Question

If $l_1,m_1,n_1$ and $l_2,m_2,n_2$ be the DC's of two concurrent lines, the direction cosines of the line bisecting the angles between them are proportional to

• A. $l_1{l}_2,m_1{m}_2,n_1,n_2$
• B. $l_1{m}_2,l_1{n}_2,l_1{n}_3$
• C. $l_1+l_2,m_1+m_2,n_1+n_2$
• D. None of these

Answer 1

Answer: (C)

$l_1+l_2,m_1+m_2,n_1+n_2$

Let direction cosines of $OA=(l_1,m_1,n_1)$ and direction cosines of $OB=(l_2,m_2,n_2)$.
Taking two points on $l$ and $m$ such that $OA=OB=r$
Let $C$ be the mid-point of $AB$. Then, $OC$ is the bisector of the angle $AOB$.
Now, coordinates of $A$ are $(l_{1}r,m_{1}r,n_{1}r)$ and coordinates of $B$ are $(l_{2}r,m_{2}r,n_{2}r)$.
$\therefore$ the coordinates of $C$ are
$\frac{(l_1+l_2)r}{2},\frac{(m_1+m_2)r}{2},\frac{(n_1+n_2)r}{2}$.
Hence, the direction cosines of $OC$ are proportional to $l_1+l_2,m_1+m_2$ and $n_1+n_2$.