ME740 Homework Set 1
Dingjiang Zhou
ENG ME 740 Homework Solution Set 1
Solutions by
Dingjiang Zhou
EXERCISE 1.1
As in figure 1, we can find the geometrical relation between [xi , yi ]T and [xe , ye ]T .
First, 6 OAQ is the right angle, such that OA = xi . In the same way, we have OI = yi ,
OD = xe and OE = ye .
Second, since 6 DOA = θ, from geometry, we have, 6 DQC = 6 IQF = 6 F OE = 6 GDO = θ.
Then
xi = OA = OB − AB = OB − CD = OD · cosθ − QD · sinθ = xe · cosθ − ye · sinθ
similarly
yi = OI = OG + GI = OG + QC = OD · sinθ + QD · cosθ = xe · sinθ + ye · cosθ.
Combine these two equation into
amatrix
form, we get
xi
cos θ − sin θ xe
=
.
Q.E.D.
yi
sin θ
cos θ
ye
EXERCISE 1.2
As in the first part, I have verified that the net composite of rigid motion in a plane can be
represented as the matrix multiplication, that is
for two steps of rigid motion
cos θ1 − sin θ1
x
T1 :
, 1
sin θ1
cos θ1
y1
and
cos θ2 − sin θ2
x
T1 :
, 2
sin θ2
cos θ2
y2
can be represented as


cos θ1 − sin θ1 x1
cos θ1
y1 
A1 =  sin θ1
0
0
1
and


cos θ2 − sin θ2 x2
cos θ2
y2 .
A2 =  sin θ2
0
0
1
And, the net composite motion can be represented definitely the same with the matrix multiplication, that is
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ME740 Homework Set 1
Dingjiang Zhou
T2 ◦ T1 = A2 · A1 .
Then, in this exercise, since
T1 ◦ (T2 ◦ T3 ) = A1 · (A2 · A3 )
and
(T1 ◦ T2 ) ◦ T3 = (A1 · A2 ) · A3
and from the associative law of matrix multiplication, that
A1 · (A2 · A3 ) = (A1 · A2 ) · A3
thus, we have
(T1 ◦ T2 ) ◦ T3 = (T1 ◦ T2 ) ◦ T3 .
A direct calculation should be like this
suppose
T1 : R1 , ~r1 ;
T2 : R2 , ~r2 ; T3 : R3 , ~r3 .
Then
R1 ~r1
R2 ~r2
R3 ~r3
T1 ◦ (T2 ◦ T3 ) =
·
·
0
1
0
1
0
1
R1 ~r1
R2 R3 R2~r3 + ~r2
=
·
0
1
0
1
R1 R2 R3 R1 (R2~r3 + ~r2 ) + ~r1
=
0
1
R1 R2 R3 R1 R2~r3 + R1~r2 + ~r1
=
0
1
and
~r1
R2
·
1
0
~r2
R3 ~r3
·
1
0
1
R1 R2 R1~r2 + ~r1
R3 ~r3
=
·
0
1
0
1
R1 R2 R3 R1 R2~r3 + R1~r2 + ~r1
=
0
1
(T1 ◦ T2 ) ◦ T3 =
R1
0
.
Which implys directly that
(T1 ◦ T2 ) ◦ T3 = (T1 ◦ T2 ) ◦ T3
EXERCISE 1.3
As in Exercise 1.2, we suppose two rigid motion
T1 : R1 , ~r1 ; T2 : R2 , ~r2 .
Then, we can calculate that R1 ~r1
R2 ~r2
R1 R2 R1~r2 + ~r1
T1 ◦ T2 =
·
=
0
1
0
1
0
1
and
R2 ~r2
R1 ~r1
R2 R1 R2~r1 + ~r2
T2 ◦ T1 =
·
=
.
0
1
0
1
0
1
From matrix multiplication, because R1 and R2 are orthogonal matrices, we have
R1 · R2 = R2 · R1
but
R1~r2 + ~r1 6= R2~r1 + ~r2
except in speical cases, thus
R1 ◦ R2 6= R2 ◦ R1
Hence, the group of rigid planar motions is not abelian.
EXERCISE 1.4
Step I, the vector AP~ rotate about ẐA by θ degrees, the rotation matrix is
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ME740 Homework Set 1
Dingjiang Zhou


