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

Answer 1

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

Given $3lm-4ln+mn=0$ ...$\left(1\right)$

and $l+2m+3n=0$ ....$\left(2\right)$

From equation $\left(2\right)$, $l=-(2m+3n)$, putting in equation $\left(1\right)$, we get

$-3(2m+3n)m+4(2m+3n)n+mn=0\Rightarrow -6m^2+12n^2=0\Rightarrow m=\pm \sqrt{2}n$

Now, $m=\pm \sqrt{2}n\Rightarrow l=-(2\sqrt{2}n+3n)=-(2\sqrt{2}+3)n$

$\therefore l:m:n=-(3+2\sqrt{2})n:\sqrt{2}n:n=-(3+2\sqrt{2}):-\sqrt{2}:1$

Also, $m=-\sqrt{2}n\Rightarrow l=-(-2\sqrt{2}+3)n$

$\therefore l:m:n=-(3-2\sqrt{2})n:-\sqrt{2}n:n=-(3-2\sqrt{2}):-\sqrt{2}:1$

$\Rightarrow \cos \theta =\frac{(3+2\sqrt{2})(3-2\sqrt{2})+\left(\sqrt{2}\right)(-\sqrt{2})+1.1}{\sqrt{{(3+2\sqrt{2})}^2+{\left(\sqrt{2}\right)}^2+1^2}\sqrt{{(3-2\sqrt{2})}^2+{(-\sqrt{2})}^2+1^2}}$

$\Rightarrow \theta =\frac{\pi }{2}$