Question

Find the angle between the lines whose direction cosines l, m, n satisfy the equations 5l + m + 3n = 0 and 5mn − 2nl + 6lm = 0.

Answer 1

Given, 5l + m + 3n = 0        ...(1)

and 5mn − 2nl + 6lm = 0     ...(2)

From (1), m = − (5l + 3n)

Putting the value of m in equation (2), we get,

−5(5l + 3n)n − 2nl − 6l(5l + 3n) = 0

∴ − 25ln − 15n² − 2nl − 30l² − 18ln = 0

∴ − 30l² − 45ln − 15n² = 0

∴ 2l² + 3ln + n² = 0

∴ 2l² + 2ln + ln + n² = 0

∴ 2l(l + n) + n(l + n) = 0

∴ (l + n)(2l + n) = 0

∴ l + n = 0    or    2l + n = 0

∴ l = − n     or   n = − 2l

Now, m = − (5l + 3n), therefore, if l = − n, 

m = − (− 5n + 3n) = 2n

∴ − l = `"m"/2 = "n"` 

∴ `"l"/-1 = "m"/2 = "n"/1`

∴ the direction ratios of the first line are

a₁ = −1, b₁ = 2, c₁ = 1

If n = − 2l, m = −(5l − 6l) = l

∴ l = m = `"n"/-2`

∴ `"l"/1 = "m"/1 = "n"/-2`

∴ the direction ratios of the second line are

a₂ = 1, b₂ = 1, c₂ = − 2

Let θ be the angle between the lines.

Then 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))|`

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

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

`= |(- 1)/6|`

= `1/6`

∴ θ = cos⁻¹ `(1/6)`