Question

lf $\theta$ is the angle between two lines whose d.c.s are $l_1,m_1,n_1$ and $l_2,m_2,n_2$, then the d.cs of one of the angular bisectors of the two lines are

• A. $\frac{l_1+l_2}{2},\frac{m_1+m_2}{2},\frac{n_1+n_2}{2}$
• B. $\frac{l_1+l_2}{2\cos \left(\frac{\theta }{2}\right)},\frac{m_1+m_2}{2\cos \left(\frac{\theta }{2}\right)},\frac{n_1+n_2}{2\cos \left(\frac{\theta }{2}\right)}$
• C. $\frac{l_1+l_2}{\cos \left(\frac{\theta }{2}\right)},\frac{m_1+m_2}{\cos \left(\frac{\theta }{2}\right)},\frac{n_1+n_2}{\cos \left(\frac{\theta }{2}\right)}$
• D. $\frac{l_I+l_2}{2\sin \left(\frac{\theta }{2}\right)}\frac{m_1+m_2}{2\sin \left(\frac{\theta }{2}\right)}\frac{n_{]}+n_2}{2\sin \left(\frac{\theta }{2}\right)}$

Answer 1

The correct option is B

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

Let the direction cosines of the line be $l_1,m_1,n_1$.

Hence, $l_1=\cos \alpha$, $m_1=\cos \beta$, $n_1=\cos \gamma$

Similarly, for another line

$l_2=\cos {\alpha }^{\prime }$, $m_2=\cos {\beta }^{\prime }$, $n_2=\cos {\gamma }^{\prime }$

Now the direction cosines of the angle bisector will be,

$\cos (\frac{\alpha -{\alpha }^{\prime }}{2}+{\alpha }^{\prime }),\cos (\frac{\beta -{\beta }^{\prime }}{2}+{\beta }^{\prime }),\cos (\frac{\gamma -{\gamma }^{\prime }}{2}+{\gamma }^{\prime })$

$=\cos \left(\frac{\alpha +{\alpha }^{\prime }}{2}\right),\cos \left(\frac{\beta +{\beta }^{\prime }}{2}\right),\cos \left(\frac{\gamma +{\gamma }^{\prime }}{2}\right)$

$=l,m,n$

Now consider $\cos \left(\frac{\alpha +{\alpha }^{\prime }}{2}\right)$

Multiplying and dividing by $2\cos \left(\frac{\alpha -{\alpha }^{\prime }}{2}\right)$, we get

$\frac{2\cos \left(\frac{\alpha -{\alpha }^{\prime }}{2}\right)\cos \left(\frac{\alpha +{\alpha }^{\prime }}{2}\right)}{2\cos \left(\frac{\alpha -{\alpha }^{\prime }}{2}\right)}$

$=\frac{\cos \alpha +\cos {\alpha }^{\prime }}{2\cos \left(\frac{\alpha -{\alpha }^{\prime }}{2}\right)}$

$=\frac{l_1+l_2}{2\cos \left(\frac{\alpha -{\alpha }^{\prime }}{2}\right)}$

$=l$

Hence, $l=\frac{l_1+l_2}{2\cos \left(\frac{\theta }{2}\right)}$

Similarly,

$m=\frac{m_1+m_2}{2\cos \left(\frac{\beta -{\beta }^{\prime }}{2}\right)}$

$n=\frac{n_1+n_2}{2\cos \left(\frac{\gamma -{\gamma }^{\prime }}{2}\right)}$

Hence, option B is correct.