Find an equation of a plane that is perpendicular to the plane x-2y+z=7
I know his has to do with vectors because the dot product must = 0....
:-/
Hello, stones44!
There is an infinite number of answers . . .Find an equation of a plane that is perpendicular to the plane $\displaystyle x-2y+z\:=\:7$
A door is perpendicular to the floor. .Swing the door
. . and you have thousands of planes perpendicular to the floor.
Our plane can go through any point, say, $\displaystyle P(5,\,6,\,7)$
Its normal vector, $\displaystyle \vec{n} \:=\:\langle a,\,b,\,c\rangle$, must be perpendicular to $\displaystyle \langle 1,\,\text{-}2,\,1\rangle$
Hence, we have: .$\displaystyle a - 2b + c \:=\:0$
. . We can use, for example: .$\displaystyle \vec{n} \:=\:\langle3,\,2,\,1\rangle$
The equation is: .$\displaystyle 3(x-5) + 2(y-6) + 1(z-7)\:=\:0$
. . . . . . . . . . . . $\displaystyle \boxed{3x + 2y + z \:=\:34}$