Question

The length of the perpendicular drawn from $(1,2,3)$ to the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$ is

• A. $4$
• B. $5$
• C. $6$
• D. $7$

Answer 1

The correct option is D $7$

Direction cosines of the given line are

$\frac{3}{\sqrt{17}},\frac{2}{\sqrt{17}},\frac{-2}{\sqrt{17}}$

$\therefore AM=|\left(6.1\right).\frac{3}{\sqrt{17}}+(7-2).\frac{2}{\sqrt{17}}+(7-3).\frac{-2}{\sqrt{17}}|=17$

$AP=\sqrt{{(16-1)}^2+{(7-2)}^2+{(7-3)}^2}$

$=\sqrt{25+25+16}=\sqrt{66}$

$\therefore$ length of the perpendicular is

$PM=\sqrt{{AP}^2-{AM}^2}$

$=\sqrt{66-17}=\sqrt{49}=7$