KCET · Maths · Probability
If a line makes an angle of with each of \(X\) and \(Y\)-axis, then the acute angle made by \(Z\)-axis is
- A \(\frac{\pi}{4}\)
- B \(\frac{\pi}{6}\)
- C \(\frac{\pi}{3}\)
- D \(\frac{\pi}{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{\pi}{4}\)
Step-by-step Solution
Detailed explanation
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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