Question

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

• A. $7$
• B. $22$
• C. $20$
• D. $\sqrt{439}$

Answer 1

The correct option is A $7$

Let $A=(1,2,3)$ and $B=(6,7,7)$
Given: $\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}=\lambda$
Let us take foot of the perpendicular on the line as $P=(3\lambda +6,2\lambda +7,-2\lambda +7)$
Direction ratio's of line which is perpendicular to given line $PA=(3\lambda +6-1,2\lambda +7-2,-2\lambda +7-3)$
i.e. D.R's of $PA=(3\lambda +5,2\lambda +5,-2\lambda +4)$
and the direction ratio's of given line are $(3,2,-2)$
These two lines are perpendicular, so
$(3\lambda +5)\times 3+(2\lambda +5)\times 2+(-2\lambda +4)\times (-2)=0$
$\Rightarrow 17\lambda +17=0$
$\therefore \lambda =-1$
So point is $P=(-3+6,-2+7,2+7)$
$P=(3,5,9)$
So, perpendicular distance $AP$ is $=\sqrt{(3-1{)}^2+(5-2{)}^2+(9-3{)}^2}=7$