Question

Find the equation of the line passing through the point (1, −1, 1) and perpendicular to the lines joining the points (4, 3, 2), (1, −1, 0) and (1, 2, −1), (2, 1, 1).

Answer 1

The direction ratios of the line joining the points (4, 3, 2), (1,$-$1, 0) and (1, 2, $-$1), (2, 1, 1) are $-$3, $-$4, $-$2 and 1, $-$1, 2, respectively.

Let:
$$\begin{gathered} \overrightarrow{b_1}=-3\hat{i}-4\hat{j}-2\hat{k} \\ \overrightarrow{b_2}=\hat{i}-\hat{j}+2\hat{k} \end{gathered}$$

Since the required line is perpendicular to the lines parallel to the vectors $\overrightarrow{b_1}=-3\hat{i}-4\hat{j}-2\hat{k}\mathrm{and}\overrightarrow{b_2}=\hat{i}-\hat{j}+2\hat{k}$, it is parallel to the vector $\overrightarrow{b}=\overrightarrow{b_1}\times \overrightarrow{b_2}$.

Now,
$$\begin{gathered} \overrightarrow{b}=\overrightarrow{b_1}\times \overrightarrow{b_2} \\ =\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -3& -4& -2\\ 1& -1& 2\end{array}\right| \\ =-10\hat{i}+4\hat{j}+7\hat{k} \end{gathered}$$

So, the direction ratios of the required line are proportional to $-$10, 4, 7.

The equation of the required line passing through the point (1, $-$1, 1) and having direction ratios proportional to $-$10, 4, 7 is $\frac{x-1}{-10}=\frac{y+1}{4}=\frac{z-1}{7}$.