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Direction Cosines and Direction Ratios — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDirection Cosines and Direction Ratios5 Marks
Question
Find the angle between the lines whose direction cosines are given by the equations:
$l + m +n = 0$ and $l^2 + m^2 + n^2 = 0$
✓

Answer

Given that, $1+m+n=0 l^2+m^2+n^2=0$ From equation
(1), $I=-(m+n)$ Put the value of $I$ in equation
(2), $[-(m+n)]^2+m^2-n^2=0(m+n)^2+m^2-n^2$
$=0 m^2+n^2+2 m n+m^2-n^2=02 m^2+2 m n$
$=02 m(m+n)=0 m$
$=0, m+n=0 m$
$=-n \text { and } m=0$
Put the value of $m=-n$ in equation, $(1) I=-(m+n)=-(0+n) I=-n$
Thus, the direction ratios are proportional to $0,-n, n$ and $-n, 0, n=0,-11$ and $-1,0,1$
So, vectors parallel to these lines are $\vec{a}=0 \times \hat{ i }-\hat{ j }+\hat{ k }$ and $\vec{b}=-\hat{ i }+0 \times \hat{ j }+\hat{ k }$ respectively.
Let, $\theta$ be the angle between the $\vec{a}$ and $\vec{b}$
$\text { So, } \cos \theta=\frac{\vec{a} \times \vec{b}}{|\vec{a}||\vec{b}|}=0 \times \hat{i}-\hat{j}+\hat{k} \text { and } \vec{b}=-\hat{i}+0 \times \hat{j}+\hat{k} \text { respectively. }$
$\cos \theta=\frac{(0 \times \hat{i}-\hat{j}+\hat{k}) \times(-\hat{i}+0 \times \hat{j}+\hat{k})}{\sqrt{0^2+(-1)^2+(1)^2} \sqrt{(-1)^2+(0)^2+(1)^2}}=\frac{(0)(-1)+(-1)(0)+(1)(1)}{\sqrt{1+1} \sqrt{1+1}}=\frac{0+0+1}{\sqrt{2} \times \sqrt{2}}$
$=\frac{1}{2} \theta=\cos ^{-1}\left(\frac{1}{2}\right) \theta=\frac{\pi}{3} \text { So, angle between the lines }=\frac{\pi}{3}$

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