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# V-P Cycles

##### Age 16 to 18 Challenge Level:

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 ${\bf a} \times {\bf b}\neq {\bf 0}$. Define a sequence of vectors ${\bf b_0},\ {\bf b_1},\ {\bf b_2}\ldots$ by ${\bf b_0}={\bf b}$ and ${\bf b_{n+1}}={\bf a}\times {\bf b_n}$

#### Part 1:

Show that ${\bf b_n} \rightarrow 0$ as $n \rightarrow \infty$ if ${\bf |a|} < 1$. \par According to the article supplied on the basics of vector multiplication, the resulting vector is perpendicular to its parents and has a magnitude of ${\bf |v_1||v_2|} \sin \theta$. In the problem posed, we are given that ${\bf |a|}$ is less than 1. The maximum value of $\sin \theta$ is 1, so the product ${\bf |a|}\sin \theta$ must be less than 1; the magnitudes of the succeeding vectors in the sequence given by ${\bf b_{n+1}}={\bf a}\times {\bf b_n}$ 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 ${\bf b_n}$ will tend to as $n$ tends to infinity.

#### Part 2:

Here $|{\bf a}|=1$ and $|{\bf b_1}|=r$. The supplied hint suggests using ${\bf a} = {\bf i}$ and ${\bf b_1}= r{\bf j}$. Doing the cross product matrix math (a $3 \times 3$ matrix with "${\bf i}, {\bf j}, {\bf k}$" on the top line, "$1,0,0$" for the ${\bf a}$ line, and "$0,r,0$" for ${\bf b_1}$, we obtain ${\bf b_2}= r{\bf k}$. Performing the next cross product, ${\bf a} \times {\bf b_2}$, we obtain $-r{\bf j}$, and, doing it again, $-r{\bf k}$. The cross product ${\bf a}\times {\bf b_4}$ produces a result of $r{\bf j}$ which is ${\bf b_1}$; therefore, we have a cycle.

If we begin our movement from the origin in 3-space, ${\bf b_1}$ tells us to advance $r$ units up the $y$-axis ${\bf j}$ direction). Then ${\bf b_2}$ says to advance $r$ units positively along the $z$-axis. Vectors ${\bf b_3}$ and ${\bf b_4}$, 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 ${\bf a}$ and ${\bf b}$; I'll do more work on it and post back more ideas. \par Addendum to previous submission- The key components determining the location of the square are ${\bf a}$, a unit vector, and $\bf {b}$, a vector of magnitude $r$. Vector ${\bf a}$ determines the plane in which the square will situate itself which has vector ${\bf a}$ perpendicular to the plane of the square; ${\bf b}$ moves within that plane. With the direction vector ${\bf a}$ pointed up relative to our perspective, the motion of the vector ${\bf b}$ is to trace the square in a counterclockwise direction, like a baseball diamond. (Yes, I'm an American!)

Thomas 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 ${\bf a}$.