Question

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

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{2}$
• D. $0$

Answer 1

The correct option is C $\frac{\pi }{2}$

$l+m+n=0$
$lm+mn-2nl=0$
$lm-2nl+mn=0$
$l(m-2n)+mn=0$
$-(m+n)(m-2n)+mn=0$
$(m+n)(2n-m)+mn=0$
$2nm-m^2+2n^2-mn+mn=0$
$2n^2+2nm-m^2=0$
$m^2-2nm-2n^2=0$
By applying quadratic formula we get
$m=\frac{2n\pm \sqrt{4n^2+8n^2}}{2}$
$m=(1+\sqrt{3})n$ and $m=(1-\sqrt{3})n$
Hence corresponding l values will be
$l=-(2+\sqrt{3})n$ and $l=-(2-\sqrt{3})n$
Taking ratios, we get the direction ratios for line one and two respectively as.
$\left(\right(1+\sqrt{3}),1,-(2+\sqrt{3}\left)\right)$ and $\left(\right(1-\sqrt{3}),1,-(2-\sqrt{3}\left)\right)$
Taking dot product we get
$(1+\sqrt{3})(1-\sqrt{3})+1+(2+\sqrt{3})(2-\sqrt{3})$
$=1-3+1+4-3$
$=6-6=0$
Hence the angle between them is $\frac{\pi }{2}$