Question

Each of the angles \( \beta \) and \( \gamma \) that a given line makes with the positive y- and z-axes, respectively, is half the angle that this line makes with the positive x-axis. Then the sum of all possible values of the angle \( \beta \) is

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

Hint:

Use the relationship between the direction cosines of a line and the angles it makes with the coordinate axes: \( l = \cos \alpha \), \( m = \cos \beta \), \( n = \cos \gamma \), and \( l^2 + m^2 + n^2 = 1 \). Substitute the given relationships between the angles and solve the resulting trigonometric equation. Remember to consider the possible range of angles with the positive axes.

Answer 1

The Correct Option is A

To resolve the problem, we examine the specified conditions regarding the angles a line forms with the axes. Let \( \alpha \) denote the angle with the x-axis, \( \beta \) with the y-axis, and \( \gamma \) with the z-axis.

The problem states:

• \( \beta = \frac{\alpha}{2} \)
• \( \gamma = \frac{\alpha}{2} \)
• The objective is to determine the sum of all valid \( \beta \) values.

The direction cosines of the line with respect to the x-, y-, and z-axes are \( \cos \alpha \), \( \cos \beta \), and \( \cos \gamma \), respectively.

The fundamental identity relating direction cosines is:

\(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1\)

Substituting the given relationships into this identity yields:

• \(\cos^2 \alpha + \left(\cos \frac{\alpha}{2}\right)^2 + \left(\cos \frac{\alpha}{2}\right)^2 = 1\)

This equation simplifies to:

\(\cos^2 \alpha + 2 \cos^2 \frac{\alpha}{2} = 1\)

Employing the half-angle identity \( \cos^2 \frac{\alpha}{2} = \frac{1 + \cos \alpha}{2} \), we substitute and obtain:

\(\cos^2 \alpha + 2 \times \frac{1 + \cos \alpha}{2} = 1\)

Further simplification results in:

\(\cos^2 \alpha + 1 + \cos \alpha = 1\)

Rearranging the terms gives:

\(\cos^2 \alpha + \cos \alpha = 0\)

Factoring the quadratic equation yields:

\(\cos \alpha (\cos \alpha + 1) = 0\)

This equation presents two possible solutions for \( \alpha \):

• \( \cos \alpha = 0 \), which implies \( \alpha = \frac{\pi}{2} \)
• \( \cos \alpha = -1 \), which implies \( \alpha = \pi \)

Substituting these values back to find the corresponding \( \beta \) values:

• When \( \alpha = \frac{\pi}{2} \): \( \beta = \frac{\pi}{4} \)
• When \( \alpha = \pi \): \( \beta = \frac{\pi}{2} \)

The sum of all possible \( \beta \) values is:

\(\frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4}\)

Consequently, the total sum of all permissible \( \beta \) values is \(\frac{3\pi}{4}\).