Question

If the shortest distance(in $\text{units}$) between the lines $2x-2=y-\lambda =z$ and $x-1=2y=z-\lambda$ is equal to $\frac{10}{\sqrt{17}},$ then the value of $\lambda$ can be

• A. $-1$
• B. $-2$
• C. $2$
• D. $1$

Answer 1

Answer: (B), (C)

$2$

Given lines :
$L_1:2x-2=y-\lambda =z$ and $L_2:x-1=2y=z-\lambda$
$\Rightarrow L_1:\frac{x-1}{1}=\frac{y-\lambda }{2}=\frac{z}{2},\Rightarrow L_2:\frac{x-1}{2}=\frac{y}{1}=\frac{z-\lambda }{2}$
$D.R^{\prime }s$ of $L_1=(a_1,b_1,c_1)=(1,2,2)$ and $D.R^{\prime }s$ of $L_2=(a_2,b_2,c_2)=(2,1,2)$
$\therefore$ shortest distance $=\left|\frac{\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|}{\sqrt{\sum (b_1{c}_2-b_2{c}_1{)}^2}}\right|\frac{10}{\sqrt{17}}=\left|\frac{\left|\begin{array}{ccc}0& -\lambda & \lambda \\ 1& 2& 2\\ 2& 1& 2\end{array}\right|}{\sqrt{(4-2{)}^2+(2-4{)}^2+(1-4{)}^2}}\right|\Rightarrow |\lambda (2-4)+\lambda (1-4)|=10\Rightarrow 5\lambda =\pm 10\Rightarrow \lambda =\pm 2$