If 11- m1- n1 and 12- m2- n2- are d-c-s of OA - OB such that - AOB - where - O - is the origin- then the d-c-s of the internal bisector of the angle - AOB areA- 11-l2-2 sin 2- m 1- m 2-2 sin-2- n 1- n 2-2 sin-2C- 11-122-2 sin-2- m1-m22-2 sin-2- n1-n22-2 sin-2D- 11-22-2 cos 2- m1-m22-2 cos-2- n1-2-2 cos 2-

The correct option is B l_1 + l_2/2 cos θ/2, m_1 + m_2/2 cos θ/2, n_1 + n_2/2 cos θ/2Let OA and OB be two lines with d.c’s l_1, m_1, n_1 and l_2, m_2, n_2,. Le ...

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Byju's Answer

Standard XII

Mathematics

Angle between Two Line Segments

If 11, m1, n1...

Question

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

\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}}

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\frac{l_{1} + l_{2}}{2 c o s \frac{θ}{2}}, \frac{m_{1} + m_{2}}{2 c o s \frac{θ}{2}}, \frac{n_{1} + n_{2}}{2 c o s \frac{θ}{2}}

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\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}}

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\frac{l_{1} - l_{2}}{2 c o s \frac{θ}{2}}, \frac{m_{1} - m_{2}}{2 c o s \frac{θ}{2}}, \frac{n_{1} - n_{2}}{2 c o s \frac{θ}{2}}

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Solution

The correct option is B $\frac{l_{1} + l_{2}}{2 c o s \frac{θ}{2}}, \frac{m_{1} + m_{2}}{2 c o s \frac{θ}{2}}, \frac{n_{1} + n_{2}}{2 c o s \frac{θ}{2}}$
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, $O C = \sqrt{{(\frac{l_{1} + l_{2}}{2})}^{2} + {(\frac{m_{1} + m_{2}}{2})}^{2} + {(\frac{n_{1} + n_{2}}{2})}^{2}} = \frac{1}{2} \sqrt{(l_{1}^{2} + m_{1}^{2} + n_{1}^{2}) + (l_{2}^{2} + m_{2}^{2} + n_{2}^{2}) + (l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2})} = \frac{1}{2} \sqrt{2 + 2 c o s θ} [Q c o s θ = l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}] = \frac{1}{2} \sqrt{2 (1 + c o s θ)} = c o s (\frac{θ}{2})$

$∴$ 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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If (l1,m1,n1) and (l2,m2,n2,) are d.c.'s of ¯¯¯¯¯¯¯¯¯¯OA, ¯¯¯¯¯¯¯¯OB such that ∠AOB=θ where ‘O’ is the origin, then the d.c.’s of the internal bisector of the angle ∠AOB are

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