226 CHAPTER 8. LINEAR TRANSFORMATIONS

Of course the matrix is an orthogonal matrix so it is easy to take the inverse by simplytaking the transpose. Then doing the computation and then some simplification yields

=

 a2 +(1− a2

)cos θ ab (1− cos θ)− c sin θ ac (1− cos θ) + b sin θ

ab (1− cos θ) + c sin θ b2 +(1− b2

)cos θ bc (1− cos θ)− a sin θ

ac (1− cos θ)− b sin θ bc (1− cos θ) + a sin θ c2 +(1− c2

)cos θ

 . (8.4)

With this, it is clear how to rotate clockwise about the unit vector, (a, b, c) . Just rotatecounter clockwise through an angle of −θ. Thus the matrix for this clockwise rotation is just

=

 a2 +(1− a2

)cos θ ab (1− cos θ) + c sin θ ac (1− cos θ)− b sin θ

ab (1− cos θ)− c sin θ b2 +(1− b2

)cos θ bc (1− cos θ) + a sin θ

ac (1− cos θ) + b sin θ bc (1− cos θ)− a sin θ c2 +(1− c2

)cos θ

 .

In deriving 8.4 it was assumed that c ̸= ±1 but even in this case, it gives the correctanswer. Suppose for example that c = 1 so you are rotating in the counter clockwisedirection about the positive z axis. Then a, b are both equal to zero and 8.4 reduces to 2.24.

8.3.2 The Euler Angles

An important application of the above theory is to the Euler angles, important in themechanics of rotating bodies. Lagrange studied these things back in the 1700’s. To describethe Euler angles consider the following picture in which x1, x2 and x3 are the usual coordinateaxes fixed in space and the axes labeled with a superscript denote other coordinate axes.Here is the picture.

ϕ

ϕ

x3 = x13

x1x11

x2

x12

θ

θ

x13x23

x11 = x21

x12

x22

ψ

ψ

x23 = x33

x21x31

x22

x32

We obtain ϕ by rotating counter clockwise about the fixed x3 axis. Thus this rotationhas the matrix  cosϕ − sinϕ 0

sinϕ cosϕ 0

0 0 1

 ≡M1 (ϕ)

Next rotate counter clockwise about the x11 axis which results from the first rotation throughan angle of θ. Thus it is desired to rotate counter clockwise through an angle θ about theunit vector  cosϕ − sinϕ 0

sinϕ cosϕ 0

0 0 1

 1

0

0

 =

 cosϕ

sinϕ

0

 .

226 CHAPTER 8. LINEAR TRANSFORMATIONSOf course the matrix is an orthogonal matrix so it is easy to take the inverse by simplytaking the transpose. Then doing the computation and then some simplification yieldsa? +(1—a?)cos@ ab(1—cos@) —csin@ ac(1—cos6) + bsin#=| ab(1—cos#)+csin@ b?+(1—6?)cos@ —be(1—cos@)—asind |. (8.4)ac(1—cos@)—bsin@ bc(1—cos@)+asin@ c? + (1—c*) cosWith this, it is clear how to rotate clockwise about the unit vector, (a,b,c). Just rotatecounter clockwise through an angle of —6. Thus the matrix for this clockwise rotation is justa? +(1—a?)cos@ ab(1—cos@)+csin@ ac(1—cos@) — bsin#d=| ab(1—cosé)—csind b?+4+(1—b?)cos@ — be(1 — cos@) + asinac(1—cos@)+bsin@ bc(1—cos@)—asin@ ce? + (1—c?) cosIn deriving 8.4 it was assumed that c 4 +1 but even in this case, it gives the correctanswer. Suppose for example that c = 1 so you are rotating in the counter clockwisedirection about the positive z axis. Then a,b are both equal to zero and 8.4 reduces to 2.24.8.3.2 The Euler AnglesAn important application of the above theory is to the Euler angles, important in themechanics of rotating bodies. Lagrange studied these things back in the 1700’s. To describethe Euler angles consider the following picture in which x1, x2 and x3 are the usual coordinateaxes fixed in space and the axes labeled with a superscript denote other coordinate axes.Here is the picture.v321 v3 2 3L3 = 23 6 Lz = £3v311 am) 3v Tvv2 x36 rf = 0} i1 1 9Ly Lyvy vyWe obtain ¢ by rotating counter clockwise about the fixed x3 axis. Thus this rotationhas the matrixcos@ —sind 0sind cosdé 0 | =M,(¢)0 0 1Next rotate counter clockwise about the x} axis which results from the first rotation throughan angle of 6. Thus it is desired to rotate counter clockwise through an angle @ about theunit vectorcos@ —singd 0 1 cos @sing cosd 0 0 |= sing0 0 1 0 0