Question

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

• A. parallel
• B. perpendicular
• C. Inclined at $co{s}^{-1}\frac{1}{6}$
• D. skew lines

Answer 1

The correct option is C Inclined at $co{s}^{-1}\frac{1}{6}$

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