Question

If a line makes angles of \( \frac{3\pi}{4} \) and \( \frac{\pi}{3} \) with the positive directions of \( x \), \( y \), and \( z \)-axes respectively, then \( \theta \) is:

• A. \( -\frac{\pi}{3} \)
• B. \( \frac{\pi}{3} \) only
• C. \( \frac{\pi}{6} \)
• D. \( \pm \frac{\pi}{3} \)

Hint:

The direction cosines of a line are always in the range \( -1 \) to \( 1 \), and the angle can have multiple solutions due to the symmetry of the coordinate axes.

Answer 1

The Correct Option is D

To determine \( \theta \), given a line forming angles \( \frac{3\pi}{4} \) and \( \frac{\pi}{3} \) with the positive \( x \)- and \( y \)-axes respectively, we utilize direction cosines. For a line with direction angles \( \alpha, \beta, \gamma \) with the \( x, y, z \)-axes, the direction cosines are \( l = \cos \alpha \), \( m = \cos \beta \), and \( n = \cos \gamma \). These satisfy the fundamental relation \( l^2 + m^2 + n^2 = 1 \).

Substituting the given angles, we get \( l = \cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2} \) and \( m = \cos \frac{\pi}{3} = \frac{1}{2} \).

Now, we solve for \( n \):

\(\left(-\frac{\sqrt{2}}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + n^2 = 1\).

This simplifies to \( \frac{2}{4} + \frac{1}{4} + n^2 = 1 \), which further reduces to \( \frac{3}{4} + n^2 = 1 \).

Solving for \( n^2 \) yields \( n^2 = 1 - \frac{3}{4} = \frac{1}{4} \), so \( n = \pm \frac{1}{2} \).

Since \( n = \cos \gamma = \cos \theta \), we have \( \cos \theta = \pm \frac{1}{2} \).

Therefore, the possible values for \( \theta \) are \( \theta = \pm \frac{\pi}{3} \).

The solution is \( \pm \frac{\pi}{3} \).