Question

If $(l_1,m_1,n_1)$ and $(l_2,m_2,n_2,)$ are d.c.'s of $\overline{OA,}$ $\overline{OB}$ such that $\angle AOB=\theta$ where ‘O’ is the origin, then the d.c.’s of the internal bisector of the angle $\angle AOB$ are

• A. $\frac{l_1+l_2}{2sin\frac{\theta }{2}},\frac{m_1+m_2}{2sin\frac{\theta }{2}},\frac{n_1+n_2}{2sin\frac{\theta }{2}}$
• B. $\frac{l_1+l_2}{2cos\frac{\theta }{2}},\frac{m_1+m_2}{2cos\frac{\theta }{2}},\frac{n_1+n_2}{2cos\frac{\theta }{2}}$
• C. $\frac{l_1-l_2}{2sin\frac{\theta }{2}},\frac{m_1-m_2}{2sin\frac{\theta }{2}},\frac{n_1-n_2}{2sin\frac{\theta }{2}}$
• D. $\frac{l_1-l_2}{2cos\frac{\theta }{2}},\frac{m_1-m_2}{2cos\frac{\theta }{2}},\frac{n_1-n_2}{2cos\frac{\theta }{2}}$

Answer 1

The correct option is B $\frac{l_1+l_2}{2cos\frac{\theta }{2}},\frac{m_1+m_2}{2cos\frac{\theta }{2}},\frac{n_1+n_2}{2cos\frac{\theta }{2}}$

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 $\angle AOB$ .Then, C is the mid point of AB and so its coordinates are $(\frac{l_1+l_2}{2},\frac{m_1+m_2}{2},\frac{n_1+n_2}{2})$.
$\therefore$ 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, $OC=\sqrt{{\left(\frac{l_1+l_2}{2}\right)}^2+{\left(\frac{m_1+m_2}{2}\right)}^2+{\left(\frac{n_1+n_2}{2}\right)}^2}=\frac{1}{2}\sqrt{(l_1^2+m_1^2+n_1^2)+(l_2^2+m_2^2+n_2^2)+(l_1{l}_2+m_1{m}_2+n_1{n}_2)}=\frac{1}{2}\sqrt{2+2cos\theta }[Qcos\theta =l_1{l}_2+m_1{m}_2+n_1{n}_2]=\frac{1}{2}\sqrt{2(1+cos\theta )}=cos\left(\frac{\theta }{2}\right)$

$\therefore$ d.c's of OC are
$\frac{l_1+l_2}{2\left(OC\right)},\frac{m_1+m_2}{2\left(OC\right)},\frac{n_1+n_2}{2\left(OC\right)}$