5. Cross Product

b. Cross Product and Triple Product

1. Algebraic Definition of Cross Products

The dot product is one way to compute the product of two vectors and it produces a scalar. The cross product is another way to compute the product of two vectors, but it produces a vector.

The cross product is only defined in 3 dimensions
(and 7 dimensions)

Why? See this   Wikipedia article   on the 7-dimensional cross product

If \(\vec u=\left\langle u_1,u_2,u_3\right\rangle\) and \(\vec v=\left\langle v_1,v_2,v_3\right\rangle\) are vectors, then their cross product or vector product is the vector: \[ \vec u\times\vec v =\left\langle u_2v_3-u_3v_2,u_3v_1-u_1v_3,u_1v_2-u_2v_1\right\rangle \]

The name cross product is used to emphasize that it is denoted by putting a cross between the two vectors.

The name vector product is used to emphasize that the cross product takes two vectors and gives back a vector, unlike the dot (or scalar) product, which takes two vectors and gives back a scalar.

Although it is possible to memorize this formula for the cross product, it is usually only used for theoretical purposes. No one actually uses this formula to compute a cross product. Rather, we use a formula based on the determinant.

Consider the "formal" determinant Why formal? expanded on the top row:

\[\begin{aligned} \det\begin{pmatrix} \hat\imath & \hat\jmath & \hat k \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{pmatrix} &=\hat\imath \begin{vmatrix} u_2 & u_3 \\ v_2 & v_3 \end{vmatrix} -\hat\jmath \begin{vmatrix} u_1 & u_3 \\ v_1 & v_3 \end{vmatrix} +\hat k \begin{vmatrix} u_1 & u_2 \\ v_1 & v_2 \end{vmatrix} \\ &=(u_2v_3-u_3v_2)\hat\imath-(u_1v_3-u_3v_1)\hat\jmath+(u_1v_2-u_2v_1)\hat k \\ &=\left\langle u_2v_3-u_3v_2,u_3v_1-u_1v_3,u_1v_2-u_2v_1\right\rangle \end{aligned}\] This is the same as the cross product. So we have:

The cross product of \(\vec u=\left\langle u_1,u_2,u_3\right\rangle\) and \(\vec v=\left\langle v_1,v_2,v_3\right\rangle\) may be computed using the determinant: \[ \vec u\times\vec v =\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix} \] which has \(\hat\imath,\hat\jmath,\) and \(\hat k\) on the first row, the vector \(\vec u\) on the second row and \(\vec v\) on the third row.

Memorize This!

Compute the cross product of the vectors \(\left\langle 1,2,1\right\rangle\) and \(\left\langle -2,5,6\right\rangle\).

The cross product is \[\begin{aligned} \left\langle 1,2,1\right\rangle\times\left\langle -2,5,6\right\rangle &=\begin{vmatrix} \hat\imath & \hat\jmath & \hat k \\ 1 & 2 & 1 \\ -2 & 5 & 6 \end{vmatrix} \\ &=\hat\imath\begin{vmatrix} 2 & 1 \\ 5 & 6 \end{vmatrix} -\hat\jmath\begin{vmatrix} 1 & 1 \\ -2 & 6 \end{vmatrix} +\hat k\begin{vmatrix} 1 & 2 \\ -2 & 5 \end{vmatrix} \\ &=(12-5)\hat\imath-(6+2)\hat\jmath+(5+4)\hat k \\ &=7\hat\imath-8\hat\jmath+9\hat k=\left\langle 7,-8,9\right\rangle \end{aligned}\]

Don't forget the minus before the \(\,\large \hat\jmath\)-term!

PY: Future Random Exercise

Here are the two most important properties of the cross product:

Since a determinant changes sign when two rows are interchanged, the cross product changes sign when the two vectors are interchanged:

The Cross Product is Anti-Commutative: \(\qquad \vec u\times\vec v=-\vec v\times\vec u\)

Since a determinant is zero when two rows are equal, the cross product is zero when the two vectors are equal:

If the two vectors are equal, their Cross Product is zero: \(\qquad \vec v\times\vec v=0\)