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

Sum

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

#### Solution

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 = - (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)

#### Notes

[Note: Answer in the textbook is incorrect.]

Concept: Scalar Triple Product of Vectors
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