Visualizing a cross-product of two vectors

Question

The book (Calculus with Analytic Geometry by Thurman S. Peterson (printed in 1960)) tells me that "the cross product $\bf A \times \bf B$ can be considered as the vector obtained by (a) projecting $\bf B$ on a plane perpendicular to $\bf A$, (b) rotating the projection $90^o$ in the positive direction about $\bf A$, and, (c) multiplying the resulting vector by $\it A$. Each of the operations changes a closed polygon into a closed polygon."

I'm pretty confused because of the way it was explained, making me think that it's just the same as opening a folder with its spine aligned on the edge of the table and its standing cover's (the folder's) edge can be visualized as the vector $\bf A \times \bf B$. Something's wrong with my logic.

What does the book means? Please explain. Thanks! By the way...illustrations (whether hand-written drawings or drawings aided by software are welcomed. They're better too!)

Answer 1

Let $A$ be a vertical flagpole, and let $B$ be a stick coming out of the ground at the same point where the flagpole is standing. Suppose $B$ is situated so that it makes a $45^\circ$ angle (or whatever angle you like) with the ground, and it is sticking out to the East. The base of the flagpole is the origin of our coordinates.

When the book says, "project $B$ onto a plane perpendicular to $A$," they're saying, "The sun is directly overhead. Look at the shadow cast by $B$ on the ground." Mark the shadow with a rope, running from the origin to the tip of the shadow. You could tie it around the flagpole base, for example.

Now, take that rope, and rotate it in the positive direction (counterclockwise) about the flagpole, keeping it on the ground. In other words, rotate it from pointing due East, to pointing due North. Note that the rope, after rotation, is perpendicular to both $A$ and $B$.

Finally, if $A$ isn't of unit length, multiply the length of the North-pointing rope by the length of $A$. Now it represents $A\times B$.

Does that help?

Answer 2

$\newcommand{\AA}{\color{blue}{\bf A}}$ $\newcommand{\BB}{\color{orange}{\bf B}}$ $\newcommand{\CC}{\color{magenta}{\bf C}}$

I will color-code the vectors: $\AA$, $\BB$ and $\CC = \AA\times\BB$, and assume that $|\AA| = 1$, note that this only sets a scale for the sketch below, so is really not that important.

The trick here is that the vector $\CC$ must be perpendicular to both $\AA$ and $\BB$, so let's make sure first that it is perpendicular to $\AA$. To do this define the plane containing all the vectors perpendicular to $\AA$ (green plane), the vector $\CC$ must be one of them!

The question is, where on the plane do we place it? Well, we already stated that it also must be perpendicular to $\BB$, so let's drop the shadow of $\BB$ onto the green plane. To recap, both $\CC$ and the shadow of $\BB$ are on the plane and, therefore, are perpendicular to $\AA$.

The only thing left is to ensure that $\CC$ is perpendicular to $\BB$, well just place $\CC$ $90^\circ$ apart from the shadow of $\BB$ and that's it