Page 119 - Numerical methods for chemical engineering

A specific example of an orthogonal matrix 105

To answer the first question, we write u [k] = Qe [k] explicitly,
 [k]   [k] [k] [k] 
e Q 11 e + Q 12 e + Q 13 e
 
Q 11 Q 12 Q 13 1 1 2 3
[k]  [k]   [k] [k] [k] 
u =  Q 21 Q 22 Q 23   e  =  Q 21 e + Q 22 e + Q 23 e  (3.4)
 2   1 2 3 
Q 31 Q 32 Q 33 [k] [k] [k] [k]
e 3 Q 31 e 1 + Q 32 e 2 + Q 33 e 3
[m]
Using e = δ mj ,wehave
j
     
Q 11 Q 12 Q 13
[1] [2] [3]
u =  Q 21  u =  Q 22  u =  Q 23  (3.5)
Q 31 Q 32 Q 33
Taking the dot products of each e [ j] with each u [k] yields
Q jk = e [ j] · u [k] (3.6)
If we want instead to rotate the coordinate system without changing the vectors, we apply the
inverse rotation Q −1 . To derive the elements of Q −1 , we have merely to swap e [ j] ⇔ u [ j]
to obtain

Q −1 = u [ j] · e [k] = e [k] · u [ j] = Q kj = Q T (3.7)
jk jk
We thus see that for this “rotation matrix” Q,

Q −1 = Q T (3.8)
As a rotation transforms one orthogonal basis into another, a matrix satisfying (3.8) is said
to be orthogonal. Note that matrices that perform improper rotations, i.e. that combine a
rotation with a mirror reflection, also map orthogonal bases into other ones, and also satisfy
(3.8).

A specific example of an orthogonal matrix

We now consider a specific Q that performs a counter-clockwise rotation (in the 1–2 plane)
by an angle α around e [3] (Figure 3.1),
 
cos α −sin α 0
Q =  sin α cos α 0  (3.9)
0 0 1

3
The effect of Q on each v ∈ is
     
cos α −sin α 0 v 1 v 1 cos α − v 2 sin α
Qv =  sin α cos α 0   v 2  =  v 1 sin α + v 2 cos α  (3.10)
0 0 1 v 3 v 3
In particular, the basis set obtained by rotating the basis (3.2) is
     
cos α −sin α 0
0
u [1] = Qe [1] =  sin α  u [2] = Qe [2] =  cos α  u [3] = Qe [3] =  
0 0 1
(3.11)