Question

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

Answer 1

5l + m + 3n = 0        .......[Given]

∴ m = − (5l + 3n)     .......(i)

5mn − 2nl + 6lm = 0    .......[Given]

∴ −5(5l + 3n)n − 2nl − 6l(5l + 3n) = 0  .......[[From (i)]

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

∴ – 30l² – 45ln – 15n² = 0

∴ 2l² + 3ln + n² = 0

∴ (2l + n)(l + n) = 0

∴ n = −2l or n = − l

If n = −2l, then from (i), we get

m = −[5l + 3(– 2l)]

∴ m = l

∴ m = l, n = −2l

∴ Direction ratios of the first line are proportional to l, l, −2l

i.e., 1, 1, −2

If n = − l, then from (i), we get

m = −[5l + 3(– l)]

∴ m = −2l

∴ m = −2l, n = −l

∴ Direction ratios of the second line are proportional to l, −2l, −l

i.e., 1, −2, −1

Let θ be the angle between the two lines.

∴ cos θ = `|(1(1) + 1(2) + (-2)(-1))/(sqrt(1^2 + 1^2 + (-2)^2)*sqrt(1^2 + (-2)^2 + (-1)^2))|`

= `|(1 - 2 + 2)/(sqrt(1 + 1 + 4)*sqrt(1 + 4 + 1))|` 

= `|1/(sqrt(6)*sqrt(6))|`

= `1/6`

∴ θ = `cos^-1(1/6)`