# Cross multiplying vectors

• 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.

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

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.

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

Quote:

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

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

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

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

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

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

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

Originally Posted by HallsofIvy
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!