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Here, we are going to solve another example which is similar to this problem.

Ex: Prove that the lines whose direction cosines are given by equations, 2l + 2m - n = 0 and mn + nl + lm = 0 are mutually

perpendicular. 

Solution:

In this type of question,

1. We have to find direction ratios of both the lines d₁ , d₂

2. Now find d₁ . d₂

If d₁ . d₂ = 0 then the lines are perpendicular to each other otherwise not.

Given, 2l + 2m - n = 0 .........1

and mn + nl + lm = 0  .........2

From equation 1, we get

n = 2l + 2m

Put value of n in equation 2, we get

=> m(2l + 2m) + (2l + 2m)l + lm = 0

=> 2lm + 2m² + 2l² + 2lm + lm = 0

=> 2m² + 2l² + 5lm = 0

=> (m + 2l)*(2m + l) = 0

=> m = -2l, -l/2

Put l = 1, we get

m = -2, -1/2

and n = -2, -1

Now, d₁ = (1, -2, -2) and d₂ = (1, -1/2, 1)

Now, d₁ . d₂ = (1, -2, -2) . (1, -1/2, 1)

                    = 1 + 1 - 2

                    = 2 - 2

                    = 0

Since d₁ . d₂ = 0

So, the two lines are perpendicular to each other.

In this way, we solve such type of problem.