# Coordinates, vectors and cross products

• Oct 15th 2010, 08:30 AM
yellowFattyBEan
Coordinates, vectors and cross products
Hi

I have 4 3D points and I have been told I need to create 2 vecors out of them. I then need to cross product these vecotrs to then get another vector.

Now with my terrible grasp of maths, I'm ahving a hard time working out this stuff.

Can anybody help me work out a solution for solving this problem?
• Oct 15th 2010, 08:56 AM
earboth
Quote:

Originally Posted by yellowFattyBEan
Hi

I have 4 3D points and I have been told I need to create 2 vecors out of them. I then need to cross product these vecotrs to then get another vector.

Now with my terrible grasp of maths, I'm ahving a hard time working out this stuff.

Can anybody help me work out a solution for solving this problem?

Which problem?

• Oct 15th 2010, 09:18 AM
yellowFattyBEan
ok

I'm working with a 3d graphics package so it's kinda hard to relate everything.

I have a 2d plane, and I've tried to mark the corners of this. (sorry for the crudeness)

(-6,0,-6) (6,0,6)
_________
/ /
/ /
/ /
/________/

(-6,0,6) (6,0,-6)

Basically, I need to take these points, convert them to vectors and find the cross product so I can then go on to prove that this is a 2d plane ranther than a 3d object.
• Oct 15th 2010, 11:20 AM
earboth
Quote:

Originally Posted by yellowFattyBEan
ok

I'm working with a 3d graphics package so it's kinda hard to relate everything.

I have a 2d plane, and I've tried to mark the corners of this. (sorry for the crudeness)

(-6,0,-6) (6,0,6)
_________
/ /
/ /
/ /
/________/

(-6,0,6) (6,0,-6)

Basically, I need to take these points, convert them to vectors and find the cross product so I can then go on to prove that this is a 2d plane ranther than a 3d object.

If I understand your question correctly you want to show that these 4 points are placed in a plane (?).

If so:

1. Three points which don't lie on a straight line allways lie in a plane. Take 3 points and determine the equation of this plane:
A(-6, 0 ,-6), B(6, 0, 6), C(-6, 0, 6)

$p: \vec r = (-6, 0, -6) + s \cdot ((6,0,6)-(-6,0,-6))+t \cdot ((-6,0,6)-(-6,0,-6))$

$p:\vec r = (-6,0-6)+s(12,0,12)+t(0,0,12)$

2. Now check if the point D(6,0,-6) belongs to the plane p:

$(6,0,-6)=(-6,0-6)+s(12,0,12)+t(0,0,12)$

This equation has the only solution s = 1 and t = -1. Therefore $D\in p$

3. Since the second coordinate of all points is zero, all points are situated in the x-z-plane.