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Math Help - Cross and Dot Products

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    Cross and Dot Products

    Find the cross product axb and verify it is orthogonal to both a and b

    a = <e^2t, 1, e^-t>
    b = <e^t, 2, e^-t>

    I calculated the cross product to be <-e^-t, e^t - 1, 2e^2t - e^t> and tried dotting it with a and b. Problem is, I don't get 0 in either case. Isn't that how you show it is orthogonal?
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    Quote Originally Posted by veronicak5678 View Post
    Find the cross product axb and verify it is orthogonal to both a and b

    a = <e^2t, 1, e^-t>
    b = <e^t, 2, e^-t>

    I calculated the cross product to be <-e^-t, e^t - 1 , 2e^2t - e^t> and tried dotting it with a and b. Problem is, I don't get 0 in either case. Isn't that how you show it is orthogonal?
    correction on your j component ...

    <-e^-t, 1 - e^t , 2e^2t - e^t>

    now find the dot products
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    I see what I did wrong. Thanks for your help!
    Last edited by veronicak5678; January 29th 2009 at 06:01 PM.
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    Quote Originally Posted by veronicak5678 View Post
    Isn't the j component negative? I originally had 1-e^t, then looked at my notes and saw it had to be negative. It didn't dot out to 0 in either case, anyway.
    j component ...

    -(e^{2t} \cdot e^{-t} - e^{-t} \cdot e^{t})<br />

    -(e^t - 1)

    1 - e^t

    dot product with a

    <e^{2t}, 1, e^{-t}> \bullet <-e^{-t}, 1 - e^t , 2e^{2t} - e^t>

    -e^t + 1 - e^t + 2e^t - 1 = 0


    dot product with b

    <e^t, 2, e^{-t}> \bullet <-e^{-t}, 1 - e^t , 2e^{2t} - e^t>

    -1 + 2 - 2e^t + 2e^t - 1 = 0
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