Question

Find the angle between the lines whose direction cosines are given by the equations: 3l + m + 5n = 0 and 6mn – 2nl + 5lm = 0.

Answer 1

Eliminating m from the given two equations, we get

⇒ 2n² + 3ln + l² = 0

⇒ (n + l)(2n + l) = 0

⇒ Either n = – l or l = – 2n

Now if l = – n, then m = – 2n

And if l = – 2n, then m = n.

Thus the direction ratios of two lines are proportional to – n, –2n, n and –2n, n, n

i.e. 1, 2, –1 and –2, 1, 1.

So, vectors parallel to these lines are `veca = hati + 2hatj - hatk` and `vecb = -2hati + hatj + hatk`, respectively.

If θ is the angle between the lines

Then cos θ = `(veca * vecb)/(|veca||vecb|)`

= `((hati + 2hatj - hatk)*(-2hati + hatj + hatk))/(sqrt(1^2 + 2^2 + (-1)^2) sqrt((-2)^2 + 1^2 + 1^2))`

= `-1/6`

Hence θ = `cos^-1 (-1/6)`.