Question

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

• A. parallel
• B. perpendicular
• C. inclined at $co{s}^{-1}\left(\frac{1}{6}\right)$
• D. skew lines

Answer 1

The correct option is C inclined at $co{s}^{-1}\left(\frac{1}{6}\right)$

$3l+m+5n=0$ .....(i)
$6mn-2nl+5ml=0$ ......(ii)
Substituting the value of n from Eq. (i) in Eq. (ii), then
$\Rightarrow 6(\frac{l}{m}{)}^2+9(\frac{l}{m})-6=0$
$\therefore \frac{l_1}{m_1}=\frac{1}{2}and\frac{l_2}{m_2}=-2$
From Eq. (i), We get,
$\frac{l_1}{n_1}=-1and\frac{l_2}{n_2}=-2\therefore \frac{l_1}{1}=\frac{m_1}{2}=\frac{n_1}{-1}=\frac{\sqrt{l_1^2+m_1^2+n_1^2}}{\sqrt{1+4+1}}=\frac{1}{\sqrt{6}}$
and $\frac{l_2}{2}=\frac{m^2}{-1}=\frac{n_2}{-1}$
$=\frac{\sqrt{l_1^2+m_1^2+n_1^2}}{\sqrt{1+4+1}}=\frac{1}{\sqrt{6}}$
If $\theta$ be the angle between the lines
$\therefore cos\theta =\left(\frac{1}{\sqrt{6}}\right)\left(\frac{2}{\sqrt{6}}\right)+\left(\frac{2}{\sqrt{6}}\right)(-\frac{1}{\sqrt{6}})+(-\frac{1}{\sqrt{6}})(-\frac{1}{\sqrt{6}})$
$=\frac{1}{6}$
$\therefore \theta =co{s}^{-1}\left(\frac{1}{6}\right)$