Question

Write the value of λ for which the lines $\frac{x-3}{-3}=\frac{y+2}{2\lambda }=\frac{z+4}{2}\mathrm{and}\frac{x+1}{3\lambda }=\frac{y-2}{1}=\frac{z+6}{-5}$ are perpendicular to each other.

Answer 1

We have

$$\begin{gathered} \frac{x-3}{-3}=\frac{y+2}{2\lambda }=\frac{z+4}{2} \\ \frac{x+1}{3\lambda }=\frac{y-2}{1}=\frac{z+6}{-5} \end{gathered}$$

The given lines are parallel to vectors $\overrightarrow{b_1}=-3\hat{i}+2\lambda \hat{j}+2\hat{k}\mathrm{and}\overrightarrow{b_2}=3\lambda \hat{i}+\hat{j}-5\hat{k}$.

For $\overrightarrow{b_1}\perp \overrightarrow{b_2}$, we must have

$$\begin{gathered} \overrightarrow{b_1}.\overrightarrow{b_2}=0 \\ \Rightarrow \left(-3\hat{i}+2\lambda \hat{j}+2\hat{k}\right).\left(3\lambda \hat{i}+\hat{j}-5\hat{k}\right)=0 \\ \Rightarrow -7\lambda -10=0 \\ \Rightarrow \lambda =-\frac{10}{7} \end{gathered}$$