Question

The direction cosines $l,m,n$ of two lines are satisfying $3l+m+5n=0$ and $6mn-2nl+5lm=0$. If $\theta$ is the angle between those lines then $|\cos \theta |=$

• A. $\frac{1}{\sqrt{6}}$
• B. $1\sqrt{2}$
• C. $\frac{1}{6}$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

Answer: (C)

We have
$3l+m+5n=0$ and $6mn-2nl+5lm=0$
$\Rightarrow 6n(-3l-5n)-2nl+5l(-3l-5n)=0$
$\Rightarrow -18nl-30n^2-2nl-15l^2-25nl=0$
$\Rightarrow -15l^2-30n^2-45nl=0$
$\Rightarrow l^2+3nl+2n^2=0$
$\Rightarrow (l+n)(l+2n)=0$
$\Rightarrow l=-n,-2n$
When $I=-n$
$-3n+m+5n=0$
$\Rightarrow m=-2n$
So, $d^{\prime }r$ are
$-n,-2n,n$ or $1,2,-1$
When $l=-2n$
$-6n+m+5n=0$
$\Rightarrow m=n$
So, $d^{\prime }r$ are
$-2n,n,n$
$-2,1,1$
$\therefore \cos \theta =\frac{1\times (-2)+2\times 1+(-1)\times 1}{\sqrt{1+4+1}\sqrt{4+1+1}}=\frac{-1}{6}$
$\therefore |\cos \theta |=\frac{1}{6}$