cross product

A x B = ( AyBz - AzBy )i +
        ( AzBx - AxBz )j +
        ( AxBy - AyBx )k

| i  j  k  |
| a1 a2 a3 |   = AxB
| b1 b2 b3 |

For some reason, the official definition of cross-product is that it is only defined for vectors of three dimensions. However, the above formula will still work for other dimensions. In fact, Mathematica's cross() function will allow vectors of dimensions above 3.

V w = v × w.

           |  0  -vz  vy |
cross(v) = |  vz  0  -vx |
           | -vy  vx  0  |.

In Physics, we use this to determine Torque and Angular Momentum. For Torque, take the cross product of the Force Vector and the Radius of the disc, or vector from the point of application to the center. (t = F x R)
The Torque vector will lie in a plane perpendicular to both the Force and Radius - so, if the Radius is straight along the X-axis and Force is along the Y, Torque can only be in the Z-axis.

To find out exactly where this vector will end up, we do one of the above operations, preferrably breaking both vectors up into i j and k components and multiplying them with one another. (Anark points out that this is called Cyclic Permutation.) It is imperative to remember that if an i term is multiplied with a j term it turns into a k term, and if a j term is multiplied with an i it becomes -k.

A simple way to remember this is that if we go forward in ijk, it stays positive, but if we go backwards it turns negative:

i * j = k
j * k = i
i * k = j

j * i = -k
k * j = -i
k * i = -j 

To save time on math finals and real-world applications (there are such things) we can simplify in most cases. Something big and ugly like:
(3 cos wti + 3 sin wtj + 4wtk) x (-3 w sin wti + 3w cos wtj + 4wtk)
w without bold denotes angle measure in this problem. What we're actually doing here, is finding the cross product between r and r' -- angular momentum.

Can be simplified by noticing that:

u = cos wti + sin wtj,
v = -sin wti + cos wtj,
w = k

form a right-hand rule set of coordinate axes -- their cross product works just like that of i, j and k

The cross product then, is:

(3u + 4tww) x (3wv + 4ww) =
9wu x v + 12wu x w + 12tww x v =
9ww - 12wv - 12twu

Yet another cross product writeup? Well, everybody has his or her own outlook on things, and in this case I felt my outlook was sufficiently different to warrant another writeup.

The cross product is the complement of the dot product. While the dot product yields a scalar and the cross product yields a vector, this difference is largely semantical. In fact, it is reasonable to define a "scalar cross product" as the magnitude of the vector cross product. I will first discuss the scalar cross product, and then show how to find the direction of the vector cross product a X b.

The dot product of a and b is the projection of a on b times the magnitude of b. The scalar cross product is simply the component of a that is not projected onto b times the magnitude of b. See the 2-d diagram below.

              _ 
              /|
             / |
         a  /  |  component of 
           /   |  a not
          /    |  projected onto
         /     |  b
        / θ    |
       o========-------------->
        projection      b
        of a 
        onto b
        

As you can see, the scalar cross product is the complement of the dot product, though the determinant formula for cross products (given above) makes it look much more complex. Notice that while the dot product of a and b is given by a^.b = abcosθ, the scalar vector product is given by absinθ.

We've defined this "scalar cross product." What's the "real" cross product? The magnitude of the real cross product is the scalar cross product, and the direction is chosen such that it is perpendicular to both a and b.

There's just one more thing. There are actually two directions that will give perpendicular cross products. In the diagram above, one direction would be out of your monitor, and the other would be into your monitor. Mathematicians arbitrarily defined the direction of a X b to be into your monitor, and the direction of b X a to be out of your monitor. This is the so-called right hand rule. To find the "correct" cross product direction, point your right hand fingers along the vector on the left side of the X, squeeze them toward the vector on the right side of the X, and note where your thumb points.

Simple geometry/trigonometry shows that this definition of the cross product is equivalent to the determinant formula, which is useful in practice but isn't all that intuitive (to me at least).

Here is the two dimensional "cross product". It is an operation on a single vector, and it simply rotates the vector by 90 degrees (so that it is perpendicular). Perhaps flatlander physicists have a use for this sort of thing.

cross u = | i  j  | = u2 i - u1 j
          | u1 u2 |
	  

                       | i  j  k  |
cross (u, v) = u x v = | u1 u2 u3 |
                       | v1 v2 v3 |

     = (u2 v3 - u3 v2) i - (u1 v3 - u3 v1) j + (u1 v2 - u2 v1) k

Certain optimisations can be made when computing this. The lower 2x2 determinants (eg. v3 w4 - v4 w3) are each used twice, so they can be computed once and then remembered the second place. Still, it is a hefty job.

(I hope this works out right! *crosses fingers*)

                  | i  j  k  l  |
cross (u, v, w) = | u1 u2 u3 u4 |
                  | v1 v2 v3 v4 |
                  | w1 w2 w3 w4 |

        /                                                            \
   =   | (v3 w4 - v4 w3) u2 - (v2 w4 - v4 w2) u3 + (v2 w3 - v3 w2) u4 | i
        \                                                            /
        
        /                                                            \
     - | (v3 w4 - v4 w3) u1 - (v1 w4 - v4 w1) u3 + (v1 w3 - v3 w1) u4 | j
        \                                                            /
        
        /                                                            \
     + | (v2 w4 - v4 w2) u1 - (v1 w4 - v4 w1) u2 + (v1 w2 - v2 w1) u4 | k
        \                                                            /
        
        /                                                            \
     - | (v2 w3 - v3 w2) u1 - (v1 w3 - v3 w1) u2 + (v1 w2 - v2 w1) u3 | l
        \                                                            /

And that is why determinants are convenient.