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Math Help - Vectors

  1. #1
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    Vectors

    There are two vectors (d and c). d dot c will equal 50. The angle between them is 60 degrees (when the two vectors are tail to tail). What is ID X CI ?
    Last edited by Clairvoyantski; June 11th 2012 at 04:37 PM.
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  2. #2
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    Re: Vectors

    In order to do this equation, you need to know that d dot c= |d||c|cos(theta) where theta is the angle between them. I started to write "to find ID X CI" you would have to know what 'ID' and 'CI' mean" when it hit me that you mean |D X C|, the length of DXC. Of course, to do that you have to know that that length is given by |D||C| sin(theta).
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  3. #3
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    Re: Vectors

    Hello, Clairvoyantski!

    I have a rather roundabout solution . . .


    \text{There are two vectors, }\vec c\,\text{ and }\,\vec d. \;\;\;\vec d \cdot \vec c \,=\,50.
    \text{The angle between them is }60^o\text{ (when the two vectors are tail to tail).}

    \text{What is }|\vec d \times \vec c|\,?

    Code:
              C * - - - - - * B
               / \         /
              /   \       /
           c /     \     /
            /       \   /
           / 60d     \ /
        A * - - - - - * D
                d
    We have parallelogram AC\!BD,
    . . where AD = \vec d,\;\;AC = \vec c.

    We know that |\vec d \times \vec c| is the area of the parallelogram.


    Formula: . \cos60^o \:=\:\frac{\vec d \cdot \vec c}{|\vec d||\vec c|}

    So we have: . \frac{1}{2} \:=\:\frac{50}{|\vec d||\vec c|} \quad\Rightarrow\quad |\vec d||\vec c| \:=\:100

    The area of \Delta AC\!D is given by: . \tfrac{1}{2}|\vec d||\vec c|\sin60^o

    The area of parallelogram AC\!BD is: . |\vec d||\vec c|\sin60^o \;=\;(100)\left(\tfrac{\sqrt{3}}{2}\right) \;=\;50\sqrt{3}

    Therefore: . |\vec d \times \vec c| \;=\;50\sqrt{3}
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