The directions cosines of two lines are connected by relations $l + m + n = 0$ and $4 l$ is the harmonic mean between m and n then

Moderate

A

$\frac{l_{1}}{l_{2}} + \frac{m_{1}}{m_{2}} + \frac{n_{1}}{n_{2}} = \frac{3}{2}$
B

$\frac{l_{1}}{l_{2}} + \frac{m_{1}}{m_{2}} + \frac{n_{1}}{n_{2}} = \frac{- 3}{2}$
C

$\frac{l_{1}}{l_{2}} + \frac{m_{1}}{m_{2}} + \frac{n_{1}}{n_{2}} = \frac{1}{2}$
D

$\frac{l_{1}}{l_{2}} + \frac{m_{1}}{m_{2}} + \frac{n_{1}}{n_{2}} = \frac{- 1}{2}$

Solution

$l + m + n = 0$
$4 l = \frac{2 m n}{m + n} \Rightarrow 2 l m - m n + 2 \ln = 0$
by eliminating n, we get
$2 {(\frac{l}{m})}^{2} - (\frac{l}{m}) - 1 = 0$
$\Rightarrow \frac{l_{1}}{m_{1}} = 1, \frac{l_{2}}{m_{2}} = \frac{- 1}{2}$ and $\Rightarrow \frac{n_{1}}{n_{2}} = - 2$
$∴ \frac{l_{1}}{l_{2}} + \frac{m_{1}}{m_{2}} + \frac{n_{1}}{n_{2}} = 1 - \frac{1}{2} - 2 = \frac{- 3}{2}$

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