You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**NoSash****Member**- Registered: 2006-02-06
- Posts: 7

See the image, show using the diagram...

I have no idea how to apply the diagram.

Offline

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 21,063

This is called the scalar triple product. It is given by [a, b, c] where a, b, and c are the three vectors. You can understand this better if you know determinants. This can be of some help. And this one too.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

There are two ways to explain this. Either geometrically or algebraically.

To do it algebraically, just do:

a = <a1, a2, a3>

b = <b1, b2, b3>

c = <c1, c2, c3>

And then compute each dot/cross product.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Offline

**NoSash****Member**- Registered: 2006-02-06
- Posts: 7

Yes I know how to prove it algebraically... just not with the diagram.

I'm going to check out your links, thanks ganesh.

*Last edited by NoSash (2006-03-18 05:00:19)*

Offline

**NoSash****Member**- Registered: 2006-02-06
- Posts: 7

eh... if anyone is interested in the solution, here it is:

It's really simple actually... Because it is a CIRCLE, 2-dimensional shape, the scalar triple products are all equal to zero...

Three vectors in a 2D plane are always linearly dependent.

Offline

**fgarb****Member**- Registered: 2006-03-03
- Posts: 89

That's a very strange diagram. The equalities you wrote down are definitely true for the case where all three vectors are in the same plane as you said, but just to make sure you understand, these forumulas are also true for any three vectors.

Offline

Pages: **1**