Question

The equation of angular bisector between the lines $\frac{x-2}{1}=\frac{y-3}{1}=\frac{4-z}{1}$ and $\frac{1-x}{1}=\frac{y-4}{2}=\frac{z-5}{1}$ is given by

• A. $\frac{3x-5}{\sqrt{2}-1}=\frac{3y-8}{2+\sqrt{2}}=\frac{3z-13}{1-\sqrt{2}}$
• B. $\frac{3x-5}{\sqrt{2}-2}=\frac{3y-8}{2+\sqrt{2}}=\frac{3z-13}{1-\sqrt{2}}$
• C. $\frac{3x-5}{\sqrt{2}+1}=\frac{3y-8}{\sqrt{2}+2}=\frac{3z-13}{1+\sqrt{2}}$
• D. $\frac{3x-5}{\sqrt{2}+1}=\frac{3y-8}{\sqrt{2}-2}=\frac{3z-13}{-1-\sqrt{2}}$

Answer 1

Answer: (A), (D)

$\frac{3x-5}{\sqrt{2}+1}=\frac{3y-8}{\sqrt{2}-2}=\frac{3z-13}{-1-\sqrt{2}}$

Given lines : $L_1:\frac{x-2}{1}=\frac{y-3}{1}=\frac{4-z}{1}=t$
$\Rightarrow (x,y,z)=(t+2,t+3,4-t)$
and $L_2:\frac{1-x}{1}=\frac{y-4}{2}=\frac{z-5}{1}=s$
$\Rightarrow (x,y,z)=(1-s,2s+4,s+5)$
For point of intesection :
$t+2=1-s,t+3=2s+4,4-t=s+5$
on sloving : $s=\frac{-2}{3},t=\frac{-1}{3}$
So, point of intersection is : $(\frac{5}{3},\frac{8}{3},\frac{13}{3})$
Now, $D.C^{\prime }s$ of $L_1=(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{-1}{\sqrt{3}})$ and
$D.C^{\prime }s$ of $L_2=(\frac{-1}{\sqrt{6}},\frac{2}{\sqrt{6}},\frac{1}{\sqrt{6}})$
So, $D.R^{\prime }s$ of angular bisector $=(l_1\pm l_2,m_1\pm m_2,n_1\pm n_2)$
$i.e.(\frac{\sqrt{2}-1}{\sqrt{6}},\frac{\sqrt{2}+2}{\sqrt{6}},\frac{1-\sqrt{2}}{\sqrt{6}})$ or $(\frac{\sqrt{2}+1}{\sqrt{6}},\frac{\sqrt{2}-2}{\sqrt{6}},\frac{-\sqrt{2}-1}{\sqrt{6}})$
$\Rightarrow D.R,s$of angular bisector is : $(\sqrt{2}-1,\sqrt{2}+2,1-\sqrt{2})$ or $(\sqrt{2}+1,\sqrt{2}-2,-\sqrt{2}-1)$
So, equation is: $\frac{x-\frac{5}{3}}{\sqrt{2}-1}=\frac{y-\frac{8}{3}}{\sqrt{2}+2}=\frac{z-\frac{13}{3}}{1-\sqrt{2}}$ or
$\frac{x-\frac{5}{3}}{\sqrt{2}+1}=\frac{y-\frac{8}{3}}{\sqrt{2}-2}=\frac{z-\frac{13}{3}}{-1-\sqrt{2}}$
$\Rightarrow \frac{3x-5}{\sqrt{2}-1}=\frac{3y-8}{2+\sqrt{2}}=\frac{3z-13}{1-\sqrt{2}}$ or $\frac{3x-5}{\sqrt{2}+1}=\frac{3y-8}{\sqrt{2}-2}=\frac{3z-13}{-1-\sqrt{2}}$