If $l_{1}, m_{1}, n_{1}$ and $l_{2}, m_{2}, n_{2}$ be the DC's of two concurrent lines, the direction cosines of the line bisecting the angles between them are proportional to

l_{1} l_{2}, m_{1} m_{2}, n_{1}, n_{2}

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l_{1} m_{2}, l_{1} n_{2}, l_{1} n_{3}

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l_{1} + l_{2}, m_{1} + m_{2}, n_{1} + n_{2}

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None of these

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Solution

The correct option is D $l_{1} + l_{2}, m_{1} + m_{2}, n_{1} + n_{2}$
Let direction cosines of $O A = (l_{1}, m_{1}, n_{1})$ and direction cosines of $O B = (l_{2}, m_{2}, n_{2})$ .
Taking two points on $l$ and $m$ such that $O A = O B = r$
Let $C$ be the mid-point of $A B$ . Then, $O C$ is the bisector of the angle $A O B$ .
Now, coordinates of $A$ are $(l_{1} r, m_{1} r, n_{1} r)$ and coordinates of $B$ are $(l_{2} r, m_{2} r, n_{2} r)$ .
$∴$ the coordinates of $C$ are
$\frac{(l_{1} + l_{2}) r}{2}, \frac{(m_{1} + m_{2}) r}{2}, \frac{(n_{1} + n_{2}) r}{2}$ .
Hence, the direction cosines of $O C$ are proportional to $l_{1} + l_{2}, m_{1} + m_{2}$ and $n_{1} + n_{2}$ .

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