Question:
Find the angle between the lines whose direction cosines are given by the equations \(l + m + n = 0\), \(l^2 + m^2 - n^2 = 0\).
Solution:
The given equations are
\(l + m + n = 0\) ......(i)
\(l^2 + m^2 - n^2 = 0\) .......(ii)
From equation (i) \(n = - (l + m)\)
Putting the value of \(n\) in equation (ii) we get
\(l^2 + m^2 + [ - (l + m)^2] = 0\)
⇒ \(l^2 + m^2 - l^2 - m^2 - 2lm = 0\)
⇒ \( - 2lm = 0\)
⇒ \(lm = 0\)
⇒ \((- m - n)m = 0\) .....[∵ \(l = - m - n\)]
⇒ \((m + n)m = 0\)
⇒ \(m = 0\) or \(m = - n\)
⇒ \(l = 0\) or \(l = - n\)
∴ Direction cosines of the two lines are
0, - n, n and - n, 0, n
⇒ 0, - 1, 1 and - 1, 0, 1
\(\therefore \cos \theta = \frac{{(0i - j + k) \cdot ( - i + 0j + k)}}{{\sqrt {{{(0)}^2} + {{( - 1)}^2} + {{(1)}^2}} \sqrt {{{( - 1)}^2} + {{(0)}^2} + {{(1)}^2}} }}\)
\(\Rightarrow \cos \theta = \frac{1}{{\sqrt 2 \sqrt 2 }}\)
\(\Rightarrow \theta = \frac{\pi }{3}\)
Thus, Required angle is \(\frac{\pi }{3}\)