Question

If the direction ratios of two lines given by $l+2m+3n=0$ and $3lm+mn-4nl=0$, then 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 $\frac{\pi }{2}$

Given, $l+2m+3n=0$, $3lm+mn-4nl=0$

$l(3m-4n)+mn=0$

$(2m+3n)(4n-3m)+mn=0$

$8mn-6m^2+12n^2-9mn+mn=0$

$12n^2=6m^2$

$\therefore m=\pm \sqrt{2}$

The corresponding values of $l$ will be $-(3+2\sqrt{2})n$ and $-(3-2\sqrt{2})n$

Hence, the direction ratios will be

$(\sqrt{2},1,-(3+2\sqrt{2}\left)\right)$ and

$(-\sqrt{2},1,-(3-2\sqrt{2}\left)\right)$

Taking dot product, we get

$-2+1+(9-8)$

$=-1+9-8$ $=0$

Hence, $\theta =\frac{\pi }{2}$