Question

The equation of a line which is parallel to the angle bisector between the lines $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{0}$ and $\frac{x-1}{0}=\frac{y-4}{-1}=\frac{z-5}{-1}$ and passing through the point $(1,-2,3)$ is

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

Answer 1

Answer: (A), (D)

$\frac{x-1}{1}=\frac{y+2}{2}=\frac{z-3}{1}$

Given lines :
$L_1:\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{0}$ and $L_2:\frac{x-1}{0}=\frac{y-4}{-1}=\frac{z-5}{-1}$
$D.R^{\prime }s$ of $L_1=(1,1,0)$
$\Rightarrow D.C^{\prime }s$ of $L_1=(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0)$ and
$D.R^{\prime }s$ of $L_2=(0,-1,-1)$
$\Rightarrow D.C^{\prime }s$ of $L_2=(0,\frac{-1}{\sqrt{2}},\frac{-1}{\sqrt{2}})$
So, $D.R^{\prime }s$ of angular bisector is $(l_1\pm l_2,m_1\pm m_2,n_1\pm n_2)=(\frac{1}{\sqrt{2}},0,\frac{-1}{\sqrt{2}})$
or $(\frac{1}{\sqrt{2}},\frac{2}{\sqrt{2}},\frac{1}{\sqrt{2}})$
$\Rightarrow (1,0,-1)$ or $(1,2,1)$
So, equation of required lines can be :
$\frac{x-1}{1}=\frac{y+2}{0}=\frac{z-3}{-1}$ or $\frac{x-1}{1}=\frac{y+2}{2}=\frac{z-3}{1}$