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QuantumJG's first year uni maths revision thread.

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QuantumJG:
How can you do this distributivity proof?

(αa + βb) x c =αa x c + βb x c

the only thing I can think of is setting,

a = (a1, a2, a3)

b = (b1, b2, b3)

c = (c1, c2, c3)

and showing that the RHS = LHS

is there a neater way? 

humph:
Depends what you've already proved/can assume beforehand, but in general no, that's the most direct (albeit lengthy) way.

/0:
Here is a proof for coplanar vectors, from Introduction to Electrodynamics - Griffiths. It also includes the source for the general case. (Which, unfortunately, I don't have)

kamil9876:
I like that geometry :P

I agree with humph, was just about to say that it depends what you have proven earlier/take as the definition of cross product. Like for instance if you know the linearity (in a single row) property of determinants and know how determinants relate to it, this becomes trivial.

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