cos θ − sin θ 0
cos θ 0.
A1 =  sin θ
0
0
1
Step II, it is subsequently rotated about
X̂
by
φ
degrees,
A

 the rotation matrix is
1
0
0
A2 = 0 cos φ − sin φ.
0 sin φ cos φ
Thus, the rotation matrix whichwill accopmlish these
given order would be
  rotations in the 
1
0
0
cos θ − sin θ 0
cos θ 0.
A = A2 · A1 = 0 cos φ − sin φ  sin θ
0 sin φ cos φ
0
0
1
EXERCISE 1.5
Step I, the vector AP~ rotate about ŶA

cos(30◦ )
0
A1 = 
− sin(30◦ )
by 30 degrees, the
rotation matrix
 is
  √3
1
0
0 sin(30◦ )
2
2 
=
1
0
 0 1 √0 .
3
0 cos(30◦ )
−1 0
2
2
Step II, it is subsequently rotated about X̂A by 45 degrees,
the rotation
 matrix is

 
1 √0
0√
1
0
0


A2 = 0 cos(45◦ ) − sin(45◦ ) = 0 √22 −√ 22 .
◦
◦
2
0 sin(45 ) cos(45 )
0 22
2
Thus, the rotation matrixwhich will accopmlish
these rotations
√
  √ in the given order
 would be
3
1
3
1
1 √0
0√
0
0
2√
√
2 

 2
 √2
2
6 .
A = A2 · A1 = 0 √22 −√ 22   0 1 √0  =  42

−
2
√
√
√4
2
2
3
1
2
2
6
0 2
−2 0 2
− 4
2
2
4
EXERCISE 1.6
This is a problem about the coordinate transformation.
The frame B is initially coincident with the frame A.
The frame BI is obtained by rotating frame B about ẐB by θ degrees, which can be represented
by the 3 × 3 matrix


cos θ − sin θ 0
cos θ 0.
A1 =  sin θ
0
0
1
The frame BII is obtained by rotating frame BI about X̂BI (in the problem, it is ”about
X̂B ”) by φ degrees, which can be represented
by the 3 × 3matrix

1
0
0
A2 = 0 cos φ − sin φ.
0 sin φ cos φ
Then, we have the coordinates transformation equation
A~
P = A1 · A2 ·B P~ .
Then, the rotation matrix which will change the description of vector from BP~ to AP~ should
be

 
 

cos θ − sin θ 0
1
0
0
cos θ − cos φ sin θ
sin φ sin θ
cos θ 0 · 0 cos φ − sin φ =  sin θ
cos φ cos θ − cos θ sin φ.
A = A1 · A2 =  sin θ
0
0
1
0 sin φ cos φ
0
sin φ
cos φ
EXERCISE 1.7
Set θ and φ in exercise 1.6 by 30◦ and 45◦ respectively, then the rotation matrix which will
change the description of vector from BP~ to AP~ should be
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ME740 Homework Set 1
Dingjiang Zhou


cos θ − cos φ sin θ
sin φ sin θ
cos φ cos θ − cos θ sin φ
A =  sin θ
0
sin φ
cos φ


◦
◦
cos(30 ) − cos(45 ) sin(30◦ )
sin(45◦ ) sin(30◦ )
=  sin(30◦ ) cos(45◦ ) cos(30◦ ) − cos(30◦ ) sin(45◦ )
0
sin(45◦ )
cos(45◦ )
√
√
√ 
3
2
−√ 42
4√

 21
6
= 2
−√ 46  .
√4
2
2
0
2
2
EXERCISE 1.8
The unit eigenvector ~v associated with the eigenvalue 1, defines a direction that
is invariant under multiplication by A
B R. Proper spatial rotation matrices always
have at least one eigenvalue equal to 1, and this defines the axis of rotation.
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