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Math Help - Cross product problem

  1. #1
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    Cross product problem

    Hello,
    I'm facing problems while trying to solve this problem

    we have u and v and we need to find b that satisfies the following equation:

    u x b = v

    I've tried to solve it and i got only one coordinates for B and it we should get infinity number of vectors satisfies this equation up there
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  2. #2
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    Re: Cross product problem

    Quote Originally Posted by ahmedzoro10 View Post
    Hello,
    I'm facing problems while trying to solve this problem

    we have u and v and we need to find b that satisfies the following equation:

    u x b = v

    I've tried to solve it and i got only one coordinates for B and it we should get infinity number of vectors satisfies this equation up there
    it would be helpful to know what u and v are ...
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  3. #3
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    Re: Cross product problem

    Quote Originally Posted by skeeter View Post
    it would be helpful to know what u and v are ...
    u= [-1 3 2] and v= [1 1 -1]
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  4. #4
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    Re: Cross product problem

    if b = (b_1,b_2,b_3) then u \times b = v means that (3b_3 - 2b_2,2b_1+b_3,-b_2-3b_1) = (1,1,-1).

    this is a system of 3 linear equations in 3 unknowns:

    \begin{bmatrix}0&-2&3\\2&0&1\\-3&-1&0\end{bmatrix} \begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix} = \begin{bmatrix}1\\1\\-1\end{bmatrix}

    can you continue?
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  5. #5
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    Re: Cross product problem

    Quote Originally Posted by Deveno View Post
    if b = (b_1,b_2,b_3) then u \times b = v means that (3b_3 - 2b_2,2b_1+b_3,-b_2-3b_1) = (1,1,-1).

    this is a system of 3 linear equations in 3 unknowns:

    {\color{red}\begin{bmatrix}0&-2&3\\2&0&1\\-3&-1&0\end{bmatrix}} \begin{bmatrix}b_1\\b_2\\b_3\end{bmatrix} = \begin{bmatrix}1\\1\\-1\end{bmatrix}
    Be careful:
    That matrix is singular.

    You might try rref.
    Last edited by Plato; December 15th 2011 at 08:44 AM.
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  6. #6
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    Re: Cross product problem

    Quote Originally Posted by Plato View Post
    Be careful:
    That matrix is singular.

    You might try rref.
    I got it . Thanks alot
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  7. #7
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    Re: Cross product problem

    I got A^(-1) = 0! is that right!
    I've used adjoint method getting the inverse of A
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  8. #8
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    Re: Cross product problem

    Quote Originally Posted by ahmedzoro10 View Post
    I got A^(-1) = 0! is that right!
    I've used adjoint method getting the inverse of A
    No, that is not right. Plato told you that A does not have an inverse.

    (And no invertible matrix has 0 as its inverse.)
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  9. #9
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    Re: Cross product problem

    Quote Originally Posted by HallsofIvy View Post
    No, that is not right. Plato told you that A does not have an inverse.

    (And no invertible matrix has 0 as its inverse.)
    oh sorry I haven't noticed his post saying "this matrices is singular"
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  10. #10
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    Re: Cross product problem

    b will equal [ t , 1-3t , 1-2t ] assuming that b1 = t ,right?
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  11. #11
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    Re: Cross product problem

    one question,please
    Why we don't just put
    3b3-2b2=1
    2b1+b3=1
    -3b1-b2=-1
    and solve for this equations
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  12. #12
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    Re: Cross product problem

    You certainly can do that. In particular, If you multiply the last equation by 2 and subtract it from the first equation, you get 6b_1+ 3b_3= 3. From the second equation, b_3= 1- 2b_1 so 6b_1+ 3b_3= 6b_1+3- 6b_1= 3 which is true for all x. Taking b_1= t gives b2= 1- 3t and b_3= 1- 2t, exactly what you give above.
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