \beta ,\cos \gamma$
Explanation: Let $\mu =3\hat{i}+o\hat{j}-4\hat{k}$be the given vector.
Then magnitude of vector $'\mu '$ is $|\vec{\mu}|=\sqrt{3^{2}+0^{2}+(-4)^{2}}=\sqrt{25}=5$
Let the direction cosines of vector are $\cos \alpha ,\cos \beta ,\cos \gamma$
We have $\begin{aligned} &\cos \alpha=\frac{\mu i}{|\vec{\mu}|}=\frac{3}{5} \\ \end{aligned}$
We have $\begin{aligned} &\cos \beta=\frac{\mu \cdot j}{|\vec{\mu}|}=\frac{0}{5} \\ \end{aligned}$
We have $\begin{aligned} &\cos \gamma=\frac{\mu \cdot k}{|\vec{\mu}|}=\frac{-4}{5} \end{aligned}$
$\therefore$ The direction cosines of given vector are$\frac{3}{5},0,\frac{-4}{5}$

Algebra of Vectors Exercise 22.9 Question 7 (i)

Answer: $\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right), \cos ^{-1}\left(\frac{-1}{\sqrt{3}}\right), \cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
Given: $\hat{i}-\hat{j}+\hat{k}$
Hint: Find $\cos \alpha ,\cos \beta ,\cos \gamma$
Explanation: Let $\mu =\hat{i}-\hat{j}+\hat{k}$ be the given vector
Then magnitude of vector $\mu$ is $|\vec{\mu}|=\sqrt{1^{2}+(-1)^{2}+(1)^{2}}=\sqrt{3}$
Let the direction angles of vector are $\alpha ,\beta ,\gamma$
We have $\begin{aligned} &\cos \alpha=\frac{\mu i}{|\vec{\mu}|}=\frac{1}{\sqrt{3}} \Rightarrow \alpha=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right) \\ \end{aligned}$
We have $\begin{aligned} &\cos \beta=\frac{\mu \cdot j}{|\vec{\mu}|}=\frac{-1}{\sqrt{3}} \Rightarrow \beta=\cos ^{-1}\left(\frac{-1}{\sqrt{3}}\right) \\ \end{aligned}$
We have $\begin{aligned} &\cos \gamma=\frac{\mu \cdot k}{|\vec{\mu}|}=\frac{1}{\sqrt{3}} \Rightarrow \gamma=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right) \end{aligned}$
$\therefore$ The angles of the given vector are $\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right), \cos ^{-1}\left(\frac{-1}{\sqrt{3}}\right), \cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)$

Algebra of Vectors Exercise 22.9 Question 7 (ii)

Answer: $\frac{\pi }{2},\frac{\pi }{4},\frac{3\pi }{4}$
Given:$\hat{j}-\hat{k}$
Hint: Find $\cos \alpha ,\cos \beta ,\cos \gamma$
Explanation: Let $\vec{u}=\hat{j}-\hat{k}$ be the given vector
The magnitude of vector $\mu$ is $|\mu|=\sqrt{0^{2}+(1)^{2}+(-1)^{2}}=\sqrt{2}$
Let the direction angle of the vector are $\alpha ,\beta ,\gamma$
We have $\cos \alpha=\frac{\mu i}{|\mu|}=\frac{0}{\sqrt{2}}=0 \Rightarrow \alpha=\cos ^{-1}(0)=\frac{\pi}{2}\left[\because \cos \frac{\pi}{2}=0\right]$
We have $\cos \beta=\frac{\mu \cdot j}{|\mu|}=\frac{1}{\sqrt{2}} \Rightarrow \beta=\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}\left[\because \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}\right]$
We have $\cos \gamma=\frac{\mu . k}{|\mu|}=\frac{-1}{\sqrt{2}} \Rightarrow \gamma=\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)=\frac{3 \pi}{4}\left[\because \cos \frac{3 \pi}{4}=\frac{-1}{\sqrt{2}}\right]$
$\therefore$ The angles of the given vector are$\frac{\pi }{2},\frac{\pi }{4},\frac{3\pi }{4}$

Algebra of Vectors Exercise 22 .9 Question 7 (iii)

