The acute angle between two lines whose direction ratios are give… - Vidyadip

MCQ

The acute angle between two lines whose direction ratios are given by $l+ m - n =0$ and $l^2+ m ^2- n ^2=0$ is

A
$\frac{\pi}{2}$
✓
$\frac{\pi}{3}$
C
$\frac{\pi}{4}$
D
$\frac{\pi}{6}$

✓

Answer

Correct option: B.

$\frac{\pi}{3}$

(B) $l+ m - n =0$ and $l^2+ m ^2- n ^2=0$
$\Rightarrow l+ m = n$ and $l^2+ m ^2= n ^2$
Putting $l+ m = n$ in $l^2+ m ^2= n ^2$, we get $l^2+ m ^2=(l+ m )^2$
$\Rightarrow 2 l m=0 \Rightarrow l=0$ or $m =0$
If $l=0$, then $m = n$
$\therefore \quad \frac{l}{0}=\frac{ m }{1}=\frac{ n }{1}$
If $m =0$, then $l= n$
$\therefore \quad \frac{l}{1}=\frac{ m }{0}=\frac{ n }{1}$
∴ the d.r.s of the lines are proportional to $0,1,1$ and $1,0,1$.
$\therefore \quad \cos \theta=\left|\frac{0(1)+1(0)+1(1)}{\sqrt{0+1+1} \sqrt{1+0+1}}\right|=\frac{1}{2}$
$\Rightarrow \theta=\cos ^{-1}\left(\frac{1}{2}\right) \Rightarrow \theta=\frac{\pi}{3}$

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