Question

If \(\alpha,2\alpha,3\alpha\) are angles made by a ray with \(OX\), \(OY\), \(OZ\) axes respectively, then all the possible values of \(\alpha\) are:

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

Hint:

If \(\alpha,\beta,\gamma\) are direction angles of a line, always use \[ \cos^2\alpha+\cos^2\beta+\cos^2\gamma=1 \] to find the required values.

Answer 1

The Correct Option is D

Step 1: Use the direction cosine property.
If a ray makes angles \[ \alpha,\ \beta,\ \gamma \] with the coordinate axes, then \[ \cos^2\alpha+\cos^2\beta+\cos^2\gamma=1 \] Given, \[ \beta=2\alpha,\quad \gamma=3\alpha \] Therefore, \[ \cos^2\alpha+\cos^22\alpha+\cos^23\alpha=1 \]

Step 2: Test the given options.
For \[ \alpha=\frac{\pi}{6}, \] \[ \cos^2\frac{\pi}{6} = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \] \[ \cos^2\frac{\pi}{3} = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \] \[ \cos^2\frac{\pi}{2} = 0 \] Thus, \[ \frac{3}{4}+\frac{1}{4}+0=1 \] Hence, \[ \alpha=\frac{\pi}{6} \] is valid.
Now check \[ \alpha=\frac{\pi}{4} \] \[ \cos^2\frac{\pi}{4} = \frac{1}{2} \] \[ \cos^2\frac{\pi}{2} = 0 \] \[ \cos^2\frac{3\pi}{4} = \left(-\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \] Thus, \[ \frac{1}{2}+0+\frac{1}{2}=1 \] Hence, \[ \alpha=\frac{\pi}{4} \] is also valid.

Step 3: Final conclusion.
Therefore, all possible values of \(\alpha\) are \[ \boxed{\frac{\pi}{6},\ \frac{\pi}{4}} \]