Question

The acute angle between two lines such that the direction cosines \(l, m, n\), of each of them satisfy the equations \(l + m + n = 0\) and \(l^2 + m^2 - n^2 = 0\) is :

• A. \(15^\circ\)
• B. \(30^\circ\)
• C. \(60^\circ\)
• D. \(45^\circ\)

Hint:

Eliminate \(n\) using \(n = -(l+m)\); find the two direction-cosine sets and take the angle between them.

Answer 1

The Correct Option is C

From \(l + m + n = 0\), \(n = -(l+m)\), so \(n^2 = (l+m)^2\).
\(l^2 + m^2 - n^2 = 0\) becomes \(l^2 + m^2 - (l+m)^2 = -2lm = 0\), so \(lm = 0\).
If \(l = 0\): \(n = -m\), and with \(m^2 + n^2 = 1\) the direction cosines are \(\left(0, \tfrac{1}{\sqrt2}, -\tfrac{1}{\sqrt2}\right)\).
If \(m = 0\): \(n = -l\), giving \(\left(\tfrac{1}{\sqrt2}, 0, -\tfrac{1}{\sqrt2}\right)\).
\(\cos\theta = 0\cdot\tfrac{1}{\sqrt2} + \tfrac{1}{\sqrt2}\cdot 0 + \left(-\tfrac{1}{\sqrt2}\right)\left(-\tfrac{1}{\sqrt2}\right) = \tfrac12\).
\(\theta = 60^\circ\).
Answer: (3) \(60^\circ\)