Appendix B:\(\quad\)Proof of the Right-Hand Rule for the Cross Product

We will prove the right-hand rule for the cross product of two vectors in \(\Real{3}\).

For any vectors v and w in \(\Real{3}\), define a new vector, \(\textbf{n}(\textbf{v},\textbf{w})\), as follows:

1. If v and w are nonzero and not parallel, and \(\theta\) is the angle between them, then \(\textbf{n}(\textbf{v},\textbf{w})\) is the vector in \(\Real{3}\) such that:

   1. the magnitude of \(\textbf{n}(\textbf{v},\textbf{w})\) is \(\norm{\textbf{v}}\,\norm{\textbf{w}}\,\sin \theta\),

   2. \(\textbf{n}(\textbf{v},\textbf{w})\) is perpendicular to the plane containing v and w, and

   3. v, w, \(\textbf{n}(\textbf{v},\textbf{w})\) form a right-handed system.

2. If v and w are nonzero and parallel, then \(\textbf{n}(\textbf{v},\textbf{w}) = \textbf{0}\).

3. If either v or w is \(\textbf{0}\), then \(\textbf{n}(\textbf{v},\textbf{w}) = \textbf{0}\).

The goal is to show that \(\textbf{n}(\textbf{v},\textbf{w}) = \Crossprod{\textbf{v}}{\textbf{w}}\) for all v, w in \(\Real{3}\), which would prove the right-hand rule for the cross product (by part 1(c) of our definition). To do this, we will perform the following steps:

Step 1: Show that \(\textbf{n}(\textbf{v},\textbf{w}) = \Crossprod{\textbf{v}}{\textbf{w}}\) if v and w are any two of the basis vectors i, j, k.

This was already shown in Example 1.11 in Section 1.4.

Step 2: Show that \(\textbf{n}(a\textbf{v},b\textbf{w}) = ab(\Crossprod{\textbf{v}}{\textbf{w}})\) for any scalars \(a\), \(b\) if v and w are any two of the basis vectors i, j, k.

If either \(a = 0\) or \(b = 0\) then \(\textbf{n}(a\textbf{v},b\textbf{w}) = \textbf{0} = ab(\Crossprod{\textbf{v}}{\textbf{w}})\), so the result holds. So assume that \(a \ne 0\) and \(b \ne 0\). Let v and w be any two of the basis vectors i, j, k. For example, we will show that the result holds for \(\textbf{v} = \textbf{i}\) and \(\textbf{w} = \textbf{k}\) (the other possibilities follow in a similar fashion).

For \(a\textbf{v} = a\textbf{i}\) and \(b\textbf{w} = b\textbf{k}\), the angle \(\theta\) between \(a\textbf{v}\) and \(b\textbf{w}\) is \(90\Degrees\). Hence the magnitude of \(\textbf{n}(a\textbf{v},b\textbf{w})\), by definition, is \(\norm{a\textbf{i}}\,\norm{b\textbf{k}}\,\sin 90\Degrees = \abs{ab}\). Also, by definition, \(\textbf{n}(a\textbf{v},b\textbf{w})\) is perpendicular to the plane containing \(a\textbf{i}\) and \(b\textbf{k}\), namely, the \(xz\)-plane. Thus, \(\textbf{n}(a\textbf{v},b\textbf{w})\) must be a scalar multiple of j. Since its magnitude is \(\abs{ab}\), then \(\textbf{n}(a\textbf{v},b\textbf{w})\) must be either \(\abs{ab}\textbf{j}\) or \(-\abs{ab}\textbf{j}\).

There are four possibilities for the combinations of signs for \(a\) and \(b\). We will consider the case when \(a > 0\) and \(b > 0\) (the other three possibilities are handled similarly). In this case, \(\textbf{n}(a\textbf{v},b\textbf{w})\) must be either \(ab\textbf{j}\) or \(-ab\textbf{j}\). Now, since i, j, k form a right-handed system, then i, k, j form a left-handed system, and so i, k, \(-\textbf{j}\) form a right-handed system. Thus, \(a\textbf{i}\), \(b\textbf{k}\), \(-ab\textbf{j}\) form a right-handed system (since \(a > 0\), \(b > 0\), and \(ab > 0\)). So since, by definition, \(a\textbf{i}\), \(b\textbf{k}\), \(\textbf{n}(a\textbf{i},b\textbf{k})\) form a right-handed system, and since \(\textbf{n}(a\textbf{i},b\textbf{k})\) has to be either \(ab\textbf{j}\) or \(-ab\textbf{j}\), this means that we must have \(\textbf{n}(a\textbf{i},b\textbf{k}) = -ab\textbf{j}\).

