Question

If a line makes an angle of \( 30^\circ \) with the positive direction of \( x \)-axis, \( 120^\circ \) with the positive direction of \( y \)-axis, then the angle which it makes with the positive direction of \( z \)-axis is:

• A. \( 90^\circ \)
• B. \( 120^\circ \)
• C. \( 60^\circ \)
• D. \( 0^\circ \)

Hint:

To find unknown angles in direction cosines, use the identity \( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \) and solve for the missing angle.

Answer 1

The Correct Option is A

Solution: Step 1: Apply the direction cosine relation.
The angles \( \alpha \), \( \beta \), and \( \gamma \) that a line makes with the positive \( x \)-, \( y \)-, and \( z \)-axes, respectively, follow the direction cosine equation: \[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1. \] Given: \[ \alpha = 30^\circ, \quad \beta = 120^\circ, \quad \gamma = ?. \] Step 2: Compute \( \cos \alpha \) and \( \cos \beta \).
We know: \[ \cos \alpha = \cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \cos \beta = \cos 120^\circ = -\frac{1}{2}. \] Thus: \[ \cos^2 \alpha = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}, \quad \cos^2 \beta = \left(-\frac{1}{2}\right)^2 = \frac{1}{4}. \] Step 3: Solve for \( \cos^2 \gamma \).
Substitute the known values into the direction cosine equation: \[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1, \] \[ \frac{3}{4} + \frac{1}{4} + \cos^2 \gamma = 1 \quad \Rightarrow \quad \cos^2 \gamma = 1 - \left(\frac{3}{4} + \frac{1}{4}\right). \] Simplify: \[ \cos^2 \gamma = 1 - 1 = 0. \] Step 4: Determine \( \gamma \).
Since \( \cos^2 \gamma = 0 \), it follows that: \[ \cos \gamma = 0. \] The angle \( \gamma \) for which \( \cos \gamma = 0 \) is: \[ \gamma = 90^\circ. \] Step 5: Final Answer.
Thus, the angle that the line makes with the positive direction of the \( z \)-axis is: \[ \boxed{90^\circ}. \]