Question

The distance (in units) of the point $(1,-5,9)$ from the plane $x-y+z=5$ measured along the line $x=y=z$ is:

• A. $\frac{20}{3}$
• B. $3\sqrt{10}$
• C. $10\sqrt{3}$
• D. $\frac{20}{\sqrt{3}}$

Answer 1

The correct option is C $10\sqrt{3}$

The equation of line through $A(1,-5,9)$ along the line $x=y=z$ is $\frac{x-1}{1}=\frac{y+5}{1}=\frac{z-9}{1}$
Any point on this line is $P(t+1,t-5,t+9)$, which lies on the plane $x-y+z=5$
Hence $t+1-(t-5)+t+9=5$
$\Rightarrow t+10=0$
$\Rightarrow t=-10$
$P(-10+1,-10-5,-10+9)$ i.e. $P(-9,-15,-1)$
Hence, required distance is $|\overrightarrow{AP}|=\sqrt{{10}^2+(-10{)}^2+{10}^2}=10\sqrt{3}$