But we know that \(\Crossprod{a\textbf{i}}{b\textbf{k}} = ab (\Crossprod{\textbf{i}}{\textbf{k}}) = ab (-\textbf{j}) = -ab\textbf{j}\). Therefore, \(\textbf{n}(a\textbf{i},b\textbf{k}) = ab(\Crossprod{\textbf{i}}{\textbf{k}})\), which is what we needed to show.
\(\therefore ~~ \textbf{n}(a\textbf{v},b\textbf{w}) = ab(\Crossprod{\textbf{v}}{\textbf{w}}) \enskip\checkmark\)

Step 3: Show that \(\textbf{n}(\textbf{u},\textbf{v} + \textbf{w}) = \textbf{n}(\textbf{u},\textbf{v}) + \textbf{n}(\textbf{u},\textbf{w})\) for any vectors u, v, w.

If \(\textbf{u} = \textbf{0}\) then the result holds trivially since \(\textbf{n}(\textbf{u},\textbf{v} + \textbf{w})\), \(\textbf{n}(\textbf{u},\textbf{v})\) and \(\textbf{n}(\textbf{u},\textbf{w})\) are all the zero vector. If \(\textbf{v} = \textbf{0}\), then the result follows easily since \(\textbf{n}(\textbf{u},\textbf{v} + \textbf{w}) = \textbf{n}(\textbf{u},\textbf{0} + \textbf{w}) = \textbf{n}(\textbf{u},\textbf{w}) = \textbf{0} + \textbf{n}(\textbf{u},\textbf{w}) = \textbf{n}(\textbf{u},\textbf{0}) + \textbf{n}(\textbf{u},\textbf{w}) = \textbf{n}(\textbf{u},\textbf{v}) + \textbf{n}(\textbf{u},\textbf{w})\). A similar argument shows that the result holds if \(\textbf{w} = \textbf{0}\).

So now assume that u, v and w are all nonzero vectors. We will describe a geometric construction of \(\textbf{n}(\textbf{u},\textbf{v})\), which is shown in the figure below. Let \(P\) be a plane perpendicular to u. Multiply the vector v by the positive scalar \(\norm{\textbf{u}}\), then project the vector \(\norm{\textbf{u}}\,\textbf{v}\) straight down onto the plane \(P\). You can think of this projection vector (denoted by \(proj_{P} \norm{\textbf{u}}\,\textbf{v}\)) as the shadow of the vector \(\norm{\textbf{u}}\,\textbf{v}\) on the plane \(P\), with the light source directly overhead the terminal point of \(\norm{\textbf{u}}\,\textbf{v}\). If \(\theta\) is the angle between u and v, then we see that \(proj_{P} \norm{\textbf{u}}\,\textbf{v}\) has magnitude \(\norm{\textbf{u}}\,\norm{\textbf{v}} \sin \theta\), which is the magnitude of \(\textbf{n}(\textbf{u},\textbf{v})\). So rotating \(proj_{P} \norm{\textbf{u}}\,\textbf{v}\) by \(90\Degrees\) in a counter-clockwise direction in the plane \(P\) gives a vector whose magnitude is the same as that of \(\textbf{n}(\textbf{u},\textbf{v})\) and which is perpendicular to \(proj_{P} \norm{\textbf{u}}\,\textbf{v}\) (and hence perpendicular to v). Since this vector is in \(P\) then it is also perpendicular to u. And we can see that u, v and this vector form a right-handed system. Hence this vector must be \(\textbf{n}(\textbf{u},\textbf{v})\). Note that this holds even if \(\textbf{u} \parallel \textbf{v}\), since in that case \(\theta = 0\Degrees\) and so \(\sin \theta = 0\) which means that \(\textbf{n}(\textbf{u},\textbf{v})\) has magnitude \(0\), which is what we would expect.

