Question

If the $d{r}^{\prime }s$ 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 $\frac{\pi }{2}$

Given, $1+2m+3n=0$ ---eq.1

$3lm+2m+3n=0$ ----eq.2

So, $l=-2m-3m=-(2m+3n)$

Putting the value in eq.2

$\Rightarrow -3(2m+3n)m+4(2m+3n)n+mn=0\Rightarrow -6m^2-9nm+8mn+12n62+mn=0\Rightarrow -6m^2+12n^2\Rightarrow 12n^2-6m^2=0\Rightarrow 6(2n^2-m^2)=0\Rightarrow 2n^2-m^2=0\Rightarrow m^2=2n^{\Rightarrow }m=\pm \sqrt{2n^2}\text{or}\sqrt{2}n,-\sqrt{2}n$

For $m=\sqrt{2}n$

$L=-(2m+3n)=-2\sqrt{2}n-3m=n(-2\sqrt{2}-3)$

For $m=-\sqrt{2}n$

$L=-(2m+3n)=2\sqrt{2}n-3m=n(2\sqrt{2}-3)$

$\therefore$ Direction ratios of two lines are proportional to

$n(-2\sqrt{2}-3),\sqrt{2}n,n$

and $n(2\sqrt{2}-3),-\sqrt{2}n,n$

So, $\overrightarrow{a}=(-2\sqrt{2}-3)\hat{i}+\sqrt{2}\hat{j}+\hat{k}$

$\overrightarrow{b}=(2\sqrt{2}-3)\hat{i}-\sqrt{2}\hat{j}+\hat{k}$

So, $\cos \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{a}|.|\overrightarrow{b}|}$

$\cos \theta =\frac{\left[\right(-2\sqrt{2}-3)\hat{i}+\sqrt{2}\hat{j}+\hat{k}].\left[\right(2\sqrt{2}-3)\hat{i}-\sqrt{2}\hat{j}+\hat{k}]}{\sqrt{8+9-12\sqrt{2}+2+1}\sqrt{8+9-12\sqrt{2}+2+1}}$

$\cos \theta =\frac{1-2+1}{\sqrt{20-12\sqrt{2}}.\sqrt{20-12\sqrt{2}}}=0\therefore \cos \theta =0\theta ={\cos }^{-1}\left(0\right)=\frac{\pi }{2}$

Correct answer is (A)