If $l_1, m_1, n_1$ and $l_2, m_2, n_2$ are direction cosines of $\mathbf{O A}$ 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

Question

If $l_1, m_1, n_1$ and $l_2, m_2, n_2$ are direction cosines of $\mathbf{O A}$ 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, $\frac{I_1+I_2}{2 \sin \frac{\theta}{2}}, \frac{m_1+m_2}{2 \sin \frac{\theta}{2}}, \frac{n_1+n_2}{2 \sin \frac{\theta}{2}}$, $\frac{l_1-l_2}{2 \cos \frac{\theta}{2}}, \frac{m_1-m_2}{2 \cos \frac{\theta}{2}}, \frac{n_1-n_2}{2 \cos \frac{\theta}{2}}$, $\frac{l_1-l_2}{2 \sin \frac{\theta}{2}}, \frac{m_1-m_2}{2 \sin \frac{\theta}{2}}, \frac{n_1-n_2}{2 \sin \frac{\theta}{2}}$, $\frac{l_1+I_2}{2 \cos \frac{\theta}{2}}, \frac{m_1+m_2}{2 \cos \frac{\theta}{2}}, \frac{n_1+n_2}{2 \cos \frac{\theta}{2}}$

MARKS App · Accepted Answer

$\because l_1 l_2+m_1 m_2+n_1 n_2=\cos \theta$ 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 $l_1+l_2, m_1+m_{2,}, n_1+n_2$ $D R$ 's of $O B$ are $l_1-l_2, m_1-m_2, n_1-n_2$ Now, $\quad \Sigma\left(l_1+l_2\right)^2=\mathbf{l}+\mathbf{l}+2 \cos \theta$ $=2(1+\cos \theta)=4 \cos ^2 \frac{\theta}{2}$ $\therefore D C$ 's of internal bisector are $\frac{l_1+l_2}{2 \cos \frac{\theta}{2}}, \frac{m_1+m_2}{2 \cos \frac{\theta}{2}}, \frac{n_1+n_2}{2 \cos \frac{\theta}{2}}$

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