Answer:$\cos ^{-1}\left(\frac{4}{9}\right), \cos ^{-1}\left(\frac{8}{9}\right), \cos ^{-1}\left(\frac{1}{9}\right)$
Given: $4\hat{i}+8\hat{j}+\hat{k}$
Hint: Find $\cos \alpha ,\cos \beta ,\cos \gamma$
Explanation: Let $\vec{u}=4\hat{i}+8\hat{j}+\hat{k}$ be the given vector
The magnitude of vector $\mu$ is $|\mu|=\sqrt{4^{2}+(8)^{2}+(1)^{2}}=\sqrt{81}=9$
Let the direction angle of the vector are $\alpha ,\beta ,\gamma$
We have $\begin{aligned} &\cos \alpha=\frac{\mu i}{|\mu|}=\frac{4}{9} \Rightarrow \alpha=\cos ^{-1}\left(\frac{4}{9}\right) \\ \end{aligned}$
We have $\begin{aligned} &\cos \beta=\frac{\mu \cdot j}{|\mu|}=\frac{8}{9} \Rightarrow \beta=\cos ^{-1}\left(\frac{8}{9}\right) \\ \end{aligned}$
We have $\begin{aligned} &\cos \gamma=\frac{\mu \cdot j}{|\mu|}=\frac{1}{9} \Rightarrow \gamma=\cos ^{-1}\left(\frac{1}{9}\right) \end{aligned}$
$\therefore$ The angles of the given vector are $\cos ^{-1}\left(\frac{4}{9}\right), \cos ^{-1}\left(\frac{8}{9}\right), \cos ^{-1}\left(\frac{1}{9}\right)$

Algebra of Vectors Exercise 22.9 Question 8

Answer: Direction cosines are equal
Given: Show that the vector $\hat{i}+\hat{j}+\hat{k}$is equally inclined with the axis OX, OY and OZ.
Hint: Find $\mid \vec{a}\mid$then direction direction cosines
Explanation: Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}=1 \hat{i}+1 \hat{j}+1 \hat{k}$
A vector is equally inclined to OX, OY, OZ
i.e. X, Y and Z axis respectively.
If its direction cosines are equal
Direction ratios of $\vec{a}$ are a=1,b=1,c=1
Magnitude of $\begin{aligned} &\vec{a}=\sqrt{1^{2}+(1)^{2}+(1)^{2}} \\ \end{aligned}$
$\begin{aligned} &|\vec{a}|=\sqrt{3} \\ \end{aligned}$
Direction cosines of $\begin{aligned} &|\vec{a}| \text { are }\left(\frac{a}{|\vec{a}|}\left|\frac{b}{\mid \vec{b}}\right|^{\prime} \mid \overrightarrow{\vec{c} \mid}\right) \\ \end{aligned}$
$\begin{aligned} &=\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \end{aligned}$
Since the direction cosines are equal
$\vec{a}=\hat{i}+\hat{j}+\hat{k}$is equally inclined to OX, OY and OZ.

Algebra of Vectors Exercise 22.9 Question 9

Answer: Direction cosines are equally inclined to axis.
Given: Show that the direction cosines of a vector equally inclined to the axis OX, OY and OZ are $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}.$
Hint: Find $\mid \vec{\mu } \mid$
Explanation: Let the required vector be $\vec{\mu } =a\hat{i}+b\hat{j}+c\hat{k}$
Direction ratios are a, b, c
Since the vector is equally inclined to axis OX, OY and OZ, thus the direction cosines are equal.
$\frac{a}{\text { magnitude } \vec{\mu}}=\frac{b}{\text { magnitude } \vec{\mu}}=\frac{c}{\text { magnitude } \vec{\mu}}$
Since $a=b=c$
$\therefore$ The vector is $\vec{\mu } =a\hat{i}+a\hat{j}+a\hat{k}$
Magnitude of $\begin{aligned} &\vec{\mu}=\sqrt{(a)^{2}+(a)^{2}+(a)^{2}} \\ \end{aligned}$
$\begin{aligned} &\Rightarrow|\vec{\mu}|=\sqrt{3 a^{2}}=\sqrt{3} a \\ \end{aligned}$
Direction cosines are $\begin{aligned} &\left(\frac{a}{\sqrt{3} a}, \frac{b}{\sqrt{3} a}, \frac{c}{\sqrt{3} a}\right) \end{aligned}$
$=\begin{aligned} &\left(\frac{a}{\sqrt{3} a}, \frac{b}{\sqrt{3} a}, \frac{c}{\sqrt{3} a}\right) \end{aligned}$
$=\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}.$
Hence Proved

