Question

The perpendicular distance (in units) from origin to the obtuse angular bisector between the lines $\frac{x-1}{-1}=\frac{y+1}{2}=\frac{z-1}{1}$ and $\frac{x-1}{1}=\frac{y+1}{-1}=\frac{z-1}{2}$ is :

• A. $\frac{\sqrt{13}}{5}$
• B. $\sqrt{\frac{13}{5}}$
• C. $\sqrt{\frac{17}{5}}$
• D. $\frac{\sqrt{17}}{5}$

Answer 1

The correct option is B $\sqrt{\frac{13}{5}}$

Given lines are $L_1:\frac{x-1}{-1}=\frac{y+1}{2}=\frac{z-1}{1}$ and $L_2:\frac{x-1}{1}=\frac{y+1}{-1}=\frac{z-1}{2}$
$D.R^{\prime }s$ of $L_1=(-1,2,1)$
$\Rightarrow D.c^{\prime }s$ of $L_1=(l_1,m_1,n_1)=(\frac{-1}{\sqrt{6}},\frac{2}{\sqrt{6}},\frac{1}{\sqrt{6}})$
$D.R^{\prime }s$ of $L_2=(1,-1,2)$
$\Rightarrow D.c^{\prime }s$ of $L_2=(l_2,m_2,n_2)=(\frac{1}{\sqrt{6}},\frac{-1}{\sqrt{6}},\frac{2}{\sqrt{6}})$
Here $l_1\cdot l_2+m_1\cdot m_2+n_1\cdot n_2=-\frac{1}{6}-\frac{2}{6}+\frac{2}{6}=-\frac{1}{6}<0$
If $l_1\cdot l_2+m_1\cdot m_2+n_1\cdot n_2<0$
So, Acute angle bisector $D.R^{\prime }s$ will be
$(l_1-l_2,m_1-m_2,n_1-n_2)$
Obtuse angle bisector $D.R^{\prime }s$ will be $(l_1+l_2,m_1+m_2,n_1+n_2)$ i.e. $(0,\frac{1}{\sqrt{6}},\frac{3}{\sqrt{6}})$
$\therefore$ Obtuse angle bisector equation is :
$\frac{x-1}{0}=\frac{y+1}{1}=\frac{z-1}{3}$
Now, let $\frac{x-1}{0}=\frac{y+1}{1}=\frac{z-1}{3}=t$
$\Rightarrow P(x,y,z)=(1,t-1,3t+1)$
$\therefore 1\cdot 0+(t-1)\cdot 1+(3t+1)\cdot 3=0$
$\Rightarrow 10t=-2\Rightarrow t=-\frac{1}{5}$
$\therefore P=(1,\frac{-6}{5},\frac{2}{5})$
So, we have $\overrightarrow{OP}=\sqrt{1+\frac{36}{25}+\frac{4}{25}}=\sqrt{\frac{13}{5}}$