Question

If \[{l_1},{\text{ }}{m_1},{\text{ }}{n_1},\]and \[{l_2},{\text{ }}{m_2},{\text{ }}{n_2},\] be the DC’s of two concurrent lines, the direction cosines of the line bisecting the angles between them are proportional to
A) \[{l_1}{l_2},{\text{ }}{m_1}{m_2},{\text{ }}{n_1}{n_2}\]
B) \[{l_1}{m_2},{\text{ }}{l_1}{n_2},{\text{ }}{l_1}{n_3}\]
C) \[{l_1} + {\text{ }}{l_2}{,_{}}{m_1} + {\text{ }}{m_2},{\text{ }}{n_1} + {\text{ }}{n_2}\]
D) None of these

Answer 1

Hint: In the question we are given, the lines are concurrent and direction cosines are also given. We will start with drawing a figure first, and then using the given information we will solve accordingly.

Complete step-by-step answer:
Let us first draw a figure, in it let the two concurrent lines be OA and OB, where O is the meeting point and C is the midpoint of A and B. And let the direction cosines of OA \[ = \] \[{l_1},{\text{ }}{m_1},{\text{ }}{n_1},\] and direction cosines of OB \[ = \] \[{l_2},{\text{ }}{m_2},{\text{ }}{n_2},\]

Now, since OA and OB are concurrent lines, so we can take, \[OA{\text{ }} = {\text{ }}OB{\text{ }} = {\text{ }}r\]
Then, C is the midpoint of AB. Then, OC is the bisector of the \[\angle AOB.\]
Now, with that we will get the coordinates of A and B which are \[\left( {{l_1}r,{m_1}r,{n_1}r} \right)\] and \[\left( {{l_2}r,{m_2}r,{n_2}r} \right)\]respectively.
So, we will get the coordinates of C which are,
\[\frac{{({l_1} + {l_2})r}}{2},\frac{{({m_1} + {m_2})r}}{2},\frac{{({n_1} + {n_2})r}}{2}\]
So, the direction cosines of OC become proportional to \[{l_1} + {\text{ }}{l_2}{,_{}}{m_1} + {\text{ }}{m_2},{\text{ }}{n_1} + {\text{ }}{n_2}\]
Thus, option (C), \[{l_1} + {\text{ }}{l_2}{,_{}}{m_1} + {\text{ }}{m_2},{\text{ }}{n_1} + {\text{ }}{n_2}\] is correct.
So, the correct answer is “Option C”.

Note: The lines which pass through the same point and are equal, are called concurrent lines. Direction cosines of a line are the cosines of the angle made by the line with the positive direction of the coordinate axes.