Proving three 3D points are collinear with Cross Product

[Masschaos]

Hi.
I've been studying for a test and have been scouring my text for methods of proving points in 3-dimensional space are collinear.
The best I can see is that the cross-product of the vectors must equal zero.
Can someone explain how to do this a little more clearly?

 

[Rasalhague]

If a, b and c are three points and there's a number t such that c = a + t(b-a), then a, b and c are collinear. Conversely, if they're collinear, then there's a number t such that c = a + t(b-a), and their triple product will equal zero: a.(b x c) = 0.

You haven't defined what you mean by the cross product of three vectors. But if you mean something like "a x (b x c)", then it's not true. For example, take

a = < 5, 5, 5 >

b = < 1, 3, 9 >

c = < -16, -8, 16 > = a + 4(b-a)

a x (b x c) = < -11, -3, 21 >

EDIT: I've just corrected a mistake I made. In the first version of this post, I wrote c = t(b-a), where I should have written c = a + t(b-a).

 

[Some Pig]

(a-b)x(b-c)=0
or axb+bxc+cxa=0

 

[Masschaos]

Ahh, I think I understand a little better.
What I understand is that if I have points A, B, and C. I can find the line AB and AC. If I do the following (AB)/(AC) and find that vectors i, j and k all have the same factor. Then is that proof that they are collinear.

Example
AB = (1,1,1)
AC = (3,3,3)
(AB)/(AC) = (1/3, 1/3, 1/3).

As AB and AC are all multiples of that base factor, they are collinear?
Or have I gone off on a tangent?

 

[Rasalhague]

Yes - although careful with that method: you might get a situation where you're trying to divide by zero, and so don't get any result that way, even though the points may be collinear. (Think of what happens when the points all lie in one of the coordinate planes.)

 

[olivermsun]

Example
AB = (1,1,1)
AC = (3,3,3)
(AB)/(AC) = (1/3, 1/3, 1/3).

As AB and AC are all multiples of that base factor, they are collinear?

Are you placing A is at the origin? (If you do then that's one way to make life a lot easier).

 

[LCKurtz]

Ahh, I think I understand a little better.
What I understand is that if I have points A, B, and C. I can find the line AB and AC. If I do the following (AB)/(AC) and find that vectors i, j and k all have the same factor. Then is that proof that they are collinear.

Example
AB = (1,1,1)
AC = (3,3,3)
(AB)/(AC) = (1/3, 1/3, 1/3).

As AB and AC are all multiples of that base factor, they are collinear?
Or have I gone off on a tangent?

I don't care for the AB/AC notation. All you need to note is AC is a constant times AB. If it wasn't, they wouldn't be collinear.

 

[Hammie]

Hi.
I've been studying for a test and have been scouring my text for methods of proving points in 3-dimensional space are collinear.
The best I can see is that the cross-product of the vectors must equal zero.
Can someone explain how to do this a little more clearly?

mainly because if the points are co-linear, you have a parellpiped with volume zero..

 

[LCKurtz]

Hi.
I've been studying for a test and have been scouring my text for methods of proving points in 3-dimensional space are collinear.
The best I can see is that the cross-product of the vectors must equal zero.
Can someone explain how to do this a little more clearly?

mainly because if the points are co-linear, you have a parellpiped with volume zero..

No. For that calculation you need three coplanar vectors, and the calculation would be the triple scalar (box) product A dot (B cross C). Not the same test as for parallel vectors.

 

[Masschaos]

Ahh thanks.
I understand much better now.

 

[paulfr]

(a-b)x(b-c)=0
or axb+bxc+cxa=0

To be a bit more precise

If a b and c are points in a plane
Then <a -b> and <b-c> are direction vectors in the plane

If the points are on the same line, then the angle between them is 0 degrees

Since <a -b> x <b-c> = || <a -b> || || <b-c> || sin x
If x = 0 then sin x = 0 so

<a -b> x <b-c> = 0