If (l1,m1,n1) and (l2,m2,n2,) are d.c.'s of ¯¯¯¯¯¯¯¯¯¯OA, ¯¯¯¯¯¯

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If

(l_{1}, m_{1}, n_{1})

and

(l_{2}, m_{2}, n_{2},)

are d.c.'s of

¯ ¯¯¯¯¯¯¯¯ ¯ O A,

¯ ¯¯¯¯¯¯ ¯ O B

such that

∠ A O B = θ

where ‘O’ is the origin, then the d.c.’s of the internal bisector of the angle

∠ A O B

are

Options:

A . l1+l22sinθ2,m1+m22sinθ2,n1+n22sinθ2

B . l1+l22cosθ2,m1+m22cosθ2,n1+n22cosθ2

C . l1−l22sinθ2,m1−m22sinθ2,n1−n22sinθ2

D . l1−l22cosθ2,m1−m22cosθ2,n1−n22cosθ2

Answer: Option B
:
B
Let OA and OB be two lines with d.c’s

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

and

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

. Let OA = OB = 1. Then, the coordinates of A and B are

(l_{1}, m_{1}, n_{1})

and

(l_{2}, m_{2}, n_{2})

, respectively. Let OC be the bisector of

∠ A O B

.Then, C is the mid point of AB and so its coordinates are

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

.

∴

d.r's of OC are

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

We have,

O C = \sqrt{{(\frac{l_{1} + l_{2}}{2})}^{2} + {(\frac{m_{1} + m_{2}}{2})}^{2} + {(\frac{n_{1} + n_{2}}{2})}^{2}} = \frac{1}{2} \sqrt{(l_{1}^{2} + m_{1}^{2} + n_{1}^{2}) + (l_{2}^{2} + m_{2}^{2} + n_{2}^{2}) + (l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2})} = \frac{1}{2} \sqrt{2 + 2 c o s θ} [Q c o s θ = l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}] = \frac{1}{2} \sqrt{2 (1 + c o s θ)} = c o s (\frac{θ}{2})

If (l1,m1,n1) And (l2,m2,n2,) Are D.c.'s Of ¯¯¯¯¯¯¯¯...

∴

d.c's of OC are

\frac{l_{1} + l_{2}}{2 (O C)}, \frac{m_{1} + m_{2}}{2 (O C)}, \frac{n_{1} + n_{2}}{2 (O C)}

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