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  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -

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Topic review (newest first)

fgarb
2006-03-23 14:50:17

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.

NoSash
2006-03-23 10:19:47

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.

NoSash
2006-03-19 03:59:54

Yes I know how to prove it algebraically... just not with the diagram.
I'm going to check out your links, thanks ganesh.

Ricky
2006-03-18 18:07:47

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.

ganesh
2006-03-18 17:39:27

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.

NoSash
2006-03-18 16:09:25

See the image, show using the diagram...
I have no idea how to apply the diagram.

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