Proving Three 3D Points Are Collinear With Cross Product

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Proving three 3D points are collinear with Cross Product

• Thread starter Thread starter Masschaos
• Start date Start date May 5, 2011
• Tags Tags 3d Cross Cross product Points Product

Click For Summary To prove that three points in 3D space are collinear, the cross product of the vectors formed by these points must equal zero. Specifically, if points A, B, and C are given, the condition c = a + t(b - a) indicates collinearity for some scalar t. The discussion clarifies that the triple scalar product a · (b x c) must also equal zero for the points to be collinear. Additionally, it is noted that using ratios of direction vectors can lead to division by zero, complicating the proof. Understanding these vector relationships is essential for confirming collinearity in three-dimensional space. Masschaos Messages 30 Reaction score 0 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? Physics news on Phys.org

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Rasalhague Messages 1,383 Reaction score 2 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). Last edited: May 5, 2011 Some Pig Messages 38 Reaction score 0 (a-b)x(b-c)=0 or axb+bxc+cxa=0 Masschaos Messages 30 Reaction score 0 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 Messages 1,383 Reaction score 2 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 Science Advisor Messages 1,280 Reaction score 141

Masschaos said: 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 Science Advisor Homework Helper Messages 9,567 Reaction score 775

Masschaos said: 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 Messages 111 Reaction score 0

Masschaos said: 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 Science Advisor Homework Helper Messages 9,567 Reaction score 775

Masschaos said: 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?

Hammie said: 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 Messages 30 Reaction score 0 Ahh thanks. I understand much better now. paulfr Messages 193 Reaction score 3

Some Pig said: (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

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