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Math Help - problem with vectors

  1. #1
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    problem with vectors

    I need some help with the the following problem:

    If vectors a and b are orthogonal, |a|=3, |b|=4.
    Calculate:
    a) p=|(a+b)x(a-b)|
    b) q=|(3a-b)x(a-2b)|
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  2. #2
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    Quote Originally Posted by sillyme View Post
    I need some help with the the following problem:
    If vectors a and b are orthogonal, |a|=3, |b|=4.
    Calculate:
    a) p=|(a+b)x(a-b)|
    b) q=|(3a-b)x(a-2b)|
    Is that a cross product u\times v?
    Or is it a dot product: u \cdot v?
    I suspect it is a dot product.
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  3. #3
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    Quote Originally Posted by Plato View Post
    Is that a cross product u\times v?
    Or is it a dot product: u \cdot v?
    I suspect it is a dot product.
    It is a cross product.
    p is the norm of the cross product between the vector sum (a+b) and the vector difference (a-b)

    p = |(a+b)\times (a-b)|
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  4. #4
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    Some useful properties:
    • (u + v) \times w = (u \times w) + (v \times w)
    • w \times (u + v) = (w \times u) + (w \times v)
    • u \times v = -(v \times u)
    • |u \times v | = |u||v|\sin \theta


    \begin{aligned} (a+b) \times (a-b) & = ( a \times (a-b)) + (b \times (a-b)) \\ & = (a \times a - (a \times b)) \ + \ ( b \times a - b \times b) \\ & = (0 - a \times b) + (-(a \times b) - 0) \\ & = -2 (a \times b) \end{aligned}

    So: p = | -2 (a \times b) | = 2 |a \times b | = 2 |a||b| \sin \theta

    But since they're orthogonal ....
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  5. #5
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    Thank you o_O.
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  6. #6
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    A very simple way to do this is to set up a coordinate system so the \vec{a}= 3\vec{i} and \vec{b}= 4\vec{j}.
    Then \vec{a}+ \vec{b}= 3\vec{i}+ 4\vec{j}, \vec{a}- \vec{b}= 3\vec{i}- 4\vec{j} , 3\vec{a}- \vec{b}= 9\vec{i}- 4\vec{j}, and \vec{a}- 2\vec{b}= 3\vec{i}-8\vec{j}.
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