Calculate and check a Cartesian cross product

Calculate and check a Cartesian cross product

Calculate the three components in order, taking particular care with the middle sign. Check that a proposed common perpendicular has zero dot product with both inputs. If a direction is required, also check that your result is nonzero.

In a right-handed orthonormal basis, the source coordinate formula and its MF27 version are the same:

$$\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}\times\begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix} =\begin{pmatrix}a_2b_3-a_3b_2\\-(a_1b_3-a_3b_1)\\a_1b_2-a_2b_1\end{pmatrix} =\begin{pmatrix}a_2b_3-a_3b_2\\a_3b_1-a_1b_3\\a_1b_2-a_2b_1\end{pmatrix}.$$

This column equals $(|\mathbf a||\mathbf b|\sin\theta)\widehat{\mathbf n}$ in the nonparallel case. The positive cyclic basis products are

$$\mathbf i\times\mathbf j=\mathbf k,\quad\mathbf j\times\mathbf k=\mathbf i,\quad\mathbf k\times\mathbf i=\mathbf j.$$

Reversed order gives

$$\mathbf j\times\mathbf i=-\mathbf k,\quad\mathbf k\times\mathbf j=-\mathbf i,\quad\mathbf i\times\mathbf k=-\mathbf j,$$

and $\mathbf i\times\mathbf i=\mathbf j\times\mathbf j=\mathbf k\times\mathbf k=\mathbf0$. The source QR exploring how the product relates to its inputs decodes to the CJC mathematics learning resource.

Derive the signs from the basis products

To derive the coordinate formula, expand the basis expressions into all nine terms before substituting the cyclic rules:

$$\begin{aligned} \mathbf a\times\mathbf b&=(a_1\mathbf i+a_2\mathbf j+a_3\mathbf k)\times(b_1\mathbf i+b_2\mathbf j+b_3\mathbf k)\\ &=a_1b_1\mathbf i\times\mathbf i+a_1b_2\mathbf i\times\mathbf j+a_1b_3\mathbf i\times\mathbf k\\ &\quad+a_2b_1\mathbf j\times\mathbf i+a_2b_2\mathbf j\times\mathbf j+a_2b_3\mathbf j\times\mathbf k\\ &\quad+a_3b_1\mathbf k\times\mathbf i+a_3b_2\mathbf k\times\mathbf j+a_3b_3\mathbf k\times\mathbf k\\ &=a_1b_1\mathbf0+a_1b_2\mathbf k+a_1b_3(-\mathbf j)\\ &\quad+a_2b_1(-\mathbf k)+a_2b_2\mathbf0+a_2b_3\mathbf i\\ &\quad+a_3b_1\mathbf j+a_3b_2(-\mathbf i)+a_3b_3\mathbf0\\ &=(a_2b_3-a_3b_2)\mathbf i-(a_1b_3-a_3b_1)\mathbf j+(a_1b_2-a_2b_1)\mathbf k\\ &=\begin{pmatrix}a_2b_3-a_3b_2\\-(a_1b_3-a_3b_1)\\a_1b_2-a_2b_1\end{pmatrix}. \end{aligned}$$

Example 17 produces a common perpendicular and checks it using two independent dot products. A zero result would pass those checks too, so verify that the requested perpendicular direction is nonzero. Self-Practice 6 applies the component formula to two pairs.