Question

If $\theta$ be the angle between the lines whose direction cosines's are given by $3l+m+5n=0$, $6mn-2nl+5lm=0$. Find value of 6cos$\theta$

Answer 1

Given, $3l+m+5n=0$ ...........$\left(1\right)$

and $6mn-2nl+5lm=0$ ..........$\left(2\right)$

From equation $\left(1\right)m=-(3l+5n)$

put the value of $m=-(3l+5n)$ in equation $\left(2\right)$ we have

$6\times -(3l+5n)n-2nl+5l\times -(3l+5n)=0$

$\Rightarrow -18ln-30n^2-2nl-15l^2-25ln=0$

$\Rightarrow -45ln-30n^2-15l^2=0$

$\Rightarrow l^2+3ln+2n^2=0$

$\Rightarrow (l+2n)(l+n)=0$

So,$l+2n=0$ ........$\left(3\right)$

and $l+n=0$ ........$\left(4\right)$

From equations $\left(1\right)$ and $\left(3\right)$

$3l+m+5n=0$

$l+2n=0$

$\Rightarrow \frac{l}{2}=\frac{m}{-1}=\frac{n}{-1}$ .......$\left(5\right)$

and from equation $\left(1\right)$ and $\left(3\right)$ i.e

$3l+m+5n=0$

$l+0m+n=0$

$\Rightarrow \frac{l}{2}=\frac{m}{-1}=\frac{n}{-1}$ .......$\left(6\right)$

Now angle between lines $\left(5\right)$ and $\left(6\right)$ is

$\cos \theta =\frac{2\times 1+(-1)\times 2+(-1)(-1)}{\sqrt{2^2+{(-1)}^2+{(-1)}^2}\sqrt{1^2+2^2+(-1)}}$

$=\frac{2-2+1}{\sqrt{4+1+1}\sqrt{1+4+1}}$

$=\frac{1}{\sqrt{6}}\times \frac{1}{\sqrt{6}}=\frac{1}{6}$

$\Rightarrow 6\cos \theta =6\times \frac{1}{6}=1$