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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):

enter image description here

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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1  
You should elaborate why you think that the above mentioned ways are insufficient. Saying that they are "poor/slow" is not descriptive enough. –  Tpofofn Mar 25 '12 at 0:49
    
The arrows in category theory encapsulate broad categorical data between various algebraic structures, topological spaces, etc. and their morphisms - it is a bit out of place to use it to describe the visual mnemonics here. –  anon Mar 25 '12 at 1:12
    
@anon: sorry? Small is beautiful: you can break things such as $-$, $*$, $\times$ and even numbers into functions (soon becoming theoretical cs though) -- and then formulate a category-theoretic description for the cross-product. Look we can find visual way but we may need to dig a bit. Although it may sound trivial, I think it could be quite educational, not sure whether need to dig into mathematical-logic or something like that, anyway the puzzle is set :) –  hhh Mar 25 '12 at 1:15

6 Answers 6

up vote 6 down vote accepted

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

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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: $$\begin{vmatrix} x_1&x_2&\bf X \\y_1&y_2&\bf X \end{vmatrix}$$

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Basically the same as pen -rule here, I like the pen -rule +1 :) –  hhh Sep 4 '12 at 22:02

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.

random picture

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].

$$\begin{array}{cccccc} & \langle & v_1, & v_2,& v_3 & \rangle \\ \times & \langle & w_1,& w_2, & w_3 & \rangle \\ \hline & \langle & v_2w_3-w_2v_3,& -(v_1w_3-w_1v_3),& v_1w_2-w_1v_2 & \rangle \end{array}$$

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Note 1: the order of the terms as written in the Sarrus rule depiction here are inapplicable when computing the curl $\nabla\times\mathbf{F}$: partial derivative operators go before the quantities they act on. (This was the basis of another one of OP's questions.) Note 2: Quaternions can in fact be written as a formal sum of a scalar and a vector; multiplication is computed as $$(a+\mathbf{b})(c+\mathbf{d})=ac-\mathbf{b\cdot d}+a\mathbf{d}+c\mathbf{b}+\mathbf{b\times d}.$$ With scalar parts zero, $a=c=0$, the resulting vector part is just the cross product. –  anon Mar 25 '12 at 1:44
    
@anon Ah, yes. Curl, a mnemonic inside a mnemonic :) –  Bill Cook Mar 25 '12 at 1:47
    
Where in the world did you ever get the name "circle of doom" for such an innocuous little mnemonic? I see you comment that you use this name for no particular reason, but still... –  KCd Jan 13 '13 at 6:14
    
@KCd I think I had a student complaining about remembering how the cross product worked. My response went something like "Not remembering the formula for the cross product will spell out your doom." But, yes, there's no good reason to call it that. :) –  Bill Cook Jan 22 '13 at 16:54

enter image description here

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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

  1. Technical problem with this answer here about plotting.
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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).

$$ \left[\begin{array}{c} x \\ y \\ z \end{array}\right] \times \left[\begin{array}{c} u \\ v \\ w \end{array}\right] = \left[\begin{array}{ccc} 0 & -z & y \\ z & 0 & -x \\ -y & x & 0 \end{array}\right] \left[\begin{array}{c} u \\ v \\ w \end{array}\right] $$

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