Objective to find visual and accessible ways to remember this formula fast
$$(x,y,z)\times(u,v,w)=(yw-zv,zu-xw,xv-yu)$$
I have used Sarrus' rule but it is slow, more here. Since it is slow, I have tried to find alternative ways such as binary-tree -visualization (but it is poor/slow until some clever ideas):
RobJohn's idea to "just cycle the permutation"
$$\begin{align}i&=j\times k\\j&=k\times i\\k&=i\times j\end{align}$$
Mariano Suárez‐Alvarez -idea
View the 3×3 matrix as the points of the affine plane over F3: then the terms in the determinant correspond to affine lines which are neither horizontal nor vertical --.
Anon's idea
Draw a triangle with vertices i, j, k with arrows i->j, j->k, k->i. Multiplying two of these units is done as follows: if they are the same, 0; if they go with the flow of the triangle, the third in the line; if the go opposite the flow, the third one as well but with a minus sign
We are discussing this issue more here.
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$\begingroup$I usually teach this (which requires no additional writing and avoids cyclic permutations, which are often confusing for students): $$ \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \times \left[\begin{array}{c} u \\ v \\ w \end{array}\right] $$
1) ignore $x$ and $u$ (that is: mentally block view of the first row), compute $2\times 2$ determinant of remaining stuff $yw-zv$
2) mentally block view of the second row, throw in a minus, compute $2\times 2$ determinant
3) mentally block view of the third row, compute $2\times 2$ determinant
instead of mentally blocking your view, you can put a pencil on that row
$\endgroup$ 0 $\begingroup$I really truly believe it's hard to beat the $3\times 3$ determinant mnemonic.
However, there is a trick for $3 \times 3$ determinants which make computing them a snap. Remember that you compute a $2 \times 2$ determinant by multiplying diagonals and off-diagonals (the upward diagonal) and taking the difference. You can do the same for $3 \times 3$'s...well, sort of. You just repeat the first two columns. Then subtract the products of the up diagonals from the products of the down diagonals.
By the way, RobJon and Anon's suggestions amount to the same thing. Commonly drawn as a circle (see below). I refer (for no apparent reason) to this as the "circle of doom". It helps you remember cross products among the standard basis as well as how to multiply quaternions (which amounts to nearly the same thing). Go clockwise and get: $ij=k$, $jk=i$, $ki=j$. Fight the circle and get negatives: $ji=-k$ etc.
Edit:@Blah's post is quite similar to what I use when teaching the cross-product. However, instead of columns, I usually write the vectors in rows and write out the cross product like an elementary school multiplication. Then the cross product is computed by ignoring the first, second, third columns in order; computing the corresponding $2 \times 2$ determinant; and negating the middle term [which really just amounts to using the determinant mnemonic, but involves less writing]. This is easier to implement than the $3 \times 3$ trick I showed above and is applicable when computing curl [see anon's comment below].
$$\begin{array}{cccccc} & \langle & v_1, & v_2,& v_3 & \rangle \\ \times & \langle & w_1,& w_2, & w_3 & \rangle \\ \hline & \langle & v_2w_3-w_2v_3,& -(v_1w_3-w_1v_3),& v_1w_2-w_1v_2 & \rangle \end{array}$$
$\endgroup$ 4 $\begingroup$Perhaps this is no more clear, but I found it easy to remember as a matrix product after the first vector is changed into a skew-symmetric matrix (see the Wikipedia here).
