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Math Help - Cross multiplying vectors

  1. #1
    Newbie Riyzar's Avatar
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    Cross multiplying vectors

    The question asks me to find the resulting vector without using determinants. Instead I am supposed to use properties of cross products.

    (j-k) x (k - i)

    I really can't even get started. A hint or general idea would be very helpful though.
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  2. #2
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    Quote Originally Posted by Riyzar View Post
    The question asks me to find the resulting vector without using determinants. Instead I am supposed to use properties of cross products.

    (j-k) x (k - i)

    I really can't even get started. A hint or general idea would be very helpful though.
    You should know that \vec{i}\times \vec{j}= \vec{k}, \vec{j}\times\vec{k}= \vec{i}, and \vec{k}\times\vec{i}= \vec{j}. And, of course, that \vec{u}\times\vec{v}= -\vec{v}\times\vec{u} which, among other things, implies \vec{u}\times\vec{u}= 0. Also that the cross product distributes over addition: (\vec{j}- \vec{k})\times (\vec{k}- \vec{i})= (\vec{j}- \vec{k})\vec{k}- (\vec{j}- \vec{k})\vec{i}= \vec{j}\times\vec{j}+ \vec{k}\vec{j}- \vec{k}\times\vec{i}+ \vec{k}\times\vec{i}
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  3. #3
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    Hello, Riyzar!

    \text{Find the resulting vector }without\text{ using determinants: }\;(\vec j-\vec k)  \times(\vec k - \vec i)

    Are those unit vectors? . \begin{Bmatrix}\vec i &=& \langle 1,0,0\rangle \\ \vec j &=& \langle 0,1,0\rangle \\ \vec k &=& \langle 0,0,1\rangle \end{Bmatrix}


    \left(\vec j - \vec k\right) \times \left(\vec k - \vec i\right) \;=\;\left(\vec j - \vec k\right )\times \vec k - \left(\vec j - \vec k\right)\times \vec i

    . . . . . . . . . . . . . . =\; \left(\vec j \times \vec k\right) - \left(\vec k \times \vec k \right) - \left(\vec j \times \vec i\right) + \left(\vec k \times \vec i\right)

    . . . . . . . . . . . . . . =\qquad \vec i \quad\;\; -\;\;\quad  \vec 0 \quad - \quad (-\vec k) \quad + \quad \vec j

    . . . . . . . . . . . . . . =\quad \vec i\;+\; \vec j\;+\;\vec k

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  4. #4
    Newbie Riyzar's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    Also that the cross product distributes over addition: (\vec{j}- \vec{k})\times (\vec{k}- \vec{i})= (\vec{j}- \vec{k})\vec{k}- (\vec{j}- \vec{k})\vec{i}= \vec{j}\times\vec{j}+ \vec{k}\vec{j}- \vec{k}\times\vec{i}+ \vec{k}\times\vec{i}
    That is what I was not aware of, thank you!
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