Derek Wan gave an excellent solution and we include his diagram of the final result at the end. Here's a solution from Thomas Lauffenberger who finally reveals that he is an American.

Suppose the vector product $a \times b \neq 0$ . Define a sequence of vectors $b_{0}, b_{1}, b_{2} \dots$ by $b_{0} = b$ and $b_{n + 1} = a \times b_{n}$

Part 1: Show that $b_{n} \to 0$ as $n \to \infty$ if $| a | < 1$ .

According to the article supplied on the basics of vector multiplication, the resulting vector is perpendicular to its parents and has a magnitude of $| v_{1} | | v_{2} | \sin θ$ . In the problem posed, we are given that $| a |$ is less than 1. The maximum value of $\sin θ$ is 1, so the product $| a | \sin θ$ must be less than 1; the magnitudes of the succeeding vectors in the sequence given by $b_{n + 1} = a \times b_{n}$ decrease as a geometric series, so they will tend to 0 as $n$ tends to infinity. The only vector with a magnitude off 0 is the zero vector, which $b_{n}$ will tend to as $n$ tends to infinity.

Part 2: Here $| a | = 1$ and $| b_{1} | = r$ . The supplied hint suggests using $a = i$ and $b_{1} = r j$ . Doing the cross product matrix math (a 3x3 matrix with " $i, j, k$ " on the top line, "1,0,0" for the $a$ line, and "0,r,0" for $b_{1}$ , we obtain $b_{2} = r k$ . Performing the next cross product, $a \times b_{2}$ , we obtain $- r j$ , and, doing it again, $- r k$ . The cross product $a \times b_{4}$ produces a result of $r j$ which is $b_{1}$ ; therefore, we have a cycle.

If we begin our movement from the origin in 3-space, $b_{1}$ tells us to advance r units up the y-axis $j$ direction). Then $b_{2}$ says to advance r units positively along the z-axis. Vectors $b_{3}$ and $b_{4}$ , respectively, move again along the y and z axes, but now r units in the negative direction. The shape that follows is a square with sides of r units, located within the yz-plane in this instance. I'm sure that much more can be said about this, regarding our choices for $a$ and $b$ ; I'll do more work on it and post back more ideas.

Addendum to previous submission== The key components determining the location of the square are $a$ , a unit vector, and $b$ , a vector of magnitude r. Vector $a$ determines the plane in which the square will situate itself which has vector $a$ perpendicular to the plane of the square; $b$ moves within that plane. With the direction vector $a$ pointed up relative to our perspective, the motion of the vector $b$ is to trace the square in a counterclockwise direction, like a baseball diamond. (Yes, I'm an American!)

Thoms is adding one vector onto the previous one to generate a square. Here is Derek Wan's sketch of all 6 vectors: Vector sequence

We can see that the vectors lie on the y-z plane perpendicular to the vector $a$ .