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Direction Cosines

If l1,m1,n1...

Question

If $l_{1}, m_{1}, n_{1}$ and $l_{2}, m_{2}, n_{2}$ are D.C.'s of the two lines inclined to each other at an angle $θ$ , then the D. C.'s of the internal and external bisectors of the angle between these lines are

A

\frac{l_{1} + l_{2}}{2 sin (θ / 2)}, \frac{m_{1} + m_{2}}{2 sin (θ / 2)}, \frac{n_{1} + n_{2}}{2 sin (θ / 2)}

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B

\frac{l_{1} + l_{2}}{2 cos (θ / 2)}, \frac{m_{1} + m_{2}}{2 cos (θ / 2)}, \frac{n_{1} + n_{2}}{2 cos (θ / 2)}

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C

\frac{l_{1} - l_{2}}{2 sin (θ / 2)}, \frac{m_{1} - m_{2}}{2 sin (θ / 2)}, \frac{n_{1} - n_{2}}{2 sin (θ / 2)}

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D

\frac{l_{1} - l_{2}}{2 cos (θ / 2)}, \frac{m_{1} - m_{2}}{2 cos (θ / 2)}, \frac{n_{1} - n_{2}}{2 cos (θ / 2)}

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Solution

The correct options are
B $\frac{l_{1} + l_{2}}{2 cos (θ / 2)}, \frac{m_{1} + m_{2}}{2 cos (θ / 2)}, \frac{n_{1} + n_{2}}{2 cos (θ / 2)}$
D $\frac{l_{1} - l_{2}}{2 sin (θ / 2)}, \frac{m_{1} - m_{2}}{2 sin (θ / 2)}, \frac{n_{1} - n_{2}}{2 sin (θ / 2)}$

If $l_{1}, m_{1}, n_{1}$ and $l_{2}, m_{2}, n_{2}$ are D.C.s of two lines then,

$l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2} = cos θ$

The direction ratios of the internal angle bisector are $l_{1} + l_{2}, m_{1} + m_{2}, n_{1} + n_{2}$ .

Hence, the direction cosines of the internal angle bisector are $\frac{l_{1} + l_{2}}{\sqrt{\sum {(l_{1} + l_{2})}^{2}}}, \frac{m_{1} + m_{2}}{\sqrt{\sum {(l_{1} + l_{2})}^{2}}}, \frac{n_{1} + n_{2}}{\sqrt{\sum {(l_{1} + l_{2})}^{2}}}$

where, $\sum {(l_{1} + l_{2})}^{2} = {(l_{1} + l_{2})}^{2} + {(m_{1} + m_{2})}^{2} + {(n_{1} + n_{2})}^{2} = ({l_{1}}^{2} + {m_{1}}^{2} + {n_{1}}^{2}) + ({l_{2}}^{2} + {m_{2}}^{2} + {n_{2}}^{2}) + 2 (l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}) = 1 + 1 + 2 cos θ = 2 + 2 cos θ = 4 {cos}^{2} \frac{θ}{2}$

Hence direction cosines of internal angle bisector are $\frac{l_{1} + l_{2}}{2 cos \frac{θ}{2}}, \frac{m_{1} + m_{2}}{2 cos \frac{θ}{2}}, \frac{n_{1} + n_{2}}{2 cos \frac{θ}{2}}$

Similarly , D.R.s of external angle bisector are $l_{1} - l_{2}, m_{1} - m_{2}, n_{1} - n_{2}$

So, the direction cosines of the external angle bisector are $\frac{l_{1} - l_{2}}{\sum {(l_{1} - l_{2})}^{2}}, \frac{m_{1} - m_{2}}{\sum {(l_{1} - l_{2})}^{2}}, \frac{n_{1} - n_{2}}{\sum {(l_{1} - l_{2})}^{2}}$

where, $\sum {(l_{1} - l_{2})}^{2} = {(l_{1} - l_{2})}^{2} + {(m_{1} - m_{2})}^{2} + {(n_{1} - n_{2})}^{2} = ({l_{1}}^{2} + {m_{1}}^{2} + {n_{1}}^{2}) + ({l_{2}}^{2} + {m_{2}}^{2} + {n_{2}}^{2}) - 2 (l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}) = 1 + 1 - 2 cos θ = 2 - 2 cos θ = 4 {sin}^{2} \frac{θ}{2}$ .

Hence, direction cosines of external angle bisector are $\frac{l_{1} - l_{2}}{2 sin \frac{θ}{2}}, \frac{m_{1} - m_{2}}{2 sin \frac{θ}{2}}, \frac{n_{1} - n_{2}}{2 sin \frac{θ}{2}}$

Hence, options 'B' and 'D' are correct.

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