# Equation to the plane.

• Sep 15th 2013, 11:17 AM
petenice
Equation to the plane.
Find an equation of the plane that passes through the points (4, 2, 1) and (-4, 7, 9) and is parallel to the z-axis.

So i know i need a point (x,y,z) and a normal vector n=<a,b,c> i have two points but how do i get a normal vector with the two points given?. I know that the final equation will be in form
ax+by+d=0 with the z axis missing because its parallel to it. Thank you. If possible can you give me the final answer so i can see what im working towards.
• Sep 15th 2013, 11:43 AM
HallsofIvy
Re: Equation to the plane.
Quote:

Originally Posted by petenice
Find an equation of the plane that passes through the points (4, 2, 1) and (-4, 7, 9) and is parallel to the z-axis.

So i know i need a point (x,y,z) and a normal vector n=<a,b,c> i have two points but how do i get a normal vector with the two points given?. I know that the final equation will be in form
ax+by+d=0 with the z axis missing because its parallel to it. Thank you. If possible can you give me the final answer so i can see what im working towards.

Yes, the equation is of the form ax+ by+ d= 0. Further, since you can divide that equation by any number without changing the line it represents, you can assume that one of a, b, and d is 1.

Since (4, 2, 1) and (-4, 7, 9) are in the plane that they must satisfy the equation. That gives you two equations to solve for the remaining two values.
• Sep 15th 2013, 11:48 AM
petenice
Re: Equation to the plane.
Quote:

Originally Posted by HallsofIvy
Further, since you can divide that equation by any number without changing the line it represents, you can assume that one of a, b, and d is 1.

Im not clear on what this mean. How does this help me get a normal vector?
• Sep 15th 2013, 12:05 PM
HallsofIvy
Re: Equation to the plane.
It doesn't it give you the pane, which is what you want!
• Sep 15th 2013, 02:41 PM
petenice
Re: Equation to the plane.
Ok got it. 5x+8y-36=0 Thank You