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Explain Solution R.D.Sharma Class 12 Chapter 22 Algebra of Vectors Exercise 22.9 Question 11 Maths Textbook Solution.

Answers (1)

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, $\begin{aligned} &x^{2}+3^{2}+0^{2}=18 \\ \end{aligned}$

$\begin{aligned} &\Rightarrow x^{2}+9=18 \\ \end{aligned}$

$\begin{aligned} &\Rightarrow x^{2}=9 \\ \end{aligned}$

$\begin{aligned} &\Rightarrow x=\pm 3 \\ \end{aligned}$

$\begin{aligned} &\therefore \vec{\mu}=3 \hat{i}+3 \hat{j} \text { or }-3 \hat{i}+3 \hat{j} \end{aligned}$

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