Question

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

Hint:

If two lines have direction cosines that are negatives of each other, the angle between them is \( 180^\circ \).

Answer 1

Let \( l, m, n \) represent the direction cosines of the line.
Step 1: Solve for \( n \) using \( l + m + n = 0 \)
From the first equation: \[ l + m + n = 0 \quad \implies \quad n = -(l + m). \]
Step 2: Substitute \( n = -(l + m) \) into the second equation:
\[ 2l^2 + 2m^2 - n^2 = 0. \]
Substitute \( n = -(l + m) \): \[ 2l^2 + 2m^2 - (-(l + m))^2 = 0. \]
Simplify: \[ 2l^2 + 2m^2 - (l^2 + 2lm + m^2) = 0. \]
\[ l^2 + m^2 - 2lm = 0. \]
Step 3: Factorize and solve:
\[ (l - m)^2 = 0 \quad \implies \quad l = m. \]
Step 4: Substitute \( l = m \) into \( l + m + n = 0 \):
\[ 2l + n = 0 \quad \implies \quad n = -2l. \]
Step 5: Determine the angle between the lines:
The direction cosines of the two lines are proportional to: \[ (l, m, n) = (1, 1, -2) \quad \text{and} \quad (-1, -1, 2). \]
Since the direction cosines are negatives of each other, the lines are antiparallel, and the angle between them is: \[ \boxed{180^\circ}. \]