Question

If the direction ratios of two lines are given by $3lm-4ln+mn=0$ and $l+2m+3n=0,$ then the angle between the lines is

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

Answer 1

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

$l+2m+3n=0\Rightarrow l=-2m-3n$
On substituting $l$ value in $3lm-4ln+nm=0,$
$3m(-2m-3n)-4n(-2m-3n)+mn=0$
$\Rightarrow 12n^2=6m^2\Rightarrow m=\pm \sqrt{2}n$
$\Rightarrow l=2\sqrt{2}n-3n$ or $l=-2\sqrt{2}n-3n$
Hence, d.r.'s of the lines are
$(-2\sqrt{2}-3,\sqrt{2},1)$ and $(2\sqrt{2}-3,-\sqrt{2},1)$
$\therefore a_1{a}_2+b_1{b}_2+c_1{c}_2=0$
We know, angle between two intersecting lines is given by$\cos \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$
Here $\cos \theta =0$
Hence, required angle $=\frac{\pi }{2}$