If the lines $\frac{2 x-4}{\lambda}=\frac{y-1}{2}=\frac{z-3}{1}$ and $\frac{x-1}{1}=\frac{3…
If the lines $\frac{2 x-4}{\lambda}=\frac{y-1}{2}=\frac{z-3}{1}$ and $\frac{x-1}{1}=\frac{3 y-1}{\lambda}=\frac{z-2}{1}$ are perpendicular to each other, then $\lambda=$
$\frac{-7}{6}$
$\frac{6}{7}$
$\frac{-6}{7}$
$\frac{7}{6}$
Solution
Lines $\frac{2(x-2)}{\lambda}=\frac{y-1}{2}=\frac{z-3}{1}$ and $\frac{x-1}{1}=\frac{3\left(y-\frac{1}{3}\right)}{\lambda}=\frac{z-2}{1}$ are perpendicular to one another.
$\begin{aligned}
& \therefore\left(\frac{\lambda}{2}\right)(1)+(2)\left(\frac{\lambda}{3}\right)+(1)(1)=0 \\
& \therefore \frac{\lambda}{2}+\frac{2 \lambda}{3}=-1 \Rightarrow \lambda=\frac{-6}{7}
\end{aligned}$