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 ×b ¹ 0. Define a sequence of vectors b₀, b₁,  b₂¼ by b₀=b and b_n+1= a×b_n

Part 1: Show that b_n ® 0 as n ® ¥ 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 |v₁||v₂| sinq. In the problem posed, we are given that |a| is less than 1. The maximum value of sinq is 1, so the product |a|sinq must be less than 1; the magnitudes of the succeeding vectors in the sequence given by b_n+1= a×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 |b₁|=r. The supplied hint suggests using a = i and b₁ = rj. 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 b₁, we obtain b₂ = r k. Performing the next cross product, a × b₂, we obtain -rj, and, doing it again, -rk. The cross product a×b₄ produces a result of rj which is b₁; therefore, we have a cycle.

If we begin our movement from the origin in 3-space, b₁ tells us to advance r units up the y-axis j direction). Then b₂ says to advance r units positively along the z-axis. Vectors b₃ and b₄, 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:

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