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Question
Find the angle between the lines whose direction cosines are given by the equations: 3l + m + 5n = 0 and 6mn – 2nl + 5lm = 0.
Solution
Eliminating m from the given two equations, we get
⇒ 2n2 + 3 ln + l2 = 0
⇒ (n + l) (2n + l) = 0
⇒ Either n = – l or l = – 2n
Now if l = – n, then m = – 2n
And if l = – 2n, then m = n.
Thus the direction ratios of two lines are proportional to – n, –2n, n and –2n, n, n,
i.e. 1, 2, –1 and –2, 1, 1.
So, vectors parallel to these lines are
`vec"a" = hat"i" + 2hat"j" - hat"k"`
And `vec"b" = -2hat"i" + hat"j" + hat"k"`, respectively.
If θ is the angle between the lines, then
`cos theta = (vec"a". vec"b")/(|vec"a"||vec"b"|)`
= `((hat"i" + 2hat"j" - hat"k")*(-2hat"i" + hat"j" + hat"k"))/(sqrt(1^2 + 2^2 + (-1)^2) sqrt((-2)^2 + 1^2 + 1^2)`
= `-1/6`
Hence θ = `cos^-1 (- 1/6)`.
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