Question

The equation(s) of angular bisector between the intersecting lines $L_1:\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ and $L_2:\frac{x}{-3}=\frac{y}{-2}=\frac{z}{-1}$ is/are

• A. $\frac{y}{2}=\frac{z}{3},x=0$
• B. $x+z=y=0$
• C. $x=y=z$
• D. $\frac{x}{1}=\frac{z}{2},y=0$

Answer 1

Answer: (B), (C)

$x=y=z$

Given lines : $L_1:\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ and $L_2:\frac{x}{-3}=\frac{y}{-2}=\frac{z}{-1}$
$\Rightarrow D.R^{\prime }s$ of $L_1=(1,2,3)$ and
$\Rightarrow D.C^{\prime }s$ of $L_1=(\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}})$
$D.R^{\prime }s$ of $L_2=(-3,-2,-1)$
$\Rightarrow D.C^{\prime }s$ of $L_2=(\frac{-3}{\sqrt{14}},\frac{-2}{\sqrt{14}},\frac{-1}{\sqrt{14}})$
We know,
equation of angle bisector is : $\frac{x-x_1}{l_1\pm l_2}=\frac{y-y_1}{m_1\pm m_2}=\frac{z-z_1}{n_1\pm n_2}$
Here, $(x_1,y_1,z_1)=(0,0,0)$
So, bisectors are : $\frac{x}{-2/\sqrt{14}}=\frac{y}{0}=\frac{z}{2/\sqrt{14}}$ or $\frac{x}{4/\sqrt{14}}=\frac{y}{4/\sqrt{14}}=\frac{z}{4/\sqrt{14}}$
$\Rightarrow x+z=y=0$ OR $x=y=z$