# What is a cross product?

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

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