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Thread: Finding a Vector

  1. Sep 28th 2008, 01:15 PM #1
    aznsushiguy
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    Finding a Vector

    Find a vector of length 3 which has:
    a) the same direction as 5i + 2j
    b) the opposite direction to 4i + 5j

    a) |5-2j| = (5^2 + (-2)^2)^.5 = (29)^.5
    then
    3[(1/(29)^.5)(5i + 2j)] = 15/(29)^.5i - 6/(29)^.5j
    the ^.5 is square root...but I don't know how to show that on here. is my math correct?

    and is b) done the same way except I would multiply the whole thing by -1?

    Thanks
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  2. Sep 28th 2008, 01:43 PM #2
    Plato
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    \left\| {5i - 2j} \right\| = \sqrt {29} to need to learn this.
    So the vector you want is \frac{{15}}{{\sqrt {29} }}i - \frac{6}{{\sqrt {29} }}j
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  3. Sep 28th 2008, 02:58 PM #3
    aznsushiguy
    aznsushiguy is offline
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    Quote Originally Posted by Plato View Post
    \left\| {5i - 2j} \right\| = \sqrt {29} to need to learn this.
    So the vector you want is \frac{{15}}{{\sqrt {29} }}i - \frac{6}{{\sqrt {29} }}j
    ah, okay, so my math was correct.
    I'll go ahead and assume the same format is used for b), but it's negative.

    Thanks
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