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Cross Products

What is a cross product?

From Wikipedia

From Wikipedia

A cross product is an operation done on two vectors that returns the vector that is perpendicular to both of them. Interestingly, the magnitude of a x b = ||a|| * ||b|| * sin(theta), which is also the area of the parallelogram created by the two vectors.

Right Hand Rule

The right hand rule is a rule that says that the cross product goes from the right vector to the left vector. So you could make this with your hand. 

From Wikipedia, again

From Wikipedia, again

 

Basically b x a = -(a x b).

How to do it

3x3 Matrix Determinant

MathJax TeX Test Page Let's say you want to find $\vec{a} \times \vec{b}$ where $\vec{a} = <1,2,3>$, and $\vec{b} = <-1,4,7>$. In order to figure out the cross product, you must take the determinant of the following matrix: $$\begin{vmatrix}i & j & k\\1 & 2 & 3\\-1 & 4 & 7\end{vmatrix}$$ Note: i,j, and k are the unit vectors in the direction of the x,y, and z axes. i is $<1,0,0>$, j is $<0,1,0>$, and k is $<0,0,1>$. $$\text{That equals } i \cdot \begin{vmatrix}2 & 3\\4 & 7\end{vmatrix} - j \cdot \begin{vmatrix}1 & 3\\-1 & 7\end{vmatrix} + k \cdot \begin{vmatrix}1 & 2\\-1 & 4\end{vmatrix}$$ $$i= \cdot (2 \cdot 7 - 3 \cdot 4) - j \cdot (7+3) + k \cdot (4 + 2) $$ $$ = 2i - 10j + 6k = <2,-10, 6>$$

Another Way

MathJax TeX Test Page This isn't necessarily an easier way; it's actually almost the same as the matrix determinant way. The only difference between the two is how they're written. Let's look at the same problem as before: $\vec{a} \times \vec{b}$ where $\vec{a} = <1,2,3>$, and $\vec{b} = <-1,4,7>$. Now, we write it like this: $$\begin{bmatrix}1 & 2 & 3 & 1 & 2 & 3\\-1 & 4 & 7 & -1 & 4 & 7\end{bmatrix}$$ What you do now is split it up into three smaller matrices, ignoring the first and last columns: $$\begin{bmatrix}2 & 3\\4 & 7\end{bmatrix} \begin{bmatrix}3 & 1\\7 & -1\end{bmatrix} \begin{bmatrix} 1 & 2\\-1 & 4\end{bmatrix}$$ The determinants of these matrices correspond to the parts of the cross products. $$\text{So that equals: } <14 - 12, -3 - 7, 4 + 2> = <2, -10, 6>$$ Once again that's literally the same as the other method, just written differently. This just takes up less space.

Here's a really good video about it. 

David Witten

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