Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Provide Solution For R.D. Sharma Maths Class 12 Chapter 22 Algebra of Vectors Exercise 22.9 Question 10 Maths Textbook Solution.

Provide Solution For  R.D. Sharma Maths Class 12 Chapter 22 Algebra of Vectors  Exercise 22.9 Question 10 Maths Textbook Solution.

Answers (1)

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 \thetawith \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 \thetais 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}

Posted by

infoexpert21

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE date sheet 2025 for Classes 10, 12 likely in November last week: Reports
anam-khan

CBSE allows Class 10, 12 students from disaffiliated schools to appear for board exam 2025 from same school
anam-khan

CBSE withdraws affiliation of 21 schools, downgrades 6 schools in two states; most in Delhi

Latest Question