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 \( l, m, n \) be the direction cosines of the line.
Step 1: Solve for \( n \) from \( l + m + n = 0 \).
From \( l + m + n = 0 \), we get \( n = -(l + m) \).

Step 2: Substitute \( n = -(l + m) \) into \( 2l^2 + 2m^2 - n^2 = 0 \).
Substituting \( n \): \( 2l^2 + 2m^2 - (-(l + m))^2 = 0 \).
Simplifying: \( 2l^2 + 2m^2 - (l^2 + 2lm + m^2) = 0 \).
This results in \( l^2 + m^2 - 2lm = 0 \).

Step 3: Factorize and solve.
Factoring gives \( (l - m)^2 = 0 \), which implies \( l = m \).

Step 4: Substitute \( l = m \) into \( l + m + n = 0 \).
This yields \( 2l + n = 0 \), so \( n = -2l \).

Step 5: Determine the angle between the lines.
The direction cosines of the two lines are proportional to \( (l, m, n) = (1, 1, -2) \) and \( (-1, -1, 2) \).
Since the direction cosines are negatives of each other, the lines are antiparallel. The angle between them is \( 180^\circ \).
\[\boxed{180^\circ}.\]