Question

The pair of lines whose direction cosines are given by the equations $3l+m+5n=0$ and $6mn-2nl+5lm=0$ are-

• A. Parallel
• B. Perpendicular
• C. Inclined at ${\cos }^{-1}\left(\frac{1}{6}\right)$
• D. None of these.

Answer 1

The correct option is C Inclined at ${\cos }^{-1}\left(\frac{1}{6}\right)$

Given, $3l+m+5n=0$ ...(1)
and $6mn-2nl+5lm=0$ ...(2)
From eqaution (1) $m=-(3l+5n)$, put it in equation (2), we have
$6(3l+5n)n-2nl+5l(-3l-5n)=0\Rightarrow -18lm-30n^2-2nl-15l^2-25ln=0\Rightarrow l^2+2n^2+3ln=0\Rightarrow (l+2n)(l+n)=0$
So $l+2n=0$ ...(3)
and $l+n=0$ ...(4)
From equation (1) and (3), we get
$\frac{l}{1}=\frac{m}{2}=\frac{n}{-1}$ ...(5)
And from (1) and (4)
$\frac{l}{2}=\frac{m}{-1}=\frac{n}{-1}$ ...(6)
Now angle between lines (5) and (6) is
$\cos \theta =\frac{2\times 1+(-1)\times 2+(-1)\times (-1)}{\sqrt{2^2+{(-1)}^2+{(-1)}^2}\sqrt{1^2+2^2+{(-1)}^2}}=\frac{2-2+1}{\sqrt{6}\sqrt{6}}$
$\Rightarrow \cos \theta =\frac{1}{6}\Rightarrow \theta ={\cos }^{-1}\frac{1}{6}$