Rotation About an Arbitrary Axis - Kennesaw State University

Chapter 9

Rotation About an

Arbitrary Axis

9.1

Quick Review

Given a point P = (x; y; z; 1) in homogeneous coordinates, let P 0 = (x0 ; y 0 ; z 0 ; 1)

be the corresponding point after a rotation around one of the coordinate axis has

been applied. You will recall the following from our studies of transformations:

1. Rotation about the x-axis by an angle x , counterclockwise (looking along

the x-axis towards the origin). Then P 0 = Rx P where the rotation matrix,

Rx ,is given by:

3

2

1

0

0

0

6 0 cos x

sin x 0 7

7

Rx = 6

4 0 sin x

cos x 0 5

0

0

0

1

2. Rotation about the y-axis by an angle y , counterclockwise (looking along

the y-axis towards the origin). Then P 0 = Ry P where the rotation matrix,

Ry ,is given by:

2

3

cos y 0 sin y 0

6

0

1

0

0 7

7

Ry = 6

4 sin y 0 cos y 0 5

0

0

0

1

3. Rotation about the z-axis by an angle z , counterclockwise (looking along

the z-axis towards the origin). Then P 0 = Rz P where the rotation matrix,

Rz ,is given by:

2

3

cos z

sin z 0 0

6 sin z

cos z 0 0 7

7

Rz = 6

4 0

0

1 0 5

0

0

0 1

117

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CHAPTER 9. ROTATION ABOUT AN ARBITRARY AXIS

Note that the above transformations also apply to vectors.

You will also recall that

Rx 1

= RxT

Ry 1

= RyT

Rz 1

= RzT

This means in particular that these matrices are orthogonal. It can also be

proven that the product of two orthogonal matrices is itself an orthogonal matrix

(see problems at the end of the chapter). So, if we combine several rotations

about the coordinate axis, the matrix of the resulting transformation is itself an

orthogonal matrix.

One way of implementing a rotation about an arbitrary axis through the

origin is to combine rotations about the z, y, and x axes. The matrix of the

resulting transformation, Rxyz , is

2

3

Cy Cz

Cy Sz

Sy

Sx Sy Sz + Cx Cz

Sx Cy 5 (9.1)

Rxyz = Rx Ry Rz = 4 Sx Sy Cz + Cx Sz

Cx Sy Cz + Sx Sz Cx Sy Sz + Sx Cz

Cx Cy

where

Ci = cos

i

and Si = sin

i

for i = x; y; z

From what we noticed above, Rxyz is an orthogonal matrix. This means that

its inverse is its transpose.

9.2

Rotation About an Arbitrary Axis Through

the Origin

Goal: Rotate a vector v = (x; y; z) about a general axis with direction vector rb

(assume rb is a unit vector, if not, normalize it) by an angle (see ¡­gure 9.1).

Because it is clear we are talking about vectors, and vectors only, we will omit

the arrow used with vector notation.

We begin by decomposing v into two components: one parallel to rb and one

perpendicular to rb. Let us denote vk the component parallel to rb and v? the

component perpendicular to rb. You will recall from our study of vectors that

vk

Similarly,

= comprbv

v rb

b

=

2r

kb

rk

= (v rb) rb since rb is a unit vector

v?

=

=

orthrbv

v (v rb) rb

9.2. ROTATION ABOUT AN ARBITRARY AXIS THROUGH THE ORIGIN119

And we have

v = v k + v?

Figure 9.1: Rotation about a general axis through the origin, showing the axis

of rotation and the plane of rotation (see [VB1])

Let T denote the rotation we are studying. We need to compute T (v).

T (v)

= T vk + v?

= T vk + T (v? )

since T is a linear transformation. Also,

T vk = vk

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CHAPTER 9. ROTATION ABOUT AN ARBITRARY AXIS

Figure 9.2: Rotation about a general axis through the origin, showing vectors

on the plane of rotation (see [VB1])

9.2. ROTATION ABOUT AN ARBITRARY AXIS THROUGH THE ORIGIN121

since vk has the same direction as rb and we are rotating around an axis with

direction vector rb(see ¡­gure 9.1). Therefore,

T (v) = vk + T (v? )

So, T (v? ) is the only quantity we need to compute. For this, we create a two

dimensional basis in the plane of rotation (see ¡­gure 9.1 and 9.2). We will use

v? as our ¡­rst basis vector. For our second, we can use

w

=

=

Looking at ¡­gure 9.2, we see that

T (v? )

=

rb

rb

(9.2)

v?

v

cos v? + sin w

=

cos v? + sin (b

r

v)

and therefore

T (v)

= vk + T (v? )

=

(v rb) rb + cos v? + sin (b

r

=

(1

=

=

v)

(v rb) rb + cos [v (v rb) rb] + sin (b

r

v)

(v rb) rb + cos v cos (v rb) rb + sin (b

r

cos ) (v rb) rb + cos v + sin (b

r

v)

v)

We would like to express this as a matrix transformation, in other words, we

want to ¡­nd the matrix R such that T (v) = Rv. For this, we ¡­rst establish

some intermediary results.

Lemma 97 If u = (ux ; uy ; uz ) and v = (vx ; vy ; vz ) then

2

3

0

uz uy

0

ux 5 v

u v = 4 uz

uy ux

0

Proof.

u

v

=

2

i

ux

vx

j

uy

vy

uy v z

= 4 u z vx

ux vy

2

0

= 4 uz

uy

k

uz

vz

3

uz vy

u x vz 5

uy v x

uz

0

ux

3

uy

ux 5 v

0

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