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Thread: 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.

    $\displaystyle (j-k)$ x $\displaystyle (k - i)$

    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.

    $\displaystyle (j-k)$ x $\displaystyle (k - i)$

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

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

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


    $\displaystyle \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 $

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

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

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

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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: $\displaystyle (\vec{j}- \vec{k})\times (\vec{k}- \vec{i})= (\vec{j}- \vec{k})\vec{k}- (\vec{j}- \vec{k})\vec{i}= \vec{j}\times\vec{j}+ \vec{k}\vec{j}- \vec{k}\times\vec{i}+ \vec{k}\times\vec{i}$
    That is what I was not aware of, thank you!
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