KCET · Maths · Three Dimensional Geometry
If a line makes an angle of \(\frac{\pi}{3}\) with each \(X\) and \(Y\) axis, then the acute angle made by \(\mathrm{Z}\)-axis is
- A \(\frac{\pi}{3}\)
- B \(\frac{\pi}{2}\)
- C \(\frac{\pi}{4}\)
- D \(\frac{\pi}{6}\)
Answer & Solution
Correct Answer
(C) \(\frac{\pi}{4}\)
Step-by-step Solution
Detailed explanation
Given, \(\alpha=\beta=\frac{\pi}{3}\)
Let acute angle made by \(Z\)-axis be \(\gamma\).
We know that
\(\begin{aligned} & \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1 \\ & \Rightarrow \quad\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2+\cos ^2 \gamma=1 \\ & \Rightarrow \quad \frac{1}{4}+\frac{1}{4}+\cos ^2 \gamma=1 \\ & \Rightarrow \quad \cos ^2 \gamma=1-\frac{1}{2} \Rightarrow \cos \gamma= \pm \frac{1}{\sqrt{2}} \\ & \Rightarrow \quad \gamma=\frac{\pi}{4}\end{aligned}\)
[ \(\because y\) is acute]
Let acute angle made by \(Z\)-axis be \(\gamma\).
We know that
\(\begin{aligned} & \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1 \\ & \Rightarrow \quad\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2+\cos ^2 \gamma=1 \\ & \Rightarrow \quad \frac{1}{4}+\frac{1}{4}+\cos ^2 \gamma=1 \\ & \Rightarrow \quad \cos ^2 \gamma=1-\frac{1}{2} \Rightarrow \cos \gamma= \pm \frac{1}{\sqrt{2}} \\ & \Rightarrow \quad \gamma=\frac{\pi}{4}\end{aligned}\)
[ \(\because y\) is acute]
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