Now apply this same geometric construction to get \(\textbf{n}(\textbf{u},\textbf{w})\) and \(\textbf{n}(\textbf{u},\textbf{v} + \textbf{w})\). Since \(\norm{\textbf{u}}\,(\textbf{v} + \textbf{w})\) is the sum of the vectors \(\norm{\textbf{u}}\,\textbf{v}\) and \(\norm{\textbf{u}}\,\textbf{w}\), then the projection vector \(proj_{P} \norm{\textbf{u}}\,(\textbf{v} + \textbf{w})\) is the sum of the projection vectors \(proj_{P} \norm{\textbf{u}}\,\textbf{v}\) and \(proj_{P} \norm{\textbf{u}}\,\textbf{w}\) (to see this, using the shadow analogy again and the parallelogram rule for vector addition, think of how projecting a parallelogram onto a plane gives you a parallelogram in that plane). So then rotating all three projection vectors by \(90\Degrees\) in a counter-clockwise direction in the plane \(P\) preserves that sum (see the figure below), which means that \(\textbf{n}(\textbf{u},\textbf{v} + \textbf{w}) = \textbf{n}(\textbf{u},\textbf{v}) + \textbf{n}(\textbf{u},\textbf{w}).\enskip\checkmark\)

Step 4: Show that \(\textbf{n}(\textbf{w},\textbf{v}) = -\textbf{n}(\textbf{v},\textbf{w})\) for any vectors v, w.

If v and w are nonzero and parallel, or if either is \(\textbf{0}\), then \(\textbf{n}(\textbf{w},\textbf{v}) = \textbf{0} = -\textbf{n}(\textbf{v},\textbf{w})\), so the result holds. So assume that v and w are nonzero and not parallel. Then \(\textbf{n}(\textbf{w},\textbf{v})\) has magnitude \(\norm{\textbf{w}}\,\norm{\textbf{v}}\,\sin \theta\), which is the same as the magnitude of \(\textbf{n}(\textbf{v},\textbf{w})\), and hence is the same as the magnitude of \(-\textbf{n}(\textbf{v},\textbf{w})\). By definition, \(\textbf{n}(\textbf{v},\textbf{w})\) is perpendicular to the plane containing w and v, and hence so is \(-\textbf{n}(\textbf{v},\textbf{w})\). Also, v, w, \(\textbf{n}(\textbf{v},\textbf{w})\) form a right-handed system, and so w, v, \(\textbf{n}(\textbf{v},\textbf{w})\) form a left-handed system, and hence w, v, \(-\textbf{n}(\textbf{v},\textbf{w})\) form a right-handed system. Thus, we have shown that \(-\textbf{n}(\textbf{v},\textbf{w})\) is a vector with the same magnitude as \(\textbf{n}(\textbf{w},\textbf{v})\) and is perpendicular to the plane containing w and v, and that w, v, \(-\textbf{n}(\textbf{v},\textbf{w})\) form a right-handed system. So by definition this means that \(-\textbf{n}(\textbf{v},\textbf{w})\) must be \(\textbf{n}(\textbf{w},\textbf{v}). \enskip\checkmark\)

Step 5: Show that \(\textbf{n}(\textbf{v},\textbf{w}) = \Crossprod{\textbf{v}}{\textbf{w}}\) for all vectors v, w.

Write \(\textbf{v} = \vecthreeijk{v}\) and \(\textbf{w} = \vecthreeijk{w}\). Then by Steps 3 and 4, we have

We can use Steps 1 and 2 to evaluate the three terms on the right side of the last equation above:

Similarly, we can calculate

Thus, putting it all together, we have

\(\therefore ~~ \textbf{n}(\textbf{v},\textbf{w}) = \Crossprod{\textbf{v}}{\textbf{w}}\) for all vectors v, w.

So since v, w, \(\textbf{n}(\textbf{v},\textbf{w})\) form a right-handed system, then v, w, \(\Crossprod{\textbf{v}}{\textbf{w}}\) form a right-handed system, which completes the proof.

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