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

Answer 1

The correct option is 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,

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

$\therefore$ $\overrightarrow{b}\times \overrightarrow{PQ}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 2& -2\\ 5& 5& 4\end{array}\right|$

$=\hat{i}(8+10)-\hat{j}(12+10)+\hat{k}(15-10)$

$=18\hat{i}-22\hat{j}+5\hat{k}$

$\Rightarrow$ $|\overrightarrow{b}\times \overrightarrow{PQ}|=\sqrt{(18{)}^2+(-22{)}^2+(5{)}^2}$

$=\sqrt{324+484+25}$

$=\sqrt{833}$

$\Rightarrow$ $d=\frac{|\overrightarrow{b}\times \overrightarrow{PQ}|}{\left|\overrightarrow{b}\right|}$

$=\frac{|\sqrt{833|}}{|\sqrt{(3{)}^2+(2{)}^2+(2{)}^2|}}$

$=\frac{\sqrt{833}}{\sqrt{17}}$

$=\sqrt{49}$

$=7$