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 solve the problem of finding \( \theta \), given a line that makes angles of \( \frac{3\pi}{4} \) and \( \frac{\pi}{3} \) with the positive \( x \)- and \( y \)-axes respectively, we start by using the property of the direction cosines. For a line making angles \( \alpha, \beta, \gamma \) with the \( x, y, z \)-axes respectively, the direction cosines \( l, m, \) and \( n \) are given by:

\( l = \cos \alpha \), \( m = \cos \beta \), and \( n = \cos \gamma \).

These satisfy the equation:

\( l^2 + m^2 + n^2 = 1 \).

Substitute the given angles:

\( l = \cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2} \).

\( m = \cos \frac{\pi}{3} = \frac{1}{2} \).

Now, solve for \( n \):

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

\(\frac{2}{4} + \frac{1}{4} + n^2 = 1\).

\(\frac{3}{4} + n^2 = 1\).

\(n^2 = 1 - \frac{3}{4} = \frac{1}{4}\).

\(n = \pm \frac{1}{2}\).

\(n = \cos \gamma = \cos \theta = \pm \frac{1}{2}\).

Thus, the values of \( \theta \) are:

\(\theta = \pm \frac{\pi}{3}\).

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