The component formula for the cross product
The geometric definition of the cross product is a good for understanding the properties of the cross product. However, the geometric definition isn’t so useful for computing the cross product of vectors. For computations, we will want a formula in terms of the components of vectors. We start by using the geometric definition to compute the cross product of the standard unit vectors.
Cross product of unit vectors
Let i, j, and k be the standard unit vectors in R3. (We define the cross product only in three dimensions.)
The parallelogram spanned by any two of these standard unit vectors is a unit square, which has area one. Hence, by the geometric definition, the cross product must be a unit vector. Since the cross product must be perpendicular to the two unit vectors, it must be equal to the other unit vector or the opposite of that unit vector. Looking at the above graph, you can use the right-hand rule to determine the following results.
| i × j | = k | ||
| j × k | = i | ||
| k × i | = j |
What about i × k? By the right-hand rule, it must be -j. I think you get the idea (after all b × a = -a × b).
| j × i | = -k | ||
| k × j | = -i | ||
| i × k | = -j |
Finally, the cross product of any vector with itself is the zero vector (a × a = 0). So the cross product of any standard unit vector with itself the zero vector.
General vectors
With the exception of the two special properties mentioned above (b×a = -a×b, and a×a = 0), the cross product behaves like regular multiplication. It obeys the following properties:
- (ya) × b = y(a × b) = a × (yb),
- a × (b + c) = a × b + a × c,
- (b + c) × a = b × a + c × a,
where a, b, and c are vectors in R3 and y is a scalar. We can use these properties, along with the cross product of the standard unit vectors, to write the formula for the cross product in terms of components.
We write the components of a and b as:
| a = (a1,a2,a3) = a1i + a2j + a3k | ||
| b = (b1,b2,b3) = b1i + b2j + b3k |
First, we’ll assume that a3 = b3 = 0. (Then, the manipulations are much easier.) We calculate:
| a × b | = (a1i + a2j) × (b1i + b2j) | ||
| = a1b1(i × i) + a1b2(i × j) + a2b1(j × i) + a2b2(j × j) |
| a × b | = (a1b2 - a2b1)k | ||
=  k. |
What about the general case where a3 and b3 might not be zero? Well, that’s too many manipulations for me to write it out here, and you probably wouldn’t read it anyway. To calculate the result, it’s just the same process as above. You can verify yourself that in general
| a × b | = (a1i + a2j + a3k) × (b1i + b2j + b3k) | ||
=  i - j +  k |
Looking at the formula for the 3 × 3 determinant, we see that the formula for a cross product looks a lot like the formula for the 3 × 3 determinant. If we allow a matrix to have the vector i, j, and k as entries (OK, maybe this doesn’t make sense, but this is just as a tool to remember the cross product), the 3 × 3 determinant gives a handy mnemonic to remember the cross product:
a × b =  |
This is the way I remember how to compute the cross product.
