Question

The normal form of $2x-2y+z=5$ is

• A. $12x-4y+3z=39$
• B. $\frac{-6}{7}x+\frac{2}{7}y+\frac{3}{7}z=1$
• C. $\frac{12}{13}x-\frac{-4}{13}y+\frac{3}{13}z=3$
• D. $\frac{2}{3}x-\frac{2}{3}y+\frac{1}{3}z=\frac{5}{3}$

Answer 1

The correct option is D

$\frac{2}{3}x-\frac{2}{3}y+\frac{1}{3}z=\frac{5}{3}$

The dr's of the normal to the plane are $(2,-2,1)$.
The dc's will be $(\frac{2}{3},\frac{-2}{3},\frac{1}{3})$
Hence, the equation of the plane in the normal form will be,
$\frac{2x}{3}-\frac{2y}{3}+\frac{z}{3}$ = $\frac{5}{3}$