Mnemonic for cross product I am wondering why nobody teached me this on universe, I always had so much trouble remembering the cross product, but now it's all so trivial, I'm talking about the following mnemonic I thought of for:

I remember the following:


*

*It is defined in vector format for x, y, z (very logical step)

*The first value (v3.x) starts with y in the calculation, then v3.y has z and v3.z follows with x.

*Then I remember the pattern in pseudocode (if you start with y): v1.y * v2.y++ - v1.y++ * v2.y, breaking into parts:


*

*The general format is 1 * 2 - 1 * 2, not hard to remember at all

*For the vertices they go with y * y++ - y++ * y, where y++ denotes in fact z.



Does this explanation even hold? Are there facts that make sense in my world, but not in the general world? I would like to know that before I start memorizing this one, which I infact already did I think.
 A: $$
\begin{align}
&(x_1,x_2,x_3)^\mathrm{T}\times(y_1,y_2,y_3)^{\mathrm{T}}\\
=&\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
x_1 & x_2 & x_3 \\
y_1 & y_2 & y_3
\end{vmatrix}\\
=&(x_2y_3-x_3y_2)\mathbf{i}+(x_3y_1-x_1y_3)\mathbf{j}+(x_1y_2-x_2y_1)\mathbf{k}
\end{align}
$$
where $(\mathbf{i},\mathbf{j},\mathbf{k})$ are orthonormal basis in $\mathbb{R}^3$.
A: A much easier way of remembering: $$u=(u_1,u_2,u_3) \\v= (v_1,v_2,v_3) \\u\times v = \left|\begin{array}{ccc} i & j & k \\u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3\end{array} \right|$$
Where $i,j,k$ are the unit vectors in which the coordinates are expressed (most of the time the usual unit vectors). This comes from a property (or possible definition) of the cross product as $u\times v = w \text{ such that } \forall a\in\mathbb R^3,\  a\cdot w = \det(a,u,v)$ (using unit vector coordinates, i.e. orthonormal basis). The order which the above expression is written (what row goes where, whether it's $u\times v$ or $v\times u$, etc.) is relevant!
A: $$
\begin{pmatrix} 
x\\
y\\
z\\
\end{pmatrix}\times \begin{pmatrix} 
x'\\
y'\\
z'\\
\end{pmatrix} \equiv
\begin{vmatrix}
\hat{\mathbf{i}}&\hat{\mathbf{j}}&\hat{\mathbf{k}}\\
x&y&z\\
x'&y'&z'\\
\end{vmatrix}=
\begin{vmatrix} 
y&z\\
y'&z'\\
\end{vmatrix}\hat{\mathbf{i}}
-\begin{vmatrix} 
x&z\\
x'&z'\\
\end{vmatrix}\hat{\mathbf{j}}
+\begin{vmatrix} 
x&y\\
x'&y'\\
\end{vmatrix}\hat{\mathbf{k}}=(yz'-y'z)\hat{\mathbf{i}}+(x'z-xz')\hat{\mathbf{j}}+(xy'-x'y)\hat{\mathbf{k}}\equiv
\begin{pmatrix} 
yz'-y'z\\
x'z-xz'\\
xy'-x'y\\
\end{pmatrix}$$
A: I think you got it right, but expressed it very complicated, let's look at it this way:
$$
\begin{pmatrix} x\\y\\z\end{pmatrix}
\times
\begin{pmatrix} x'\\y'\\z'\end{pmatrix}
=
\begin{pmatrix} yz' - zy' \\ zx' - xz' \\ xy' - yx'\end{pmatrix}.
$$
For the first row, you just remember that you have a cross $\times$ of the other two rows, starting from the top left, just as you read a book. This gives you $yz'-zy'$. The other two rows are cyclic permutations of this, so for the second row, $y$ becomes $z$, $z$ becomes $x$.
