Question

A ray makes angles \( \alpha, 2\alpha, 3\alpha \) with the coordinate axes OX, OY, OZ respectively. Then possible values of \( \alpha \) are:

• A. \( \frac{\pi}{6}, \frac{\pi}{12} \)
• B. \( \frac{\pi}{2}, \frac{\pi}{3} \)
• C. \( \frac{\pi}{4}, \frac{3\pi}{3} \)
• D. \( \frac{\pi}{6}, \frac{\pi}{4} \)

Hint:

The sum of squares of direction cosines must equal 1. Try values that simplify trigonometric computations.

Answer 1

The Correct Option is D

Direction cosines of a vector making angles \( \theta_1, \theta_2, \theta_3 \) with the axes satisfy: \[ \cos^2 \theta_1 + \cos^2 \theta_2 + \cos^2 \theta_3 = 1 \] Let \( \theta_1 = \alpha,\ \theta_2 = 2\alpha,\ \theta_3 = 3\alpha \), then: \[ \cos^2 \alpha + \cos^2 2\alpha + \cos^2 3\alpha = 1 \] Try \( \alpha = \frac{\pi}{6} \): \[ \begin{align} \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2},\quad \cos 2\alpha = \cos \frac{\pi}{3} = \frac{1}{2},\quad \cos 3\alpha = \cos \frac{\pi}{2} = 0 \Rightarrow \left( \frac{\sqrt{3}}{2} \right)^2 + \left( \frac{1}{2} \right)^2 + 0 = \frac{3}{4} + \frac{1}{4} = 1 \] Now try \( \alpha = \frac{\pi}{4} \): \[ \begin{align} \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}},\quad \cos \frac{\pi}{2} = 0,\quad \cos \frac{3\pi}{4} = -\frac{1}{\sqrt{2}}
\Rightarrow \left( \frac{1}{\sqrt{2}} \right)^2 + 0 + \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{1}{2} + 0 + \frac{1}{2} = 1 \] Hence, both values satisfy the condition.