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

#### Solution

Given 6mn - 2nl + 5lm = 0 ....(1)

3l + m + 5n = 0. ...(2)

From (2), m = - 3l - 5n

Putting the value of m in equation (1), we get,

⇒ 6n(- 3l - 5n) - 2nl + 5l(- 3l - 5n) = 0

⇒ - 18nl - 30n^{2} - 2nl - 15l^{2} - 25nl = 0

⇒ - 30n^{2} - 45nl - 15l^{2} = 0

⇒ 2n^{2} + 3nl + l^{2} = 0

⇒ 2n^{2} + 2nl + nl + l^{2} = 0

⇒ (2n + l)(n + l) = 0

∴ 2n + l = 0 OR n + l = 0

∴ l = - 2n OR l = - n

∴ l = - 2n

From (2), 3l + m + 5n = 0

∴ - 6n + m + 5n = 0

∴ m = n

i.e. (- 2n, n, n) = (-2, 1, 1)

∴ l = - n

∴- 3n + m + 5n = 0

∴ m = - 2n

i.e. (-n, - 2n, n) = (1, 2, -1)

(a_{1}, b_{1}, c_{1}) = (-2, 1, 1) and (a_{2}, b_{2}, c_{2}) = (1, 2, -1)

cos θ = `|("a"_1"a"_2 + "b"_1"b"_2 + "c"_1"c"_2)/(sqrt("a"_1^2 + "b"_1^2 + "c"_1^2).sqrt ("a"_2^2 + "b"^2_2 + "c"_2^2))|`

`= |((2)(1) + (-1)(2) + (-1)(-1))/(sqrt((2)^2 + 1^2 + 1^2).sqrt(1^2 + 2^2 + (1)^2))|`

`= |(2 - 2 + 1)/(sqrt6.sqrt6)|`

`= |- 1/6| = 1/6`

`θ = "cos"^-1 (1/6)`