Question

If $l_{1}, m_{1}, n_{1}$ and $l_{2}, m_{2}, n_{2}$ are direction cosines of $O A$ and $O B$ such that $\angle A O B=\theta$, where $O$ is the origin, then the direction cosines of the internal angular bisector of $\angle A O B$ 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 $Q O$ produced so that $O R=r$

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 $P Q$ and $P R$, then $O A$ and $O B$ are along the bisectors of the lines direction
ratios of $O A$ are $l_{1}+l_{2}, m_{1}+m_{2,}, n_{1}+n_{2}$
DR's of $O B$ 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
$\frac{l_{1}+l_{2}}{2 \cos \frac{\theta}{2}}, \frac{m_{1}+m_{2}}{2 \cos \frac{\theta}{2}}, \frac{n_{1}+n_{2}}{2 \cos \frac{\theta}{2}}$