Question

The angle between the lines whose de's satisfy the equation $l+m+m=0$ and $l^2+m^2-n^2=0$ is

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

Answer 1

The correct option is C $\frac{\pi }{3}$

Given that the equations

$l+m+n=0$ ………..$\left(1\right)$

$l+m=-n$

$\Rightarrow -(l+m)=n$

and

$l^2+m^2+n^2=0$ ……….$\left(2\right)$

Put the value of n in equation $\left(2\right)$

$l^2+m^2+n^2=0$

$\Rightarrow l^2+m^2-(-(l+m){)}^2=0$

$\Rightarrow l^2+m^2-(l^2+m^2-2ml)=0$

$\Rightarrow l^2+m^2-l^2-m^2+2ml=0$

$\Rightarrow 2ml0$

$\Rightarrow ml=0$

$\Rightarrow m=0,l=0$

Let us put $m=0$ in equation $\left(3\right)$

$l+o+n=0$

$l=-n$

Hence, direction rates$(l,m,o)=(1,0,-1)$

Let us put $l=0$, we get $m=-n$

Here, direction ratios $(l,m,n)=(0,1,-1)$

we know that,

$\cos \theta =\frac{\overrightarrow{b_1}\cdot \overrightarrow{b_2}}{|\overrightarrow{b_1}||\overrightarrow{b_2}|}$

$=\frac{(1,0,-1)\cdot (0,1,-1)}{\sqrt{1^2+0^2+(-1{)}^2}\sqrt{0^2+1^2+(-1{)}^2}}$

$=\frac{1}{\sqrt{2}\sqrt{2}}$

$\cos \theta =\frac{1}{2}$

$\cos \theta =\cos \frac{\pi }{3}$

$\theta =\frac{\pi }{3}$

Hence, this is the answer.