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If a line makes an angle of with each of $X$ and $Y$-axis, then the acute angle made by $Z$-axis is
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Verified Answer
The correct answer is:
$\frac{\pi}{4}$
Given, $l=m=\cos \frac{\pi}{3}=\frac{1}{2}$
$\begin{array}{lc}
\therefore & l^{2}+m^{2}+n^{2}=1 \\
\Rightarrow & \left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}+n^{2}=1 \\
\Rightarrow & \frac{1}{4}+\frac{1}{4}+n^{2}=1 \\
\Rightarrow & n^{2}=\frac{1}{2} \Rightarrow n=\pm \frac{1}{\sqrt{2}}
\end{array}$
Then the acute angle made by $Z$-axis is $\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}$
$\begin{array}{lc}
\therefore & l^{2}+m^{2}+n^{2}=1 \\
\Rightarrow & \left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}+n^{2}=1 \\
\Rightarrow & \frac{1}{4}+\frac{1}{4}+n^{2}=1 \\
\Rightarrow & n^{2}=\frac{1}{2} \Rightarrow n=\pm \frac{1}{\sqrt{2}}
\end{array}$
Then the acute angle made by $Z$-axis is $\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}$
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