Find the angle between the lines whose direction cosines

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.