Question

Let two lines $L_1:\frac{x-3}{3}=\frac{y-2}{-4}=\frac{z}{-1}$ and $L_2:\frac{x-3}{4}=\frac{y-2}{1}=\frac{z}{-3}$ and $\frac{x-3}{a}=\frac{y-2}{6}=\frac{z}{b},(a,b\in R)$ is the acute angle biscetor between lines $L_1$ and $L_2.$ Then the value of $|a-b|=$

• A. 22.00
• B. 22.0
• C. 22

Answer 1

Answer: (A), (B), (C)

Given lines $L_1:\frac{x-3}{3}=\frac{y-2}{-4}=\frac{z}{-1}$
$\Rightarrow D.R^{\prime }s$ of $L_1=(3,-4,-1)$
$\Rightarrow D,C^{\prime }s=(\frac{3}{\sqrt{26}},\frac{-4}{\sqrt{26}},\frac{-1}{\sqrt{26}})$
and $L_2:\frac{x-3}{4}=\frac{y-2}{1}=\frac{z}{-3}$
$\Rightarrow D.R^{\prime }s$ of $L_2=(4,1,-3)$
$\Rightarrow D.C^{\prime }s$ of $L_2=(\frac{4}{\sqrt{26}},\frac{1}{\sqrt{26}},\frac{-3}{\sqrt{26}})$
As, $l_1{l}_2+m_1{m}_2+n_1{n}_2>0$
Acute angle bisector is :
$\frac{x-x_1}{l_1+l_2}=\frac{y-y_1}{m_1+m_2}=\frac{z-z_1}{n_1+n_2};(x_1,y_1,z_1)=(3,2,0)$
$\Rightarrow \frac{x-3}{7/\sqrt{26}}=\frac{y-2}{-3/\sqrt{26}}=\frac{z}{-4/\sqrt{26}}\Rightarrow \frac{x-3}{-14}=\frac{y-2}{6}=\frac{z}{8}$
comparing with : $\frac{x-3}{a}=\frac{y-2}{6}=\frac{z}{b}$
$\Rightarrow a=-14,b=8$
$\Rightarrow |a-b|=22$