# More Planes

• Jan 12th 2008, 12:01 PM
stones44
More Planes
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....
:-/
• Jan 12th 2008, 02:27 PM
ThePerfectHacker
There is more than one answer to this problem.
• Jan 12th 2008, 02:33 PM
Soroban
Hello, stones44!

Quote:

Find an equation of a plane that is perpendicular to the plane $\displaystyle x-2y+z\:=\:7$
There is an infinite number of answers . . .
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: .$\displaystyle 3(x-5) + 2(y-6) + 1(z-7)\:=\:0$

. . . . . . . . . . . . $\displaystyle \boxed{3x + 2y + z \:=\:34}$