Question:
Find the angle between the lines whose direction cosines are given by the equations l + m + n = 0, l2 + m2 – n2 = 0.
Solution:
Given equations are,
l + m + n = 0 ….. (i)
l2 + m2 – n2 = 0 …. (ii)
From equation (i), we have n = – (l + m)
Putting the value of n is equation (ii), we get
l2 + m2 + [-(l + m)]2 = 0
l2 + m2 – l2 – m2 – 2lm = 0
-2lm = 0
lm = 0 ⇒ (- m – n)m = 0 [Since, l = – m – n]
(m + n)m = 0 ⇒ m = 0 or m = -n
⇒ l = 0 or l = -n
Now, the direction cosines of the two lines are
0, -n, n and -n, 0, n ⇒ 0, -1, 1 and -1, 0, 1
$\cos \theta=\frac{(0 \hat{i}-\hat{j}+\hat{k}) \cdot(-\hat{i}+0 \hat{j}+\hat{k})}{\sqrt{(-1)^{2}+(1)^{2}} \sqrt{(-1)^{2}+(1)^{2}}}=\frac{1}{\sqrt{2} \cdot \sqrt{2}}=\frac{1}{2}$
$\theta=\frac{\pi}{3}$
Thus, the required angle π/3.