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