Question

Find the equations of the line passing through the point (2, 1, 3) and perpendicular to the lines
$\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}\mathrm{and}\frac{x}{-3}=\frac{y}{2}=\frac{z}{5}$

Answer 1

Let:
$$\begin{gathered} \overrightarrow{b_1}=\hat{i}+2\hat{j}+3\hat{k} \\ \overrightarrow{b_2}=-3\hat{i}+2\hat{j}+5\hat{k} \end{gathered}$$

Since the required line is perpendicular to the lines parallel to the vectors $\overrightarrow{b_1}=\hat{i}+2\hat{j}+3\hat{k}\mathrm{and}\overrightarrow{b_2}=-3\hat{i}+2\hat{j}+5\hat{k}$, it is parallel to the vector $\overrightarrow{b}=\overrightarrow{b_1}\times \overrightarrow{b_2}$.

Now,
$$\begin{gathered} \overrightarrow{b}=\overrightarrow{b_1}\times \overrightarrow{b_2} \\ =\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 3\\ -3& 2& 5\end{array}\right| \\ =4\hat{i}-14\hat{j}+8\hat{k} \\ =2\left(2\hat{i}-7\hat{j}+4\hat{k}\right) \end{gathered}$$

Thus, the direction ratios of the required line are proportional to 2, $-$7, 4.

The equation of the required line passing through the point (2, 1, 3) and having direction ratios proportional to 2, $-$7, 4 is $\frac{x-2}{2}=\frac{y-1}{-7}=\frac{z-3}{4}$.