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.

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?

Re: Equation to the plane.

It **doesn't** it give you the pane, which is what you want!

Re: Equation to the plane.

Ok got it. 5x+8y-36=0 Thank You