Equation of Plane: Finding a Normal Vector for Three Given Points

[shards5]

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]

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...