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Question
Find the angle between the lines whose direction cosines are given by the equations l + m + n = 0, l2 + m2 – n2 = 0
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Solution
We have, l + m + n = 0, l2 + m2 – n2 = 0.
Eliminating n form both the equation. we have
l2 + m2 – (l + m)2 = 0
⇒ l2 + m2 – l2 – m2 – 2ml = 0
⇒ 2lm = 0
⇒ lm = 0
⇒ l = 0 or m = 0
If l = 0, we have m + n = 0 and m2 – n2 = 0
⇒ l = 0, m = λ, n = λ
If m = 0, we have l + m = 0 and l2 – m2 = 0
⇒ l = – λ, m = 0, n = λ
So, the vector parallel to these given lines are
`veca = hatj - hatk` and `vecb = -hati + hatk`
If angle between the lines is 'θ', then
`cos theta = (|veca * vecb|)/(|veca||vecb|) = 1/(sqrt(2)*sqrt(2))`
⇒ `cos theta = 1/2`
∴ θ = `pi/3`
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