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

Sum

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 - 30n2 - 2nl - 15l2 - 25nl = 0

⇒ - 30n2 - 45nl - 15l2 = 0

⇒ 2n2 + 3nl + l2 = 0

⇒ 2n2 + 2nl + nl + l2 = 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)

(a1, b1, c1) = (-2, 1, 1) and (a2, b2, c2) = (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)

Concept: Vectors and Their Types
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