# Cross multiplying vectors

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• Sep 3rd 2010, 04:05 PM
Riyzar
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.
• Sep 3rd 2010, 04:52 PM
HallsofIvy
Quote:

Originally Posted by Riyzar
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: $\displaystyle (\vec{j}- \vec{k})\times (\vec{k}- \vec{i})= (\vec{j}- \vec{k})\vec{k}- (\vec{j}- \vec{k})\vec{i}= \vec{j}\times\vec{j}+ \vec{k}\vec{j}- \vec{k}\times\vec{i}+ \vec{k}\times\vec{i}$
• Sep 3rd 2010, 05:57 PM
Soroban
Hello, Riyzar!

Quote:

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

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

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

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

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

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

• Sep 3rd 2010, 08:07 PM
Riyzar
Quote:

Originally Posted by HallsofIvy
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!