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Math Help - proof of two cross products.

  1. #1
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    proof 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.
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  2. #2
    o_O
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    Let a = \left< a_1, a_2, a_3\right> and b = \left< b_1, b_2, b_3 \right> .

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

    Applying this here:
    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 - (b \times a) defined? Is this what we have? Yep.

    _____________

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

    Start by letting: a = \left< a_1, a_2, a_3\right> and b = \left< b_1, b_2, b_3 \right> and c = \left<c_1, c_2, c_3\right>
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  3. #3
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    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).
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