Question

If $l_1,m_1,n_1$ and $l_2,m_2,n_2$ are direction cosines of $OA$ and $OB$ such that $\angle AOB=\theta$, where $O$ is the origin, then the direction cosines of the internal angular bisector of $\angle AOB$ are

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

Answer 1

Answer: (D)

$\because l_1{l}_2+m_1{m}_2+n_1{n}_2=\cos \theta$
Through origin $O$ draw two lines parallel to given lines and take two points on each at a distance $r$ from $O$ and a point $R$ on $QO$ produced so that $OR=r$

Then, the coordinates of $P,Q$ and $R$ are
$\left(l_{1}r,m_{1}r,n_{1}r\right),\left(l_{2}r,m_{2}r,n_{2}r\right)$ and $\left(-l_{2}r,-m_{2}r,-n_{2}r\right)$ respectively.
If $A,B$ be the mid-points of $PQ$ and $PR$, then $OA$ and $OB$ are along the bisectors of the lines direction
ratios of $OA$ are $l_1+l_2,m_1+m_{2,},n_1+n_2$
DR's of $OB$ are $l_1-l_2,m_1-m_2,n_1-n_2$
Now, $\Sigma {\left(l_1+l_2\right)}^2=1+1+2\cos \theta$
$=2(1+\cos \theta )=4{\cos }^2\frac{\theta }{2}$
$\therefore D{C}^{\prime }$ s of internal bisector are
$\frac{l_1+l_2}{2\cos \frac{\theta }{2}},\frac{m_1+m_2}{2\cos \frac{\theta }{2}},\frac{n_1+n_2}{2\cos \frac{\theta }{2}}$