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**renjer****Member**- Registered: 2006-04-29
- Posts: 51

Prove that if norm(u X v)=0 then u.v=a norm(v)^2 where a is any constant.

I know that both u and v are parallel vectors and that norm(u X v) sin x = 0

Then I got stuck.

Anyone knows any good resources for me to learn linear algebra (preferably from the basics, with exercises and answers)? I'm finding it v hard to understand this topic.

Thanks.

*Last edited by renjer (2006-05-29 03:40:21)*

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

Do you mean:

If the normal of u cross v is 0, then u dot v equals a times the normal of v squared.

Is that right?

"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..."

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**renjer****Member**- Registered: 2006-04-29
- Posts: 51

Let me rephrase that, I made some mistakes in my original post.

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

Since the vectors are parrallel, cos θ = 1. So:

u.v = |u|*|v|

And thus, a = |u| / |v| so:

"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..."

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

The key is to prove the two definations of dot product are equivalent.

Def1: dot product is the sum of corresponding cartesian co-ordinates.

Def2: dot product is the product of two vector lenths, and the cosine of their angle.

Usually this is proved within a triangle consisting of vector OA, OB, and AB=OB-OA

In a catesian system, distance formula is true based on Pythagoras' Theorem.

So if you define A(x[sub]1[/sub],y[sub]1[/sub],z[sub]1[/sub]) B(x[sub]2[/sub] ,y[sub]2[/sub],z[sub]2[/sub]), you get

meanwhile, Cosine Theorem

then below must be valid

proven

*Last edited by George,Y (2006-05-30 19:54:41)*

**X'(y-Xβ)=0**

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

I assumed that you can just assume that. Good proof George.

"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..."

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