Show that the straight lines whose direction cosines are given by 2l + 2m – n = 0 and mn + nl + lm = 0 are at right angles.

Given,

2l + 2m – n = 0 [1]

⇒ n = 2(l + m) [2]

and

mn + nl + lm = 0

⇒ 2m(l + m) + 2(l + m)l + lm = 0

⇒ 2lm + 2m² + 2l² + 2lm + lm = 0

⇒ 2m² + 5lm + 2l² = 0

⇒ 2m² + 4lm + lm + 2l² = 0

⇒ (2m + l)(m + 2l) = 0

So, we have two cases,

l = -2m

⇒ -4m + 2m – n = 0 [From 1]

⇒ n = 2m

Hence, direction ratios of one line is proportional to -2m, m, -2m or direction ratios are (2, 1, -2)

Another case is,

m = -2l

⇒ 2l + 2(-2l) – n = 0

⇒ 2l – 4l = n

⇒ n = -2l

Hence, direction ratios of another lines is proportional to l, -2l, -2l or direction ratios are (1, -2, -2)

Therefore, direction vectors of two lines are and

Also,

Angle between two lines, and is given by

Now,

⇒ cos θ = 0

⇒ θ = 90°

Hence, angle between given two lines is 90°