Question

If $l_1,m_1,n_1$ and $l_2,m_2,n_2$ are DCs of the two lines inclined to each other at an angle $\theta$, then the DCs of the internal bisector of the angle between these lines 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

The correct option is B $\frac{l_1+l_2}{2\cos \frac{\theta }{2}},\frac{m_1+m_2}{2\cos \frac{\theta }{2}},\frac{n_1+n_2}{2\cos \frac{\theta }{2}}$

Let $OA$ and $OB$ be two lines with $D{C}^{\prime }s{l}_1,m_1,n_1$ and $l_2,m_2,n_2.$

Let $OA=OB=1.$

Then co-ordinates of $A$ and $B$ are $(l_1,m_1,n_1)$ and $(l_2,m_2,n_2)$respectively.

Let $OCZ$ be the bisector of $\angle AOB$ such that $C$ is the mid-point of $AB$ and so its co-ordinates are

$(\frac{l_1+l_2}{2},\frac{m_1+m_2}{2},\frac{n_1+n_2}{2})$

$\therefore D{R}^{\prime }s$ of $OC$ are $\frac{l_1+l_2}{2},\frac{m_1+m_2}{2},\frac{n_2+n_2}{2}$

$\therefore$ We have

$OC=\sqrt{{\left(\frac{l_1+l_2}{2}\right)}^2+{\left(\frac{m_1+m_2}{2}\right)}^2+{\left(\frac{n_1+n_2}{2}\right)}^2}$

$=\frac{1}{2}\sqrt{(l_1^2+m_1^2+n_1^2)+(l_2^2+m_2^2+n_2^2)+2(l_1{l}_2+m_1{m}_2+n_1{n}_2)}$

$=\frac{1}{2}\sqrt{2+2\cos \theta }[\therefore \cos \theta =l_1{l}_2+m_1{m}_2+n_1{n}_2]$

$=\frac{1}{2}\sqrt{2(1+\cos \theta )}=\cos \left(\frac{\theta }{2}\right).$

$\therefore DCs$ of $\overrightarrow{OC}$ are $\frac{l_1+l_2}{2\left(OC\right)},\frac{m_1+m_2}{2\left(OC\right)},\frac{n_1+n_2}{2\left(OC\right)}$

i.e., $\frac{l_1+l_2}{\frac{2\cos \theta }{2}},\frac{m_1+m_2}{\frac{2\cos \theta }{2}},\frac{n_1+n_2}{\frac{2\cos \theta }{2}}$