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Thread: Cross product problem

  1. #1
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    Unhappy Cross product problem

    Its actually not so much the cross product itself that's giving me trouble but sorting out the algebra that follows. The problem is as follows:

    There are two vectors U and V such that UxV = -30i + 40j. V= 4i -2j +3k, find U.

    Using U = Uxi + Uyj + Uzk, I get: UxV= (3Uy + 2Uz)i - (3Ux - 4Uz)j + (-2Ux - 4Uy)k,

    and, therefore:

    3Uy + 2Uz = -30 (eqn 1)

    -3Ux + 4Uz = 0 (eqn 2)

    -2Ux - 4Uy = 40 (eqn 3)

    So far so good, I hope. No is where I come into trouble. Instead of solving the three eqns for the three unknowns, I end up with 0=0. Which, though I'm glad is true, is of no use to me. My math is as follows:

    From eqn 2, Ux = 4/3 Uz
    From previous and eqn 3, -8/3 Uz -4Uy = 40 => Uz = -3/2 Uy - 15
    From previous and eqn 1, 3Uy -3Uy -30 = -30 => 0=0

    If you see my error, please let me know.

    Thanks much.

    j

  2. #2
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    Quote Originally Posted by mathj View Post
    There are two vectors U and V such that UxV = -30i + 40j. V= 4i -2j +3k, find U.

    Using U = Uxi + Uyj + Uzk, I get: UxV= (3Uy + 2Uz)i - (3Ux - 4Uz)j + (-2Ux - 4Uy)k,
    and, therefore:

    3Uy + 2Uz = -30 (eqn 1)

    -3Ux + 4Uz = 0 (eqn 2)

    -2Ux - 4Uy = 40 (eqn 3)

    You have those constants backward.
    [TEX]-3U_x+4U_z=40~\&~-2U_x-4U_y=0[/TEX]
    “A professor is someone who talks in someone else’s sleep”
    W.H. Auden

  3. #3
    Quote Originally Posted by mathj View Post
    Its actually not so much the cross product itself that's giving me trouble but sorting out the algebra that follows. The problem is as follows:

    There are two vectors U and V such that UxV = -30i + 40j. V= 4i -2j +3k, find U.

    Using U = Uxi + Uyj + Uzk, I get: UxV= (3Uy + 2Uz)i - (3Ux - 4Uz)j + (-2Ux - 4Uy)k,

    and, therefore:

    3Uy + 2Uz = -30 (eqn 1)

    -3Ux + 4Uz = 0 (eqn 2)

    -2Ux - 4Uy = 40 (eqn 3)

    So far so good, I hope. No is where I come into trouble. Instead of solving the three eqns for the three unknowns, I end up with 0=0. Which, though I'm glad is true, is of no use to me. My math is as follows:

    From eqn 2, Ux = 4/3 Uz
    From previous and eqn 3, -8/3 Uz -4Uy = 40 => Uz = -3/2 Uy - 15
    From previous and eqn 1, 3Uy -3Uy -30 = -30 => 0=0

    If you see my error, please let me know.

    Thanks much.

    j
    Any vector that is :

    < x, x/2 , -15-0.75 x > certifies that

    Plug in any x

  4. #4
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    Quote Originally Posted by mathj View Post
    Its actually not so much the cross product itself that's giving me trouble but sorting out the algebra that follows. The problem is as follows:

    There are two vectors U and V such that UxV = -30i + 40j. V= 4i -2j +3k, find U.

    Using U = Uxi + Uyj + Uzk, I get: UxV= (3Uy + 2Uz)i - (3Ux - 4Uz)j + (-2Ux - 4Uy)k,

    j
    U X V = (Uxi + Uyj + Uzk) X (4i - 2j + 3k) = (-2Uxk -3Uxj) + (-4Uyk + 3Uyi) + (4Uzj + 2Uzi) = (3Uy + 2Uz)i + (4Uz -3Ux)j + (-2Ux - 4Uy)k →

    3Uy + 2Uz = - 30

    4Uz - 3Ux = 40

    2Ux + 4Uy = 0

    There is no solution to this system. If you try to solve this system - say by Gauss elimination - you will get an answer like 0 = 100. This indicates NO SOLUTION. When you got 0=0, that indicated infinite number of solutions.
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  5. #5
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    Thanks everyone for your input.

    Good catch pka. Unfortunately, my error was in writing the problem, UxV = -30i + 40k.

    Additional input: I know the (or at least an) answer, its more finding where my problem with solving it lies that I'm looking for. I've plugged in various components for the vectors (in order to make up new similar problems to solve) and I seem to always reduce the system of equations to a trivial solution. I assume my issue is somewhere within the solving the system of equations or setting it up and not with the cross product part.

    Again, thanks for any help or advice you can offer.


    pka Originally Posted by mathj
    There are two vectors U and V such that UxV = -30i + 40j. V= 4i -2j +3k, find U.

    Using U = Uxi + Uyj + Uzk, I get: UxV= (3Uy + 2Uz)i - (3Ux - 4Uz)j + (-2Ux - 4Uy)k,
    and, therefore:

    3Uy + 2Uz = -30 (eqn 1)

    -3Ux + 4Uz = 0 (eqn 2)

    -2Ux - 4Uy = 40 (eqn 3)






    You have those constants backward.

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