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  • 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}

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