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Homework Help: Cross Product Vector Proof

  1. Oct 2, 2011 #1
    Not sure if this is the correct section. I apologize if it's not.

    1. The problem statement, all variables and given/known data
    For any vectors [itex]\vec{a}, \vec{b}, \vec{c}[/itex] show that:
    [itex](\vec{a} \times \vec{b} ) \times \vec{c}[/itex]
    lies in the plane of [itex]\vec{a}[/itex] and [itex]\vec{b}[/itex]

    2. Relevant equations

    3. The attempt at a solution

    I assigned [itex]\vec{a} = (e,f,g), \vec{b} = (x,y,z), \vec{c} = (s,t,u)[/itex]

    then I used the cross product formula and got:

    (u(gx-ez)-t(ey-fx), s(ey-fx)-u(fz-gy), t(fz-gy)-s(gx-ez))

    which expanded comes to:

    (gux-ezu-tey+tfx, sey-sfx-fuz+yug, zft-tgy-sgx+sez)

    I'm not sure if that helps...
  2. jcsd
  3. Oct 2, 2011 #2


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    Hint: If a vector lies in the plane of A and B, you can express it as a linear combination of the two vectors. So see if you can write your result in the form

    (something)x(e, f, g) + (something)x(x, y, z)
  4. Oct 2, 2011 #3


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    Try using what you should know about the geometry of the cross product. axb is perpendicular to a and b, right? So it's a normal to the plane of a and b. Anything that is perpendicular to that normal is in the plane of a and b. Is (axb)xc perpendicular to axb?
  5. Oct 2, 2011 #4
    Yes, that makes sense. I get it :D But is it possible to prove it algebraically?

    I don't understand :S Could you explain it? Do you mean:
    (something)*(e, f, g) + (something)*(x, y, z)

  6. Oct 2, 2011 #5


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    Sure, that's what vela meant. And you can find those two somethings with a little work. Write (axb)xc=i*a+j*b and try to find i and j.
  7. Oct 2, 2011 #6
    Ah... I see...
    So it would be like:

    (ae+bx, af+by, ag+bz) where a and b are the somethings... right?

    The goal is to get something like this:
    (gux-ezu-tey+tfx, sey-sfx-fuz+yug, zft-tgy-sgx+sez)


    the constants s, t, u could be formed by multiplying the a and b "something" right?

    Am I on the right track?


    Is that the long way of doing it? Is there an easier way? I feel that there is an easier way of solving it algebraically...
  8. Oct 3, 2011 #7


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    You already used 'a' and 'b' for the vectors, so I like (axb)xc=i*a+j*b better. But, yes, that's the general idea. That's three linear equations in the two unknown constants i and j. You can solve it systematically. There's a lot of letters flying around but everything is a constant except for i and j. I like the geometrical approach better, but you can pull it off algebraically.
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