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Math Help - another cross product

  1. #1
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    another cross product

    (tsint i + tcost j) x (( sint + tcost )i + (cost -tsint)j) =-t^2(sin^2 t+ cos^2t)k

    Could you please teach me how to get this answer on the RHS? Thank you very much.
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    Quote Originally Posted by Jenny20 View Post
    (tsint i + tcost j) x (( sint + tcost )i + (cost -tsint)j) =-t^2(sin^2 t+ cos^2t)k

    Could you please teach me how to get this answer on the RHS? Thank you very much.
    <tsin(t), t cos(t), 0> \times <sin(t) + tcos(t), cos(t) - tsin(t), 0>

     = \left | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ tsin(t) & tcos(t) & 0 \\ sin(t) + tcos(t) & cos(t) - tsin(t) & 0 \end{array} \right |

    Expanding the determinant across the top row:
     = [0 \cdot tcos(t) - 0 \cdot (cos(t) - tsin(t)] \hat{i} - [0 \cdot tsint(t) - 0 \cdot (sin(t) + tcos(t)] \hat{j}  + [tsin(t) \cdot (cos(t) - tsin(t)) - tcos(t) \cdot (sin(t) + tcos(t)] \hat{k}

    The \hat{i} and \hat{j} coefficients are obviously 0, so we get:
     = [tsin(t) \cdot (cos(t) - tsin(t)) - tcos(t) \cdot (sin(t) + tcos(t)] \hat{k}

     = [tsin(t)cos(t) - t^2sin^2(t) - tsin(t)cos(t) - t^2cos^2(t)] \hat{k}

     = -t^2 \hat{k}

    -Dan
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