just wanted to see 2 proofs if anyone knows them or where I can point my browser to see them please,

just proofs for these cross products where a and b are vectors and c is scalar.

just a x b = -b x a

and

a x (b + c) = a x b + a x c

thankyou.

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- Feb 23rd 2009, 09:24 PMrcmangoproof of two cross products.
just wanted to see 2 proofs if anyone knows them or where I can point my browser to see them please,

just proofs for these cross products where a and b are vectors and c is scalar.

just a x b = -b x a

and

a x (b + c) = a x b + a x c

thankyou. - Feb 23rd 2009, 10:11 PMo_O
Let $\displaystyle a = \left< a_1, a_2, a_3\right>$ and $\displaystyle b = \left< b_1, b_2, b_3 \right>$ .

Recall that if you have a matrix $\displaystyle A$ and you interchange two of its rows to form a matrix $\displaystyle B$, then $\displaystyle \det(B) = -\det(A)$.

Applying this here:

$\displaystyle a \times b = \left| \begin{matrix} i & j & k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{matrix} \right| = -\left| \begin{matrix} i & j & k \\ b_1 & b_2 & b_3 \\ a_1 & a_2 & a_3 \end{matrix} \right|$

But this is what we were looking for! How is $\displaystyle - (b \times a)$ defined? Is this what we have? Yep.

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For the second one, there is always the tedious route of calculating both sides and showing that they are indeed equal.

Start by letting: $\displaystyle a = \left< a_1, a_2, a_3\right>$ and $\displaystyle b = \left< b_1, b_2, b_3 \right>$ and $\displaystyle c = \left<c_1, c_2, c_3\right>$ - Feb 23rd 2009, 10:26 PMrcmango
i found something on wiki, this was more of what I was looking for, but is what you have defined above an easier way to do this than what i have found here?

of course this is just part of the proof, there needs to be factoring further.

Its just alot of work it seems, to prove it correctly.

a × b = (a1i + a2j + a3k) × (b1i + b2j + b3k)

a × b = a1i × (b1i + b2j + b3k) + a2j × (b1i + b2j + b3k) + a3k × (b1i + b2j + b3k)

a × b = (a1i × b1i) + (a1i × b2j) + (a1i × b3k) + (a2j × b1i) + (a2j × b2j) + (a2j × b3k) + (a3k × b1i) + (a3k × b2j) + (a3k × b3k).