Section 4.2 Second Definition of the Cross Product

Writing the vectors in components gives another definition for the cross product.

Definition 4.10 The Cross Product (Second Definition).

If \(\mathbf u = \langle u_1, u_2, u_3\rangle\) and \(\mathbf v = \langle v_1, v_2, v_3\rangle\text{,}\) then

\begin{equation} \mathbf u \times \mathbf v = \langle u_2v_3 - u_3v_2,\; u_3v_1 - u_1v_3,\; u_1v_2 - u_2v_1\rangle.\tag{4.4} \end{equation}

One way to remember the second definition of the cross product is to write it as a determinant:

\begin{align} \mathbf u \times \mathbf v \amp = \begin{vmatrix} \mathbf i \amp \mathbf j \amp \mathbf k \\ u_1 \amp u_2 \amp u_3 \\ v_1 \amp v_2 \amp v_3 \end{vmatrix}\notag\\ \amp = \begin{vmatrix} u_2 \amp u_3 \\ v_2 \amp v_3 \end{vmatrix}\mathbf i - \begin{vmatrix} u_1 \amp u_3 \\ v_1 \amp v_3 \end{vmatrix}\mathbf j + \begin{vmatrix} u_1 \amp u_2 \\ v_1 \amp v_2 \end{vmatrix}\mathbf k,\tag{4.5} \end{align}

where

\begin{equation} \begin{vmatrix} a \amp b \\ c \amp d \end{vmatrix} = ad - bc.\tag{4.6} \end{equation}

Example 4.11 Computing a cross product.

Calculate the cross product \(\mathbf u \times \mathbf v\) if \(\mathbf u = \langle 1,-1,2\rangle\) and \(\mathbf v = \langle 3,-2,1\rangle\text{.}\)

Solution.

\begin{align*} \mathbf u \times \mathbf v \amp= \begin{vmatrix} \mathbf i \amp \mathbf j \amp \mathbf k \\ 1 \amp -1 \amp 2 \\ 3 \amp -2 \amp 1 \end{vmatrix}\\ \amp= \begin{vmatrix} -1 \amp 2 \\ -2 \amp 1 \end{vmatrix}\mathbf i - \begin{vmatrix} 1 \amp 2 \\ 3 \amp 1 \end{vmatrix}\mathbf j + \begin{vmatrix} 1 \amp -1 \\ 3 \amp -2 \end{vmatrix}\mathbf k\\ \amp= (-1+4)\mathbf i - (1-6)\mathbf j + (-2+3)\mathbf k = 3\mathbf i + 5\mathbf j + \mathbf k = \langle 3,5,1\rangle. \end{align*}

Example 4.12 Computing a vector perpendicular to two other vectors.

Consider the points \(A(1,-1,2)\text{,}\) \(B(2,0,-1)\text{,}\) and \(C(0,-2,3)\text{.}\) Find a unit vector which is orthogonal to both \(\vec{AB}\) and \(\vec{AC}\text{.}\)

Solution.

We first compute the two vectors:

\begin{align*} \vec{AB} \amp= \langle 1,1,-3\rangle\\ \vec{AC} \amp= \langle -1,-1,1\rangle. \end{align*}

The cross product \(\vec{AB} \times \vec{AC}\) is orthogonal to both vectors \(\vec{AB}\) and \(\vec{AC}\text{:}\)

\begin{align*} \vec{AB} \times \vec{AC} \amp= \begin{vmatrix} \mathbf i \amp \mathbf j \amp \mathbf k \\ 1 \amp 1 \amp -3 \\ -1 \amp -1 \amp 1 \end{vmatrix}\\ \amp= \begin{vmatrix} 1 \amp -3 \\ -1 \amp 1 \end{vmatrix}\mathbf i - \begin{vmatrix} 1 \amp -3 \\ -1 \amp 1 \end{vmatrix}\mathbf j + \begin{vmatrix} 1 \amp 1 \\ -1 \amp -1 \end{vmatrix}\mathbf k\\ \amp= (1-3)\mathbf i - (1-3)\mathbf j + (-1+1)\mathbf k = -2\mathbf i + 2\mathbf j = \langle -2,2,0\rangle. \end{align*}

The unit vector in question is then computed as follows:

\begin{equation*} \begin{aligned} \mathbf n \amp = \frac{\vec{AB} \times \vec{AC}}{\left\|\vec{AB} \times \vec{AC}\right\|}\\ \amp = \frac{1}{\sqrt 8}\langle -2,2,0\rangle = \left\langle -\frac{1}{\sqrt 2},\frac{1}{\sqrt 2},0\right\rangle = -\frac{1}{\sqrt 2}\mathbf i + \frac{1}{\sqrt 2}\mathbf j. \end{aligned} \end{equation*}

Figure 4.13. The unit vector \(\mathbf n\) is orthogonal to both \(\vec{AB}\) and \(\vec{AC}\text{,}\) and hence to the plane through the points \(A\text{,}\) \(B\text{,}\) and \(C\text{.}\)

Example 4.14 Computing the area of a triangle.

Find the area of the triangle with the vertices \(A(1,-1,2)\text{,}\) \(B(2,0,-1)\text{,}\) and \(C(0,-2,3)\text{.}\) Note that these are the same points that we had in the previous example.

Solution.

To find the area of the triangle, we first compute the area of the parallelogram built on \(\vec{AB}\) and \(\vec{AC}\text{,}\) shown in Figure 4.15, and then divide by \(2\text{:}\)

\begin{align*} \text{Area of the parallelogram} \amp= \left\|\vec{AB} \times \vec{AC}\right\| = \sqrt 8 = 2\sqrt 2\\ \text{Area of the triangle} \amp= \frac12 \left\|\vec{AB} \times \vec{AC}\right\| = \sqrt 2. \end{align*}

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Figure 4.15. The triangle built on the vectors \(\vec{AB}\) and \(\vec{AC}\) is half of the corresponding parallelogram.

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