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Need Solution for R.D.Sharma Maths Class 12 Chapter 22 Algebra of Vectors  Exercise very Short Answer type Question 47 Maths Textbook Solution.

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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 \thetawith 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}

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