Distance Between Point and Plane

$\pi :Ax+By+Cz+D=0$ $p:(a,b,c)$ $d=\frac{|Aa+Bb+Cc+D|}{\sqrt{A^2+B^2+C^2}}$ Alternatively, with Normal Vector $n=(A,B,C)$ $d=\frac{n\cdot p}{||n|{|}^2}$

Example

• Given a plane as $\pi :x+y+z=0$
• Given a vector $v=(1,2,3)$
• We get the Normal Vector from Cartesian Equation of Plane as $n=(1,1,1)$
• Then, $n\cdot v=1+2+3=6$
• Then, $||n|{|}^2=\sqrt{3}$
• Thus, $d=\frac{6}{\sqrt{3}}=2\sqrt{3}$