$$ \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \times \left[\begin{array}{c} u \\ v \\ w \end{array}\right] = \left[\begin{array}{ccc} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{array}\right] \left[\begin{array}{c} u \\ v \\ w \end{array}\right] $$
$\endgroup$ 2 $\begingroup$This works wonderfully for me:
Say ${\bf x}=(x_1,x_2,x_3)$ and ${\bf y}=(y_1,y_2,y_3)$. Then
$${\bf x}\times {\bf y}=\left(\begin{vmatrix} x_2&x_3 \\y_2&y_3 \end{vmatrix},-\begin{vmatrix} x_1&x_3 \\y_1&y_3 \end{vmatrix},\begin{vmatrix} x_1&x_2 \\y_1&y_2 \end{vmatrix}\right)$$
You can also remember it by a cover method of the arrangement
$$\begin{vmatrix} x_1&x_2&x_3 \\y_1&y_2&y_3 \end{vmatrix}$$
First coordinate: $$\begin{vmatrix} \bf X&x_2&x_3 \\\bf X&y_2&y_3 \end{vmatrix}$$Second coordinate: $$\begin{vmatrix} x_1&\bf X&x_3 \\y_1&\bf X&y_3 \end{vmatrix}$$Third coordinate: $$\begin{vmatrix} x_1&x_2&\bf X \\y_1&y_2&\bf X \end{vmatrix}$$
$\endgroup$ 1 $\begingroup$I am not sure whether this way could deduce the more advanced case but I will elaborate the RobJohn
$$ \begin{align} \color{Red}{i}&=\color{Blue}{j}\times \color{Green}{k}\\ \color{Blue}{j}&=\color{Green}{k}\times \color{Red}{i}\\ \color{Green}{k}&=\color{Red}{i}\times \color{Blue}{j}. \end{align}$$
Now the "bit -shift" is equivalent to cross-product by the left-hand-side -element. For example, to get from $ijk$ to $jki$ -- we product by the LHS element i.e. here $k$
$$\begin{align}k\times (\color{Red}{i})&=k\times (j\times k ) \\ j&=k\times \color{Red}{i}.\end{align}$$
Cross-product order with toruses
1. Torus
$$\begin{pmatrix} ... & ... & ... & ... & ... & ... & ... & ... \\ ... & \color{Red}{i} & \color{Blue}{j} & \color{Green}{k} & i & j & k & ... \\ ... & i & \color{Blue}{j} & \color{Green}{k} & \color{Red}{i} & j & k & ... \\ ... & i & j & \color{Green}{k} & \color{Red}{i} & \color{Blue}{j} & k & ... \\ ... & ... & ... & ... & ... & ... & ... & ... \\ \end{pmatrix}$$
2. Torus cut
$$\begin{pmatrix} ... & \color{Red}{i} & \color{Blue}{j} & \color{Green}{k} & i & \color{Blue}{j} & \color{Green}{k} & \color{Red}{i} & j & \color{Green}{k} & \color{Red}{i} & \color{Blue}{j} & k & & ... \end{pmatrix} $$
3. Simplified Torus
Basically you need only to remember the order $i,j,k$, then just a few repetition to get $i,j,k,i,j$. Read from left to right to get $ijk$, $jki$ and $kij$ -- we got our cyclic permutations!
Perhaps Related
- Technical problem with this answer here about plotting.
This method is how I memorize it. Personally, I don't really need a visual way if there is an easier method of simply memorizing it. Here goes:
Cross product of vector a=(x1, y1, z1) and b=(x2, y2, z2)
First write down i, j, k, with their respective parentheses, but in reverse order:
a x b = k( ... ) + j( ... ) + i( ... )
Now, just remember xyz twice, (XY-ZX-YZ). Basically, two letters go into the parentheses, in order:
$a \times b = k(xy ... ) + j(zx ... ) + i(yz ... )$
Next just write the same term again, but with a subtraction sign:
$a \times b = k(xy-xy) + j(zx-zx) + i(yz-yz)$
Finally, label each part by writing 1-2-2-1 in the subscript (I sometimes write the numbers on top of the letters just for convenience):
$a \times b = k(x_1y_2-x_2y_1) + j(z_1x_2-z_2x_1) + i(y_1z_2-y_2z_1)$
This is just one method that I figured out myself. Sorry I'm not too good at explaining this, but I hope it helps! I like this method because I don't need to take an extra part of the paper to draw out any diagrams or anything like that.
If you are planning to share this on another source, feel free to, but please comment below! (I just want to see if others are finding this useful)
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