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Please Solve R.D.Sharma class 12 Chapter 22 Algbra of Vectors Exercise 22.9 Question 3 Maths Textbook Solution.

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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} .

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