Graphics Reference
In-Depth Information
Let's choose another matrix and repeat the above:
⎡
⎤
1
0
0
⎣
⎦
.
R
α,x
=
0
cos
α
−
sin
α
0 in
α
cos
α
This time, let
α
=
45°:
⎡
⎤
1
0
0
√
2
/
2
−
√
2
/
2
⎣
⎦
R
45
°
,x
=
0
√
2
/
2
√
2
/
2
0
√
2
Tr
(
R
45
°
,x
)
=
1
+
therefore,
√
2
45°
.
So we now have a mechanism to extract the axis and angle of rotation from a
rotation matrix. However, the algorithm for identifying the axis is far from satisfac-
tory, and later on we will discover that there is a similar technique which is readily
programable.
For completeness, let's identify the axis and angle of rotation for the matrix (
9.6
):
α
=
arccos
((
1
+
−
1
)/
2
)
=
⎡
⎤
001
0
⎣
⎦
.
R
90
°
,x
R
90
°
,y
R
90
°
,z
=
10
100
−
Once more, we begin by writing the characteristic equation for the matrix:
0
−
λ
0
1
0
−
1
−
λ
0
=
0
.
(9.8)
1
0
0
−
λ
Expanding (
9.8
) using the top row we have
λ
+
1
=
−
1
−
λ
0
0
−
1
−
λ
−
0
0
−
λ
1
0
−
λ
−
λ
+
λ
2
+
1
−
λ
=
0
λ
2
λ
3
−
+
1
−
λ
=
0
λ
3
λ
2
−
+
−
λ
+
1
=
0
λ
3
λ
2
−
+
λ
=
1
.
Again, there is a single real root:
λ
1, and substituting this in the original equations
associated with (
9.8
) to reveal the eigenvector, we have
⎧
⎨
⎩
=
−
x
+
0
y
+
z
=
0
0
x
−
2
y
+
0
z
=
0
0
.
It is obvious from the 1st and 3rd equations that
x
=
z
, and from the 2nd equation
that
y
x
+
0
y
−
z
=
=
0, which implies that the associated eigenvector is of the form
[
k
0
k
]
,
which is correct.