Algebra of Vectors Exercise 22.9 Question 10

Answer:$\theta =\frac{\pi }{3}$, components of $\vec{a}$ are $\frac{1}{2},\frac{1}{\sqrt{2}},\frac{1}{2}$
Given: A unit vector $\vec{a}$ makes an angle $\frac{\pi }{3}$with$\hat{i},\frac{\pi }{3}$with $\hat{j}$and an acute angle $\theta$with $\hat{k}$, then find $\theta$ and hence the component of $\vec{a}$
Hint: Find x, y, z
Explanation: Let us take a unit vector $\vec{a}=x\hat{i}+y\hat{j}+z\hat{k}$
So magnitude of $\vec{a}=\mid \vec{a}\mid =1$
Angle $\vec{a}$ with $\hat{i}=\frac{\pi }{3}$
$\begin{aligned} &\vec{a} \cdot \hat{i}=|\vec{a}||\vec{i}| \cos \frac{\pi}{3} \\ &(x \hat{i}+\hat{y j}+z \hat{k}) \cdot \hat{i}=1 \times 1 \times \frac{1}{2} \\ &(x \hat{i}+\hat{y j}+z \hat{k})(1 \hat{i}+0 \hat{j}+0 \hat{k})=\frac{1}{2} \\ &(x \times 1)+(y \times 0)+(z \times 0)=\frac{1}{2} \\ &x+0+0=\frac{1}{2} \\ &x=\frac{1}{2} \end{aligned}$
Angle of $\vec{a}$ with $\hat{j}=\frac{\pi }{4}$
$\begin{aligned} &\vec{a} \cdot \hat{j}=|\vec{a}||\vec{j}| \cos \frac{\pi}{4} \\ &(x \hat{i}+\hat{j}+z \hat{k}) \cdot \hat{j}=1 \times 1 \times \frac{1}{\sqrt{2}} \\ &(x \hat{i}+\hat{j}+z \hat{k}) \cdot(0 \hat{i}+1 \hat{j}+0 \hat{k})=\frac{1}{\sqrt{2}} \\ &(x \times 0)+(y \times 1)+(z \times 0)=\frac{1}{\sqrt{2}} \\ &0+y+0=\frac{1}{\sqrt{2}} \end{aligned}$
$y=\frac{1}{\sqrt{2}}$
Also,
Angle of $\vec{a}$ with $\vec{k}=0$
$\begin{aligned} &\vec{a} \cdot \hat{k}=|\vec{a}||\vec{k}| \times \cos \theta \\ &(x \hat{i}+\hat{j}+z \hat{k}) \cdot(0 \hat{i}+0 \hat{j}+1 \hat{k})=1 \times 1 \times \cos \theta \\ &(\mathrm{x} \times 0)+(\mathrm{y} \times 0)+(\mathrm{z} \times 1)=\cos \theta \\ &0+0+\mathrm{z}=\cos \theta \\ &\mathrm{Z}=\cos \theta \end{aligned}$
Now,
Magnitude of $\vec{a}=\sqrt{\left ( x \right )^{2}+\left ( y \right )^{2}+\left ( z \right )^{2}}$
$\begin{aligned} &1=\sqrt{\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}+\cos ^{2}} \theta \\ &1=\sqrt{\left(\frac{1}{4}\right)+\left(\frac{1}{2}\right)+\cos ^{2}} \theta \\ &1=\sqrt{\frac{3}{4}+\cos ^{2}} \theta \\ \end{aligned}$

Squaring on both sides

$\left ( \sqrt{\frac{3}{4}}+\cos ^{2}\theta ^{2} \right )=1^{2}$
$\begin{aligned} &\frac{3}{4}+\cos ^{2} \theta=1 \\ &\cos ^{2} \theta=1-\frac{3}{4} \\ &\cos ^{2} \theta=\frac{1}{4} \\ &\cos \theta=\pm \frac{1}{2} \end{aligned}$
Since $\theta$ is an acute angle
So,$\theta < 90^{o}$
$\therefore \theta$ is in 1st Quadrant
And $\cos \theta$is positive in 1st Quadrant
So, $\cos \theta =\frac{1}{2}$
$\theta =60^{o}=\frac{\pi }{3}$
And $z=\cos \theta =\cos 60^{o}=\frac{1}{2}$
Hence $x=\frac{1}{2},y=\frac{1}{2},z=\frac{1}{2}$
The required vector $\vec{a}$ is $\frac{1}{2}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{2}\hat{k}$
So, components of $\vec{a}$ are $\frac{1}{2},\frac{1}{\sqrt{2}} and \frac{1}{2}$

