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Math Help - Proof (vectors)

  1. #1
    Mtl
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    Proof (vectors)

    OK, so I have a proof which I have am having a hard time prooving. Well here it is |a cross b| = ((a dot a) (b dot b) - (a dot b)sqrd) to the power of 1/2


    (Also, it seems i have posted in the wrong section, if moving the thread to the Euclidean Geometry section is at all possible please let me know
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  2. #2
    Eater of Worlds
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    This looks very similar to Lagrange's Identity. Which states:

    \Vert{a\times{b}}\Vert^{2}=\Vert{a}\Vert^{2}\Vert{  b}\Vert^{2}-(a\cdot{b})^{2}

    For 3-space, you could use \langle{a_{1},a_{2},a_{3}}\rangle

    Expand out and confirm the equality.

    \Vert{a\times{b}}\Vert may be written as:

    (a_{2}b_{3}-a_{3}b_{2})+(a_{3}b_{1}-a_{1}b_{3})+(a_{1}b_{2}-a_{2}b_{1})
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  3. #3
    Mtl
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    " <br />
(a_{2}b_{3}-a_{3}b_{2})+(a_{3}b_{1}-a_{1}b_{3})+(a_{1}b_{2}-a_{2}b_{1})<br />
"

    Should not each term be squared and the term as a whole be "squarerooted"
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  4. #4
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    There are many ways to do this, no one is correct (except the one your instructor wants). Here is one way using basic definitions and equivalents.
    \left\| a \right\|\left\| b \right\|\sin \left( \theta  \right) = \left\| {a \times b} \right\|\,\& \,\left\| a \right\|\left\| b \right\|\cos \left( \theta  \right) = a \cdot b.

    Therefore,
    \begin{array}{rcl}<br />
 \left\| {a \times b} \right\|^2 & = & \left[ {\left\| a \right\|\left\| b \right\|\sin \left( \theta  \right)} \right]^2  \\ <br />
  & = & \left\| a \right\|^2 \left\| b \right\|^2 \sin ^2 \left( \theta  \right) \\ <br />
  & = & \left\| a \right\|^2 \left\| b \right\|^2  - \left( {a \cdot b} \right)^2  \\ <br />
 \end{array}
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