Graphics Reference
In-Depth Information
1
1
+
2
(
e
12
−
e
31
−
e
23
+
1
)
=
2
(
1
1
+
2
)
1
1
1
1
2
1
+
2
e
12
−
2
e
23
−
2
e
31
+
=
√
3
√
3
2
+
√
3
6
√
3
6
√
3
6
=
e
12
−
e
23
−
e
31
which makes
√
3
2
−
√
3
6
√
3
6
√
3
6
R
†
120
°
=
e
12
+
e
23
+
e
31
.
But does it work? Well, let's find out by forming the product
1
√
2
(
e
2
+
1
12
√
2
(
3
e
3
)
R
†
R
120
°
120
°
=
+
e
12
+
e
32
+
e
13
)
×
(
e
2
+
e
3
)(
3
−
e
12
−
e
32
−
e
13
)
1
6
√
2
(
e
1
+
=
e
2
+
2
e
3
)(
3
−
e
12
−
e
32
−
e
13
)
1
6
√
2
(
6
e
1
+
=
6
e
3
)
1
√
2
(
e
1
+
=
e
3
)
which is correct.
Finally, let's employ the rotor matrix. But remember that its definition of
R
θ
has
a negative bivector term, which means that we have to switch the bivector terms in
R
120
°
or use the bivector terms from
R
†
120
°
:
√
3
2
−
√
3
6
√
3
6
√
3
6
R
†
120
°
=
e
12
+
e
23
+
e
31
where
√
3
2
√
3
6
√
3
6
√
3
6
s
=
,x
=
,y
=
,z
=−
⎡
⎤
⎡
⎤
2
(y
2
z
2
)
1
−
+
2
(xy
−
sz)
2
(xz
+
sy)
v
1
v
2
v
3
R
120
°
vR
†
⎣
⎦
⎣
⎦
2
(x
2
z
2
)
120
°
=
2
(xy
+
sz)
1
−
+
2
(yz
−
sx)
2
(x
2
y
2
)
2
(xz
−
sy)
2
(yz
+
sx)
1
−
+
⎡
⎣
⎤
⎦
⎡
⎤
2
(
12
+
1
2
(
12
+
3
1
3
−
−
12
+
1
12
)
12
)
2
(
12
)
v
1
v
2
v
3
2
(
12
−
3
2
(
12
+
1
1
3
⎣
⎦
=
12
)
1
−
12
)
2
(
−
12
−
12
)
1
3
1
3
2
(
12
+
1
2
(
−
12
−
12
)
2
(
−
12
+
12
)
1
−
12
)
⎡
⎤
⎡
⎤
2
3
2
3
1
3
v
1
v
2
v
3
⎣
⎦
⎣
⎦
.
1
3
2
3
2
3
=
−
−
2
3
1
3
2
3
−
