# V-P Cycles

Form a sequence of vectors by multiplying each vector (using vector
products) by a constant vector to get the next one in the
seuence(like a GP). What happens?

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 ${\bf b_{n+1}}={\bf a}\times {\bf b_n}$

Show that ${\bf b_n} \rightarrow 0$ as $n\rightarrow \infty$ if the length $|{\bf a}|$ is less than one.

If $|{\bf a}|=1$ and $|{\bf b_1}|=r$ find the directions of the first six vectors in the sequence in relation to the vector ${\bf a}$ and draw a diagram showing these vectors. What happens to the sequence? Describe the surface on which the sequence of vectors from ${\bf b_1}$ onwards lies.

Note: You need to know that the vector product ${\bf a} \times {\bf b}$ is the product of the magnitudes of the vectors times the sine of the angle between the vectors and it is a vector perpendicular to ${\bf a}$ and ${\bf b}$.

This is a bit like a geometric sequence except that the
elements are vectors, the 'common ratio' is vector ${\bf a}$ and
the multiplication is vector products.

In the first part of the question where $|{\bf a}|< 1$ use
the fact that $|\sin \theta | < 1$ to show that the magnitudes
of the vectors in the sequence decrease.

In the second half of the question take ${\bf a}={\bf i}$ and
${\bf b_1} =r{\bf j}$

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:

Image

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

The first part of the question focusses on the magnitude of a vector product and simply requires you to show that the magnitudes of the vectors in the sequence are decreasing. The second part focusses on the fact that the vector arising from a vector product is perpendicular to the two vectors and so the sequence of vectors will cycle through perpendicular directions.