If a line makes an angle of \(\frac{\pi}{3}\) with each of the x and y-axis, then we need to find the acute angle made by the line with the z-axis.
Let \(\alpha, \beta, \gamma\) be the angles the line makes with the x, y, and z-axis respectively. Then \(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1\).
We are given that \(\alpha = \frac{\pi}{3}\) and \(\beta = \frac{\pi}{3}\). Therefore:
\(\cos^2(\frac{\pi}{3}) + \cos^2(\frac{\pi}{3}) + \cos^2 \gamma = 1\)
\((\frac{1}{2})^2 + (\frac{1}{2})^2 + \cos^2 \gamma = 1\)
\(\frac{1}{4} + \frac{1}{4} + \cos^2 \gamma = 1\)
\(\frac{1}{2} + \cos^2 \gamma = 1\)
\(\cos^2 \gamma = 1 - \frac{1}{2} = \frac{1}{2}\)
\(\cos \gamma = \pm \sqrt{\frac{1}{2}} = \pm \frac{1}{\sqrt{2}}\)
Since we want the acute angle, we take the positive value:
\(\cos \gamma = \frac{1}{\sqrt{2}}\)
\(\gamma = \cos^{-1}(\frac{1}{\sqrt{2}}) = \frac{\pi}{4}\)
Therefore, the acute angle made by the line with the z-axis is \(\frac{\pi}{4}\).
Thus, the correct option is (A) \(\frac{\pi}{4}\).
List - I | List - II | ||
(P) | γ equals | (1) | \(-\hat{i}-\hat{j}+\hat{k}\) |
(Q) | A possible choice for \(\hat{n}\) is | (2) | \(\sqrt{\frac{3}{2}}\) |
(R) | \(\overrightarrow{OR_1}\) equals | (3) | 1 |
(S) | A possible value of \(\overrightarrow{OR_1}.\hat{n}\) is | (4) | \(\frac{1}{\sqrt6}\hat{i}-\frac{2}{\sqrt6}\hat{j}+\frac{1}{\sqrt6}\hat{k}\) |
(5) | \(\sqrt{\frac{2}{3}}\) |