Question

Direction ratios of the line bisecting the acute angle between lines whose direction ratios are $(1,2,-1)$ and $(2,-1,-1),$ is

• A. $(\frac{3}{\sqrt{6}},-\frac{1}{\sqrt{6}},-\frac{2}{\sqrt{6}})$
• B. $(\frac{1}{\sqrt{6}},-\frac{3}{\sqrt{6}},-\frac{2}{\sqrt{6}})$
• C. $(\frac{2}{\sqrt{6}},-\frac{1}{\sqrt{6}},-\frac{3}{\sqrt{6}})$
• D. $(\frac{3}{\sqrt{6}},\frac{1}{\sqrt{6}},-\frac{2}{\sqrt{6}})$

Answer 1

The correct option is D $(\frac{3}{\sqrt{6}},\frac{1}{\sqrt{6}},-\frac{2}{\sqrt{6}})$

Given : $D.r^{\prime }s$ of line $L_1=(a_1,b_1,c_1)=(1,2,-1)$
$\Rightarrow D.c^{\prime }s$ of $L_1=(l_1,m_1,n_1)=(\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}},-\frac{1}{\sqrt{6}})$
and $D.r^{\prime }s$ of line $L_2=(a_2,b_2,c_2)=(2,-1,-1)$
$\Rightarrow D.c^{\prime }s$ of $L_2=(l_2,m_2,n_2)=(\frac{2}{\sqrt{6}},-\frac{1}{\sqrt{6}},-\frac{1}{\sqrt{6}})$
As, $l_1{l}_2+m_1{m}_2+n_1{n}_2>0$
So,
$D.r^{\prime }s$ of acute angle bisector is : $(l_1+l_2,m_1+m_2,n_1+n_2)=(\frac{3}{\sqrt{6}},\frac{1}{\sqrt{6}},-\frac{2}{\sqrt{6}})$