If the lines $\frac{1-x}{2}=\frac{y-8}{\lambda}=\frac{z-5}{2}$ and…
If the lines $\frac{1-x}{2}=\frac{y-8}{\lambda}=\frac{z-5}{2}$ and $\frac{x-11}{5}=\frac{y-3}{3}=\frac{z-1}{1}$ are perpendicular, then $\lambda=$
4
$-4$
$\frac{8}{3}$
$\frac{-8}{3}$
Solution
We have $\frac{1-x}{2}=\frac{y-8}{\lambda}=\frac{z-5}{2} \Rightarrow \frac{x-1}{-2}=\frac{y-8}{\lambda}=\frac{z-5}{2}$
Direction ratio of given lines are $-2, \lambda, 2$ and $5,3,1$.
Given lines are $\perp$ er. $\therefore-2(5)+\lambda(3)+2(1)=0$
$-10+3 \lambda+2=0 \Rightarrow \lambda=\frac{8}{3}$