Question

Find the perpendicular distance of the point (3, −1, 11) from the line $\frac{x}{2}=\frac{y-2}{-3}=\frac{z-3}{4}.$

Answer 1

Let the point (3, $-$1, 11) be P and the point through which the line passes be Q (0, 2, 3).
The line is parallel to the vector $\overrightarrow{b}=2\hat{i}-3\hat{j}+4\hat{k}$.

Now,

$\overrightarrow{PQ}=-3\hat{i}+3\hat{j}-8\hat{k}$
$$\begin{gathered} \therefore \overrightarrow{b}\times \overrightarrow{PQ}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -3& 4\\ -3& 3& -8\end{array}\right| \\ =12\hat{i}+4\hat{j}+15\hat{k} \\ \Rightarrow \left|\overrightarrow{b}\times \overrightarrow{PQ}\right|=\sqrt{{12}^2+4^2+{15}^2} \\ =\sqrt{144+16+225} \\ =\sqrt{385} \\ d=\frac{\left|\overrightarrow{b}\times \overrightarrow{PQ}\right|}{\left|\overrightarrow{b}\right|} \\ =\frac{\sqrt{385}}{\sqrt{29}} \\ =\sqrt{\frac{385}{29}} \end{gathered}$$

Disclaimer: The answer given in the book is incorrect. This solution is created according to the question given in the book.