Question

The acute angle bisector between the intersecting lines $\frac{x-1}{1}=\frac{y}{-1}=\frac{z-1}{3}$ and $\frac{x-1}{-3}=\frac{y}{-1}=\frac{z-1}{1}$ is

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

Answer 1

The correct option is B $\frac{x-1}{-1}=\frac{y}{-1}=\frac{z-1}{2}$

$L_1:\frac{x-1}{1}=\frac{y}{-1}=\frac{z-1}{3}$ and $L_2:\frac{x-1}{-3}=\frac{y}{-1}=\frac{z-1}{1}$
$D.R^{\prime }s$ of $L_1=(1,-1,3)$
$D.R^{\prime }s$ of $L_2=(-3,-1,1)$
$\Rightarrow D.C^{\prime }s$ of $L_1=(\frac{1}{\sqrt{11}},\frac{-1}{\sqrt{11}},\frac{3}{\sqrt{11}})$
and $D.C^{\prime }s$ of $L_2=(\frac{-3}{\sqrt{11}},\frac{-1}{\sqrt{11}},\frac{1}{\sqrt{11}})$
since $l_1{l}_2+m_1{m}_2+n_1{n}_2>0$
So, Acute angle bisector is : $\frac{x-x_1}{l_1+l_2}=\frac{y-y_1}{m_1+m_2}=\frac{z-z_1}{n_1+n_2}\Rightarrow \frac{x-1}{-2/\sqrt{11}}=\frac{y}{-2/\sqrt{11}}=\frac{z-1}{4/\sqrt{11}}\Rightarrow \frac{x-1}{-1}=\frac{y}{-1}=\frac{z-1}{2}$