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

Question

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

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Answer

Given that,$ l + m + n = 0 l^2 + m^2 + n^2 = 0$ From equation $(1),$
$l = -(m + n$) Put the value of l in equation $(2),$
$[-(m + n)]^2 + m^2 - n^2 $
$= 0 (m + n)^2 + m^2 - n^2 $
$= 0 m^2+ n^2 + 2mn + m^2 - n^2 $
$= 0 2m^2 + 2mn = 0 2m(m + n) $
$= 0 m = 0, m + n $
$= 0 m = -n$ and $m = 0$
Put the value of $m = -n$ in equation $(1) l = -(m +n) = -(0 + n) l = -n$
Thus, the direction ratios are proportional to $0, -n, n$ and $-n, 0, n \Rightarrow 0, -1 1$ and $-1, 0, 1$
So, vectors parallel to these lines are
$\vec{a}=0\times\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\vec{b}=-\hat{\text{i}}+0\times\hat{\text{j}}+\hat{\text{k}}$ respectively.
Let, $\theta$ be the angle between the $\vec{a}$ and $\vec{b}$
So, $\cos\theta=\frac{\vec{a}\times\vec{b}}{\big|\vec{a}\big|\big|\vec{b}\big|}$
$\vec{a}=0\times\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ and $\vec{b}=-\hat{\text{i}}+0\times\hat{\text{j}}+\hat{\text{k}}$ respectively.
$\cos\theta=\frac{(0\times\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})\times(-\hat{\text{i}}+0\times\hat{\text{j}}+\hat{\text{k}})}{\sqrt{0^2+(-1)^2+(1)^2}\sqrt{(-1)^2+(0)^2+(1)^2}}$
$=\frac{(0)(-1)+(-1)(0)+(1)(1)}{\sqrt{1+1}\sqrt{1+1}}$
$=\frac{0+0+1}{\sqrt{2}\times\sqrt{2}}$
$=\frac{1}{2}$
$\theta=\cos^{-1}\Big(\frac{1}{2}\Big)$
$\theta=\frac{\pi}{3}$
So, angle between the lines $=\frac{\pi}{3}$.

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