19-August-2008

Distributive Law for Rotations 

Recall (r₁,q₁)·(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¢  = (1,q)· P and  Q¢ = (1,q)Q be rotations of P and Q, respectively, through the angle q .  Then  P¢+Q¢ = (1,q)· P + (1,q)Q can be calculated using rectangular coordinates.  

Figure 2

We would like to show in the following diagram that    

 P¢+Q¢ =  (1,q)(P+Q) 

and hence that 

(1,q)·(P+Q) = P¢+Q¢ = (1,q)·P + (1,q)·Q

or at last that

(1,q)·(P+Q) = (1,q)·P + (1,q)·Q

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¢ and  S¢  = P¢+Q¢ .

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)· 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¢ and  S¢  = P¢+Q¢ .

Step 2:  Show  S' = (1,q)· S.

Proof: Let P =  (r₁,q₁).  Then  P'  = (1,q)· 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)· (R, argument (S))
= (1,q)· S

That completes the proof.

Now recall S' = (1,q)· S is equivalent to writing 

             P'+Q' =  (1,q)· (P+Q)

or 

             (1,q)·P + (1,q)·Q =  (1,q)· (P+Q)

The foregoing shows that rotation (multiplication by (1,q))  distributes over addition.

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