Question

The length (in units) of the perpendicular from the point $(-1,3,9)$ to the line $\frac{x-13}{5}=\frac{y+8}{-8}=\frac{z-31}{1}$ is:

• A. $21$
• B. $22$
• C. $20$
• D. $\sqrt{439}$

Answer 1

The correct option is A $21$

Let $A=(-1,3,9)$
Given line is $\frac{x-13}{5}=\frac{y+8}{-8}=\frac{z-31}{1}=t$
Any point $P$ on the line is $(13+5t,-8-8t,31+t)$
Let $P$ be the foot of the perpendicular of $A$
$\Rightarrow$ D.r's of $AP=(14+5t,-11-8t,22+t)$
$AP$ is perpendicular to given line.
$\Rightarrow a_1{a}_2+b_1{b}_2+c_1{c}_2=0$
$\Rightarrow 5(14+5t)-8(-11-8t)+1(22+t)$
$\Rightarrow 70+25t+88+64t+22+t=0$
$\Rightarrow 180+90t=0$
$\Rightarrow t=-2$
So, we have $P=(3,8,29)$ and$AP=(4,5,20)$
Length of $AP=\sqrt{(4{)}^2+(5{)}^2+(20{)}^2}$
Hence, length of perpendicular $=|AP|=21$