# Equation of the Plane

## Homework Statement

Find an equation of a plane containing the three points (1, -1, 0), (5, 4, 1), (5, 5, 3) in which the coefficient of x is 9. What does it mean when it says the coefficient of x is 9?

## Homework Equations

a(x-x0) + b(y-y0) + c(z-z0) = 0

## The Attempt at a Solution

Since the equation of the plane needs a normal vector I can get the normal vector using the following steps:
ab = (5, 4, 1) - (1, -1, 0) = (4,5,1)
ac = (5, 5, 3) - (1, -1, 0) = (4,6,3)
Normal Vector = (4,5,1) x (4,6,3) = <9,-8,4>
So I would get 9(x-1) - 8(y+1) +4(z-0) = 0 but where does the coefficient of x is 9 come in?

Mark44
Mentor

## Homework Statement

Find an equation of a plane containing the three points (1, -1, 0), (5, 4, 1), (5, 5, 3) in which the coefficient of x is 9. What does it mean when it says the coefficient of x is 9?

## Homework Equations

a(x-x0) + b(y-y0) + c(z-z0) = 0

## The Attempt at a Solution

Since the equation of the plane needs a normal vector I can get the normal vector using the following steps:
ab = (5, 4, 1) - (1, -1, 0) = (4,5,1)
ac = (5, 5, 3) - (1, -1, 0) = (4,6,3)
Normal Vector = (4,5,1) x (4,6,3) = <9,-8,4>
So I would get 9(x-1) - 8(y+1) +4(z-0) = 0 but where does the coefficient of x is 9 come in?
The coefficient of x in your plane equation is 9 (9x - 9 - 8y - 8 + 4z = 0). As it turned out, 9 was the x coordinate of the normal.

If you had come up with a normal of, say, <3, -8/3, 4/3>, your equation would have been 3(x - 1) - 8/3(y + 1) + 4/3(z - 0) = 0. You could have adjusted the equation to get an x coefficient of 9 by multiplying both sides of the equation by 3.

I guess that's where they were going with this problem...