19-August-2008
Distributive Law for Rotations
The more recent
account may be easier to follow.
Recall (r1,q1)路(r2,q2) =
(r1r2,q1+q2)
We now consider multiplications by points
(r1,q1) =
(1,q). This corresponds to a rotation.
A parallelogram corresponding to the map addition of the points P
= (a,b) and Q = (c,d) is indicated
below. Here 0 = (0,0) indicates the origin.
Figure 1.
Let P垄 = (1,q)路P and Q垄 =
(1,q)Q be rotations of P and Q, respectively,
through the angle q . Then P垄+Q垄 = (1,q)路 P + (1,q)Q can be
calculated using rectangular coordinates.
Figure 2
We would like to show in the following diagram that
P垄+Q垄
= (1,q)(P+Q)
and hence that
(1,q)路(P+Q) =
P垄+Q垄
= (1,q)路P + (1,q)路Q
or at last that
(1,q)路(P+Q) = (1,q)路P + (1,q)路Q
The latter says that multiplication by the factor
(1,q) distributes over the addition of points.
Step 1: Isometry of two triangles:
Claim: Triangle OPS with vertices O, P and S = P+Q is isometric to
the triangle OP'S' with vertices O, P垄
and S垄 = P垄+Q垄 .
Proof: See Figure 3. The distance of point P' to origin is the
same as that of P to the origin as it the image of P under a rotation
through angle q. Thus the side OP' has the
same length as OP.
Figure 3.
Likewise, the distance of point Q' to origin is the same as that of Q
to the origin as it the image of Q under a rotation through angle
q. Thus the side OQ' has the same length as
OQ.
Recall the addition of points forms parallelograms. It follows that
OQSP and OQ'S'P' are quadrilaterals with many pairs of opposite sides
equal in length as indicated in the above diagram.
Let P and Q have polar coordinates (r1,q1) and (r2,q2) respectively. Then P' = (1,q)路 P and Q' =
(1,q)Q have polar coordinates
(r1, q + q1) and (r2,q + q2). Thus the
angle QOP and angle Q'OP' both equal q2 - q2. The side angle side criteria now implies
triangles OPQ and angle OP'Q' are isometric.
Figure 3 (duplicate)
Now
- triangle S'Q'P' is isometric to OP'Q' by side-side-side isometry
criteria.
- triangles OP'Q' and triangle OPQ are isometric from above.
- triangle SQP is isometric to OPQ by side-side-side isometry
criteria.
Thus all the triangles with side QP or Q'P' are isometric.
Equality of angle measures follows as indicated in Figure 4.
Figure 4. Equality of Angle Measures.
The SAS criteria now impliesthe triangle OPS with vertices O, P and S =
P+Q is isometric to the triangle OP'S' with vertices O,
P垄 and S垄
= P垄+Q垄 .
Step 2: Show S' = (1,q)路S.
Proof: Let P = (r1,q1). Then P' =
(1,q)路 P = (
r1,q +q1)
The triangle OPS is isometric to the triangle OP'S'. Therefore OS and
OS' have a common length R.
Let t = angle SOP. Then isometry implies
t = angle S'OP'. Now let argument (P) = the
polar coordinate angle of P. The above diagram suggests
argument (S')
= argument (P') + t
= (q + q1) + t
= q + (q1+ t)
= q + (argument (P) + t )
== q + argument (S)
Hence
S'= (R, argument(S'))
= (R, q + argument (S))
= (1,q)路 (R, argument (S))
= (1,q)路 S
That completes the proof.
Now recall S' = (1,q)路 S is equivalent to
writing
P'+Q' = (1,q)路 (P+Q)
or
(1,q)路P + (1,q)路Q =
(1,q)路(P+Q)
The foregoing shows that rotation (multiplication by (1,q)) distributes over addition.
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