# A plane through three points

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1. Jul 27, 2016

### Heatherfield

Hi,

I'm currently reading Calc III by Marsden & Weinstein. One of the examples shows a plane being drawn through three points. While I understand their solutiom, I'm very curious as to why my solutiom doesn't work.

1. The problem statement, all variables and given/known data

Write the equatiom for a plane through A = (1, 1, 1), B = (2, 0, 0) and C = (1, 1, 0).

2. Relevant equations
For a plane through Point P = (Px, Py, Pz) orthogonal to n = <A, B, C> we write the equation:

A(X - Px) + B(Y - Py) + C(Z - Pz)

3. The attempt at a solution
The book solves the problem by filling the points into the more general formula:

Ax + By + Cz + D = 0

Which leaves us with three equations and four unknowns, which is adequate to come up with a solution (there is an infinite amount of solutions).

I tried to solve it by trying to find a normal vector n = <nx, ny, nz> that is orthogonal to the vectors AB, AC and BC.

n ⋅ AB = 0
n ⋅ AC = 0
n ⋅ BC = 0

nx - ny - nz = 0
-nz = 0
-nx + ny = 0

Solving this system leads to twice the expression nx = ny and once nz = 0. Thus I took <1, 1, 0> as a possible n.

Filling this in alongside A into the relevant equation leads to the equation

(x - 1) + (y - 1) = 0

Which simplifies to

x + y = 2

Obviously this is not a solution and the fact that the final answer incorporates a nomzero coeficient for z only proves that point. What did I do wrong?

- HF

2. Jul 27, 2016

### Stephen Tashi

What final answer are you referring to ? The final answer given in the book ?

3. Jul 28, 2016

### ehild

Why do you think your solution is wrong? Do coordinates of A, B, C fulfill the equation x+y=2?

4. Jul 28, 2016

### Stephen Tashi

Have you've visualized that the answer to the problem is a plane that is perpendicular to the xy-plane, so it "sticks straight up in the z-direction"? (The answer must be a plane containing the vertical line segment (1,1,0) to (1,1,1). ) The lack of the "z" variable in "x + y = 2" is what permits the z-coordinate to take arbitrary values.