Algebra of Vectors Exercise 22.9 Question 11

Answer:$\vec{\mu }=3\hat{i}+3\hat{j} or -3\hat{i}+3\hat{j}$
Given: Find a vector $\vec{\mu }$ of magnitude $3\sqrt{2}$ units which makes an angle of angle of $\frac{\pi }{4}$and $\frac{\pi }{2}$ with y and z axis respectively.
Hint: Use given conditions to find $\mid \vec{r}\mid$
Explanation: let $\begin{aligned} &\vec{\mu}=x \hat{i}+y \hat{j}+z \hat{k} \\ \end{aligned}$
$\begin{aligned} &|\vec{u}|=3 \sqrt{2} \\ \end{aligned}$
$\begin{aligned} &\sqrt{x^{2}+y^{2}+z^{2}}=3 \sqrt{2} \\ \end{aligned}$
Squaring on both sides
$\begin{aligned} &x^{2}+y^{2}+z^{2}=18 \end{aligned}$
It makes $\frac{\pi }{4}$with y-axis
So, $\begin{aligned} &\vec{\mu} \cdot \hat{j}=|\vec{u}| \cos \frac{\pi}{4} \\ \end{aligned}$
$\begin{aligned} &\Rightarrow y=\frac{3 \sqrt{2}}{\sqrt{2}}=3\left[\because \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}\right] \\ \end{aligned}$
Also it makes $\frac{\pi }{2}$with z-axis
So,$\begin{aligned} &\vec{\mu} \cdot \hat{k}=|\vec{u}| \cos \frac{\pi}{2} \\ \end{aligned}$
$\begin{aligned} &Z=0\left[\because \cos \frac{\pi}{2}=0\right] \\ \end{aligned}$
So,$\begin{aligned} &x^{2}+3^{2}+0^{2}=18 \\ \end{aligned}$
$\begin{aligned} &\Rightarrow x^{2}+9=18 \\ \end{aligned}$
$\begin{aligned} &\Rightarrow x^{2}=9 \\ \end{aligned}$
$\begin{aligned} &\Rightarrow x=\pm 3 \\ \end{aligned}$
$\begin{aligned} &\therefore \vec{\mu}=3 \hat{i}+3 \hat{j} \text { or }-3 \hat{i}+3 \hat{j} \end{aligned}$

Algebra of Vectors Exercise 22.9 Question 12

Answer: $\vec{r}$= $\pm 2\left ( \hat{i}+\hat{j}+\hat{k} \right )$
Given: A vector $\vec{r}$ is inclined at equal angles to the three axis. If the magnitude of $\vec{r}$ is $2\sqrt{3}$,find $\vec{r}$
Hint: Use $\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1$
Explanation: Let$\alpha ,\beta ,\gamma$ be the angles inclined to the three axis.
Since $\vec{r}$ is inclined equal angle to axis
$\begin{aligned} &\therefore \alpha=\beta=\gamma \\ &\Rightarrow 3 \cos ^{2} \alpha=1 \\ &\therefore \cos \alpha=\pm \frac{1}{\sqrt{3}} \\ \end{aligned}$
So,
$\begin{aligned} &\vec{r}=\pm \frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k}) \\ &\therefore \vec{r}=|\vec{r}| \hat{r} \\ &=2 \sqrt{3} \times\left(\pm \frac{1}{\sqrt{3}}\right)(\hat{i}+\hat{j}+\hat{k}) \\ &=\pm 2(\hat{i}+\hat{j}+\hat{k}) \\ &\therefore \vec{r}=\pm 2(\hat{i}+\hat{j}+\hat{k}) \end{aligned}$

RD Sharma class 12th exercise 22.9 is prepared specifically for students who want to gain more knowledge on the core concepts of the subject. To strengthen their basics and score good marks in exams, students can refer to this material. It is prepared by a group of subject experts who have years of experience with CBSE exam papers.

Moreover, RD Sharma class 12 solution chapter 22 exercise 22.9 solutions contain step-by-step answers which the students can refer to if they find any question difficult. As this material is updated to the latest version of the book, students can prepare for their exams and finish their homework.

RD Sharma class 12th exercise 22.9 material is the best alternative for students as it contains questions and answers in one place. Therefore making it easier to revise and prepare. However, as mathematics includes hundreds of questions, it gets difficult for students to remember all concepts.

This material is made specifically to ease the burden on students. They can prepare systematically and quickly come back to any problem which they do not understand. Every answer from this material goes through a series of quality checks to ensure that the best information is available for students.

Additionally, the RD Sharma class 12th exercise 22.9 material is available on Career360's website free of cost. Students can access them through any device with an internet connection. This means that they can use their mobile or laptop to study from their homes.