Question

The perpendicular distance of the point P (1, 2, 3) from the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$ is
(a) 7
(b) 5
(c) 0
(d) none of these

Answer 1

(a) 7

We have

$\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$

Let point (1, 2, 3) be P and the point through which the line passes be Q (6, 7, 7). Also, the line is parallel to the vector $\overrightarrow{b}=3\hat{i}+2\hat{j}-2\hat{k}$.

Now,

$\overrightarrow{PQ}=5\hat{i}+5\hat{j}+4\hat{k}$

$$\begin{gathered} \therefore \overrightarrow{b}\times \overrightarrow{PQ}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 2& -2\\ 5& 5& 4\end{array}\right| \\ =18\hat{i}-22\hat{j}+5\hat{k} \\ \Rightarrow \left|\overrightarrow{b}\times \overrightarrow{PQ}\right|=\sqrt{{18}^2+{\left(-22\right)}^2+5^2} \\ =\sqrt{324+484+25} \\ =\sqrt{833} \\ \therefore d=\frac{\left|\overrightarrow{b}\times \overrightarrow{PQ}\right|}{\left|\overrightarrow{b}\right|} \\ =\frac{\sqrt{833}}{\sqrt{17}} \\ =\sqrt{49} \\ =7 \end{gathered}$$