Find the angle between the lines whose direction cosines are give…

Question

Find the angle between the lines whose direction cosines are given by the equations:
$2l - m + 2n = 0$ and $mn + nl + lm = 0$

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Answer

Given that,
$2l - m + 2n = 0 .....(1)$
$mn + nl + lm = 0 .....(2)$
From equation $(1),$
$2l - m + 2n = 0$
$m = 2l + 2n $
Put the value of $m$ in equation $(2),$
$mn + nl + lm = 0$
$(2l + 2n) n + nl + l(2l + 2n) = 0$
$2ln + 2n^2 +nl +2l^2 + 2ln = 0$
$2l^2 + 5ln + 2n^2 = 0$
$2l^2 + 4ln + ln + 2ln^2 = 0$
$2l (l + 2n) + n(l + 2n) = 0$
$(1 + 2n) (2l = n) = 0$
$l + 2n = 0$ or $2l + n = 0$
$l = -2n$ or $\text{l}=-\frac{\text{n}}{2}$
Put the value of $l = -2n$ in equation $(1)$
$2l - m + 2n = 0$
$2 (-2n) - m + 2n = 0$
$-4n - m + 2n = 0$
$-2n - m = 0$
$-2n = m$
$m = -2n$
Again, put the value of $\text{l}=-\frac{1}{2}$ in equation $(1)$
$2l - m + 2n = 0$
$2\Big(-\frac{1}{2}\text{n}\Big)-\text{m}+2\text{n}=0$
$-n - m + 2n = 0$
$-m + n = 0$
$-m = -n$
$m = n$
So, direction cosines of the lines are given by,
$-2n, -2n, n$ or $-\frac{1}{2},\text{n},\text{n},\text{n}$
$-2, -2, 1$ or $-\frac{1}{2},1,1$
So, vectors parallel to these lines
$\vec{\text{a}}=2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=-\frac{1}{2}\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ respectively.
Let, $\theta$ be the angle between the $\vec{\text{a}}$ and $\vec{\text{b}},$
$\cos\theta=\frac{\vec{\text{a}}\times\vec{\text{b}}}{\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|}$
$=\frac{\big(-2\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\big)\times\Big(-\frac{1}{2}\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\Big)}{\sqrt{(-2)^2+(-2)^2+(1)^2}\sqrt{\Big(-\frac{1}{2}^2\Big)+(1)^2+(1)^2}}$
$=\frac{(-2)\big(-\frac{1}{2}\big)+(-2)(1)+(1)(1)}{\sqrt{4+4+1}\sqrt{\frac{1}{4}1+1}}$
$=\frac{1-2+1}{\sqrt{9}\sqrt{\frac{9}{4}}}$
$=\frac{0}{3\times\frac{3}{2}}$
$\cos\theta=0$
$\theta=\cos^{-1}(0)$
$\theta=\frac{\pi}{2}$
So, angle between the lines $=\frac{\pi}{2}$.