Question

If the d.rs of two lines are given by the equations $l+m+n=0$ and $2lm-mn+2nl=0$, then the angle between the two lines is

• A. ${120}^o$
• B. ${45}^o$
• C. ${90}^o$
• D. ${30}^o$

Answer 1

The correct option is A ${120}^o$

Given, $l+m+n=0$ and $2lm-mn+2nl=0$
$2lm+2nl-mn=0$
$2l(m+n)-mn=0$
$-2(m+n)(m+n)-mn=0$
$2(m+n{)}^2+mn=0$
$2m^2+5mn+2n^2=0$
Hence by applying quadratic formula, we get
$m=\frac{-5n\pm 3n}{4}$

$m=\frac{-n}{2}$ and $m=-2n$
Therefore, the corresponding values of l are $n$ and $\frac{-n}{2}$
Hence, the drs of one line is (-2n,n,n)
And d.r.s of other is $(\frac{-n}{2},n,\frac{-n}{2})$

Hence, the drs will be in the ration
$(-2,1,1)$ and $(-1,2,-1)$
Taking dot product, we get
$(-2)(-1)+\left(1\right)\left(2\right)+\left(1\right)(-1)=\sqrt{6}\sqrt{6}cos\theta$
$3=6\cos \theta$
$\theta ={60}^0$
Hence, angle between the lines will be ${60}^0$ and ${180}^0-{60}^0$
$={120}^0$