The component formula for the cross product

Recall the geometric definition of the cross product as given in an earlier reading.

Cross product of unit vectors

First, we derive the cross product of the unit vectors, 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, the cross product must be a 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

This little cycle diagram can help you remember these results.

What about i × k? By the right-hand rule, it must be - j. I think you get the idea.

General vectors

In an earlier reading, we mentioned the special properties of the cross product: b × a = - a × b, and a × a = 0. With these exceptions, the cross product behaves like regular multiplication, obeying these 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:

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)

Since we know that i × i = 0 = j × j and that i × j = k = - j × i, this quickly simplifies to

a × b = (a₁b₂ - a₂b₁)k

$$\left\vert\vphantom{ \begin{array}{cc} a_1 & a_2 b_1 & b_2 \end{array} }\right. \begin{array}{cc} a_1 & a_2 b_1 & b_2 \end{array} \left.\vphantom{ \begin{array}{cc} a_1 & a_2 b_1 & b_2 \end{array} }\right\vert$$

Writing the result as a determinant, as we did in the last step, is a handy way to remember the result.

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)

$$\left\vert\vphantom{ \begin{array}{cc} a_2 & a_3 b_2 & b_3 \end{array} }\right. \begin{array}{cc} a_2 & a_3 b_2 & b_3 \end{array} \left.\vphantom{ \begin{array}{cc} a_2 & a_3 b_2 & b_3 \end{array} }\right\vert \left\vert\vphantom{ \begin{array}{cc} a_1 & a_3 b_1 & b_3 \end{array} }\right. \begin{array}{cc} a_1 & a_3 b_1 & b_3 \end{array} \left.\vphantom{ \begin{array}{cc} a_1 & a_3 b_1 & b_3 \end{array} }\right\vert \left\vert\vphantom{ \begin{array}{cc} a_1 & a_2 b_1 & b_2 \end{array} }\right. \begin{array}{cc} a_1 & a_2 b_1 & b_2 \end{array} \left.\vphantom{ \begin{array}{cc} a_1 & a_2 b_1 & b_2 \end{array} }\right\vert$$

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:

$$\left\vert\vphantom{ \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} a_1 & a_2 & a_3 b_1 & b_2 & b_3 \end{array} }\right. \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} a_1 & a_2 & a_3 b_1 & b_2 & b_3 \end{array} \left.\vphantom{ \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} a_1 & a_2 & a_3 b_1 & b_2 & b_3 \end{array} }\right\vert$$

This is the way I remember how to compute the cross product.