Question

The angle between the lines whose direction cosines satisfy the equations: \[ l + m + n = 0 \quad \text{and} \quad m^2 + n^2 - l^2 = 0 \] Find the angle between the two lines.

• A. 30°
• B. 45°
• C. 60°
• D. 90°

Hint:

When dealing with direction cosines, use the relations between \( l \), \( m \), and \( n \) to simplify the problem, and apply the formula for the angle between two lines.

Answer 1

The Correct Option is C

We are given the following two equations that describe the direction cosines of two lines: 1) \( l + m + n = 0 \) 2) \( m^2 + n^2 - l^2 = 0 \) We are asked to find the angle between these two lines. The direction cosines of the lines are represented by \( l \), \( m \), and \( n \), where: - \( l \) is the cosine of the angle between the line and the \( x \)-axis, - \( m \) is the cosine of the angle between the line and the \( y \)-axis, and - \( n \) is the cosine of the angle between the line and the \( z \)-axis.
Step 1: Use the first equation to express \( n \) From the first equation \( l + m + n = 0 \), we can express \( n \) in terms of \( l \) and \( m \): \[ n = -l - m \]
Step 2: Substitute into the second equation Substitute this expression for \( n \) into the second equation \( m^2 + n^2 - l^2 = 0 \): \[ m^2 + (-l - m)^2 - l^2 = 0 \] Expanding the terms: \[ m^2 + (l^2 + 2lm + m^2) - l^2 = 0 \] Simplifying: \[ m^2 + l^2 + 2lm + m^2 - l^2 = 0 \] \[ 2m^2 + 2lm = 0 \]
Step 3: Factor the equation Factor the equation: \[ 2m(m + l) = 0 \] Thus, either \( m = 0 \) or \( m = -l \).
Step 4: Solve for the angle between the lines Let's consider the case where \( m = -l \). Substituting \( m = -l \) into \( n = -l - m \): \[ n = -l - (-l) = 0 \] Thus, the direction cosines of the lines are: \[ l, -l, 0 \] The formula for the angle \( \theta \) between two lines in terms of their direction cosines \( l_1, m_1, n_1 \) and \( l_2, m_2, n_2 \) is given by: \[ \cos \theta = l_1 l_2 + m_1 m_2 + n_1 n_2 \] Substituting \( l_1 = l, m_1 = -l, n_1 = 0 \) and \( l_2 = l, m_2 = -l, n_2 = 0 \): \[ \cos \theta = l^2 + (-l)^2 + 0 = 2l^2 \] For \( \cos \theta = \frac{1}{2} \), we get \( l^2 = \frac{1}{4} \), and thus \( l = \frac{1}{2} \). Finally, we find that the angle \( \theta \) is: \[ \theta = \cos^{-1} \left( \frac{1}{2} \right) = 60^\circ \] Thus, the angle between the lines is \( \boxed{60^\circ} \).