If $l_{1}, m_{1}, n_{1}$ and $l_{2}, m_{2}, n_{2}$ are the direction cosines of the two lines inclined to each other at an angle $θ$ , then the direction cosines of the external angular bisector of the angle between these lines are $\frac{l_{1} - l_{2}}{2 \sin α}, \frac{m_{1} - m_{2}}{2 \sin α}, \frac{n_{1} - n_{2}}{2 \sin α}$ where $2 α =$

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$θ$

$\frac{θ}{2}$

$\frac{θ}{4}$

None

answer is A.

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Detailed Solution

Given that the direction cosines of two lines are $l_{1}, m_{1}, n_{1}$ and $l_{2}, m_{2}, n_{2}$

Given $θ$ is the acute angle between the lines then $\cos θ = l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}$

The direction ratios of angular bisector of the two lines whose direction cosines are $l_{1}, m_{1}, n_{1}$ and $l_{2}, m_{2}, n_{2}$ are $〈l_{1} - l_{2}, m_{1} - m_{2}, n_{1} - n_{2}〉$

Consider the value of

$\begin{matrix} {(l_{1} - l_{2})}^{2} + {(m_{1} - m_{2})}^{2} + {(n_{1} - n_{2})}^{2} = {l_{1}}^{2} + {m_{1}}^{2} + {n_{1}}^{2} + {l_{2}}^{2} + {m_{2}}^{2} + {n_{2}}^{2} + 2 l_{1} l_{2} + 2 m_{1} m_{2} + 2 n_{1} n_{2} \\ = 2 - 2 (\cos θ) \\ = 2 (2 \sin^{2} \frac{θ}{2}) \\ = 4 \sin^{2} \frac{θ}{2} \end{matrix}$

It implies that $\sqrt{{(l_{1} - l_{2})}^{2} + {(m_{1} - m_{2})}^{2} + {(n_{1} - n_{2})}^{2}} = 2 \sin \frac{θ}{2}$

The direction cosines of angular bisector of given two lines are $\frac{l_{1} - l_{2}}{2 s i n \frac{θ}{2}} = \frac{m_{1} - m_{2}}{2 s i n \frac{θ}{2}} = \frac{n_{1} - n_{2}}{2 s i n \frac{θ}{2}}$

Comparing the above direction cosines with $\frac{l_{1} - l_{2}}{2 \sin α}, \frac{m_{1} - m_{2}}{2 \sin α}, \frac{n_{1} - n_{2}}{2 \sin α}$ , we have $α = \frac{θ}{2}$ .