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Math Help - Vectors again..

  1. #1
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    Vectors again..

    Can u check my answers:
    For the two vectors in this figure,
    A = 12m W
    B = 18m 37 ^\circ N of E
    a.) Find the magnitude and direction of the vector product \vec{A} X \vec{B}
    solutions:
    \vec{A} X \vec{B} = AB\sin{\phi} = (-12.0m)(18.0m)\sin{143}
    \vec{A} X \vec{B} = -129.99
    but im not sure but it also asks for a direction

    b.) Find the magnitude and direction \vec{B} X \vec{A}

    same as part a and the rule says \vec{B} X \vec{A} = - \vec{A} X \vec{B}
     129.99 ?


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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Can u check my answers:
    For the two vectors in this figure,
    A = 12m W
    B = 18m 37 ^\circ N of E
    a.) Find the magnitude and direction of the vector product \vec{A} X \vec{B}
    solutions:
    \vec{A} X \vec{B} = AB\sin{\phi} = (-12.0m)(18.0m)\sin{143}
    \vec{A} X \vec{B} = -129.99
    but im not sure but it also asks for a direction

    b.) Find the magnitude and direction \vec{B} X \vec{A}

    same as part a and the rule says \vec{B} X \vec{A} = - \vec{A} X \vec{B}
     129.99 ?


    For a) in the formula \vec{A} \times \vec{B} we have that
    |\vec{A} \times \vec{B}| = | \vec{A} || \vec{B} | sin( \theta_{AB})

    So your working is incorrect. The negative sign in front of A (which really depends on your coordinate system anyway) is not there.

    There are two ways to get the direction. One is the "right hand rule" (which I'll suggest you do a web search on) or simply work out the determinant:
    \vec{A} \times \vec{B} = \left | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{array} \right |

    Since your vectors are both in the xy plane (most people would choose +x in the E and +y in the N directions) your cross product will end up in the z direction. (I get the - z direction, to be precise.)

    -Dan
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  3. #3
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    my mistake is that a resultant vector has no negative...

    therefore is my answer reversed?

    so A is 129.99

    B is -129.99
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    my mistake is that a resultant vector has no negative...

    therefore is my answer reversed?

    so A is 129.99

    B is -129.99
    A vector is neither positive nor negative. It has a "size" (in whatever units it has) in a certain direction.

    In this case, choosing +x to be E and +y to be N:
    \vec{A} \times \vec{B} = \left | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 0 & -12 & 0 \\ 18cos(37) & 18 sin(37) & 0 \end{array} \right | = \hat{k}(0 - (-12)(18 cos(37))) = 172.505 \hat{k}

    Which is 172.505 ~ m^2 in the +z direction. (Perpendicular to both E and N.)

    And as you correctly noted:
    \vec{B} \times \vec{A} = - \vec{A} \times \vec{B} = -172.505 \hat{k}
    or 172.505 ~ m^2 in the -z direction.

    -Dan
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  5. #5
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    Talking

    thanks!!!
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