# Thread: Points on a plane

1. ## Points on a plane

Using basic linear algebra (Gauss elimination, Gauss-Jordan elimination, et cetera), how can we solve the problem; with four points in tridimentional space;

- Show if the four points are on the same plane
- Find the equation of the plane

I can actually solve this problem, but I'm not sure how using only matrices and Gauss-Jordan elimination...

2. Originally Posted by AnnM
Using basic linear algebra (Gauss elimination, Gauss-Jordan elimination, et cetera), how can we solve the problem; with four points in tridimentional space;

- Show if the four points are on the same plane
- Find the equation of the plane

I can actually solve this problem, but I'm not sure how using only matrices and Gauss-Jordan elimination...
Let your four points be A, B, C, D. To prove they are coplanar, find equations for lines AB, AC and AD. They will be of the form:

$\displaystyle AB = a\vec{i} + b \vec{j} + c\vec{k}$

$\displaystyle AC = d\vec{i} + e \vec{j} + f\vec{k}$

$\displaystyle AD = g\vec{i} + h \vec{j} + i\vec{k}$

Which can be written in matrix form:
$\displaystyle \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \begin{pmatrix} \vec{i} \\ \vec{j} \\ \vec{k} \end{pmatrix}$ (Not sure if that's much use though.

Points are coplanar if the triple product is 0.

Triple product involves the cross product of two of the above:

$\displaystyle AB \times AC = det \begin{pmatrix} \vec{i} & \vec{j} & \vec{k} \\ d & e & f \\ g & h & i \end{pmatrix}$

3. Originally Posted by AnnM
Using basic linear algebra (Gauss elimination, Gauss-Jordan elimination, et cetera), how can we solve the problem; with four points in tridimentional space;

- Show if the four points are on the same plane
- Find the equation of the plane

I can actually solve this problem, but I'm not sure how using only matrices and Gauss-Jordan elimination...
Dear AnnM,

You say "Show if the four points are on the same plane", but then what?

Or do you mean "Find the equation of the plane given four points that lie on the same plane" ?

Write the four points (in column vector form) as column vectors to get a 3 x 4 matrix, call it P. Then the plane is the range space of P. In other words, the plane is the set $\displaystyle \{Px : \forall x \in \mathbb{R}^4\}$

4. Originally Posted by Isomorphism
Dear AnnM,

You say "Show if the four points are on the same plane", but then what?

Or do you mean "Find the equation of the plane given four points that lie on the same plane" ?

Write the four points (in column vector form) as column vectors to get a 3 x 4 matrix, call it P. Then the plane is the range space of P. In other words, the plane is the set $\displaystyle \{Px : \forall x \in \mathbb{R}^4\}$
You can't get a single x a member of a set of Euclidean n-space, with n>1. It must be an ordered n-tuple. For n = 4, it's (w,x,y,z) where w, x, y and z are members of R. Anyway, it's 3-dimensional space.

5. But that's the problem, as I said, I know how to find the solution, I just don't know how to do it with the tools I have;

- Range space of P = That stuff is covered much later in the course
- We haven't even seen determinant (OK, I know how to do it, but the point is; I should be able to solve this question with very little; Gauss-Jordan elimination and... not much else...).

6. Originally Posted by Mush
Let your four points be A, B, C, D. To prove they are coplanar, find equations for lines AB, AC and AD. They will be of the form:

$\displaystyle AB = a\vec{i} + b \vec{j} + c\vec{k}$

$\displaystyle AC = d\vec{i} + e \vec{j} + f\vec{k}$

$\displaystyle AD = g\vec{i} + h \vec{j} + i\vec{k}$
Which can be written in matrix form:
$\displaystyle \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \begin{pmatrix} \vec{i} \\ \vec{j} \\ \vec{k} \end{pmatrix}$ (Not sure if that's much use though.

Points are coplanar if the triple product is 0.

Triple product involves the cross product of two of the above:

$\displaystyle AB \times AC = det \begin{pmatrix} \vec{i} & \vec{j} & \vec{k} \\ d & e & f \\ g & h & i \end{pmatrix}$
If $\displaystyle \vec{u}= (a_1, b_1, c_1)$, $\displaystyle \vec{v}= (a_2, b_2, c_2)$, and $\displaystyle \vec{w}= (a_3, b_3, c_3)$, then the triple product, $\displaystyle \vec{u}\cdot (\vec{v}\times\vec{w})$ is just the determinant
$\displaystyle \left|\begin{array}{ccc} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right|$

7. we suppose four points: P1,P2,P3,P4
let matrix A = [P1-P2 P2-P3 P3-P4 P4-P1], belong to R(3x4) space
if two vectors are not parallel, then they are linear independent.
So we only find how many vectors are linear independent.
we do rref(reduced row-echelon form) operation.
get rank(A):
if = 0, all points are equal.
if = 1, all points line on one line.
if = 2, construct one plane.
if = 3, construct 3D space, not in a plane.
if > 3, ha!, your logic is mistake.