Question

If the direction cosines of two lines are such that $l+m+n=0,l^2+m^2-n^2=0$, then the angle between them is

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (C)

Given that,
$l+m+n=0\dots (i)$
$\Rightarrow l+m=-n$
$\Rightarrow -(l+m)=n$
and $(l^2+m^2-n^2=0\dots (ii)$
Let us substitute per 'n' in Eq. (ii), we get
$\Rightarrow l^2+m^2-l^2-m^2-2ml=0$
$\Rightarrow 2ml=0$
i.e. either $l=0$ or $m=0$
Put $m=0$ in Eq. (i).
If $m=0$, then $l=-n$
Direction ratios $(l,m,n)=(1,0,-1)$
Put $l=0$ in Eq. (i), we get $m=-n$
Direction ratios $(l,m,n)=(0,1,-1)$
Here, $b_1\cdot b_2=(1,0,-1)\cdot (0,1,-1)=0+0+1$
$\therefore \left(b_1\right)=\sqrt{0^2+1^2+1^2}=\sqrt{2}$
and $\left(b_2\right)=\sqrt{0^2+1^2+1^2}=\sqrt{2}$
$\therefore \cos \theta =\frac{b_1\cdot b_2}{\left(b_1\right)\left(b_2\right)}$
$\Rightarrow \frac{1}{\sqrt{2}\cdot \sqrt{2}}=\frac{1}{2}$
$\Rightarrow \theta =\frac{\pi }{3}$