178 Answers
7.2. The plane passing through three points x
2
, x
3
, and x
4
is g iven by p
2
∧ p
3
∧ p
4
, and
the equatio n of this plane is p ∧ (p
2
∧ p
3
∧ p
4
) = 0. Hence, point x
1
is o n this line if
and only if p
1
∧ (p
2
∧ p
3
∧p
4
) = 0.
7.3. If we let x = x
0
e
0
+ x
1
e
1
+ x
2
e
2
+ x
3
e
3
and y = y
0
e
0
+ y
1
e
1
+ y
2
e
2
+ y
3
e
3
, their outer
product is
x ∧ y = (x
0
y
1
− x
1
y
0
)e
0
∧e
1
+ (x
0
y
2
− x
2
y
0
)e
0
∧ e
2
+ (x
0
y
3
− x
3
y
0
)e
0
∧ e
3
+(x
2
y
3
− x
3
y
2
)e
2
∧ e
3
+ (x
3
y
1
− x
1
y
3
)e
3
∧ e
1
+ (x
2
y
3
− x
3
y
2
)e
2
∧ e
3
.
Hence, factorizing L is equivalent to finding x
0
, x
1
, x
2
, x
3
, y
0
, y
1
, y
2
, and y
3
such that
m
1
= x
0
y
1
− x
1
y
0
, m
2
= x
0
y
2
− x
2
y
0
, m
3
= x
0
y
3
− x
3
y
0
,
n
1
= x
2
y
3
− x
3
y
2
, n
2
= x
3
y
1
− x
1
y
3
, n
3
= x
2
y
3
− x
3
y
2
,
for given m
1
, m
2
, m
3
, n
1
, n
2
, and n
3
. If we let
m = m
1
e
1
+ m
2
e
2
+ m
3
e
3
, n = n
1
e
1
+ n
2
e
2
+ n
3
e
3
,
x = x
1
e
1
+ x
2
e
2
+ x
3
e
3
, y = y
1
e
1
+ y
2
e
2
+ y
3
e
3
,
this is equivalent to finding x
0
, y
0
, x, and y such that
m = x
0
y − y
0
x, n = x × y,
for given m and n. Evidently, hm, ni = 0 holds if such x
0
, y
0
, x, and y exist. Con-
versely, suppose hm, ni = 0. We ca n choose two vectors x and y that are orthogonal
to n such that n = x × y. Since m is orthogonal to n, we can express m as a linear
combination of such x and y in the form m = αx + βy. Then, we can choose x
0
and
y
0
to be x
0
= −α and y
0
= β. Hence, L is factorized if and only if hm, ni = 0.
7.4. Let n = n
1
e
1
+ n
2
e
2
+ n
3
e
3
. Choose two vectors a and b that are orthogonal to n
such that
n = a × b.
Define vectors x = x
1
e
1
+x
2
e
2
+x
3
e
3
, y = y
1
e
1
+y
2
e
2
+y
3
e
3
, and z = z
1
e
1
+z
2
e
2
+z
3
e
3
as follows:
x =
h
knk
n, y = x + a, z = x + b.
Then, we see that
z ∧ x = b ∧ x, x ∧ y = x ∧ a,
y ∧ z = x ∧ b + a ∧ x + a ∧ b = x ∧ (b − a) + a ∧ b.
Hence, the following holds:
y ∧ z + z ∧ x + y ∧ z = a ∧ b = n
1
e
2
∧ e
3
+ n
2
e
3
∧ e
1
+ n
3
e
1
∧ e
2
.
We also obtain the following:
x ∧ y ∧ z = x ∧ (x + a) ∧ (x + b) = x ∧ a ∧ b = |x, a, b|e
1
∧ e
2
∧e
3
= h
h
knk
n, a × bie
1
∧e
2
∧e
3
= h
h
knk
n, nie
1
∧ e
2
∧e
3
= he
1
∧ e
2
∧ e
3
.
Answers 179
Hence, if we let x = e
0
+ x, y = e
0
+ y, and z = e
0
+ z, we obtain the following
equality:
x ∧ y ∧ z = (e
0
+ x) ∧ (e
0
+ y) ∧ (e
0
+ z)
= e
0
∧ (y ∧ z + z ∧ x + x ∧ y) + x ∧ y ∧z
= n
1
e
0
∧ e
2
∧ e
3
+ n
2
e
0
∧e
3
∧ e
1
+ n
3
e
0
∧ e
1
∧ e
2
+ he
1
∧ e
2
∧ e
3
= Π.
Thus, an arbitrary trivector Π is always factorized.
7.5. (1) The following holds:
L ∧ L
′
= (m
1
e
0
∧ e
1
+ m
2
e
0
∧ e
2
+ m
3
e
0
∧ e
3
+ n
1
e
2
∧ e
3
+ n
2
e
3
∧e
1
+ n
3
e
1
∧ e
2
)
∧(m
′
1
e
0
∧ e
1
+ m
′
2
e
0
∧ e
2
+ m
′
3
e
0
∧ e
3
+ n
′
1
e
2
∧ e
3
+ n
′
2
e
3
∧ e
1
+ n
′
3
e
1
∧e
2
)
= m
1
n
′
1
e
0
∧ e
1
∧ e
2
∧ e
3
+ m
2
n
′
2
e
0
∧ e
2
∧ e
3
∧ e
1
+ m
3
n
′
3
e
0
∧ e
3
∧ e
1
∧ e
2
+n
1
m
′
1
e
2
∧ e
3
∧e
0
∧ e
1
+ n
2
m
′
2
e
3
∧ e
1
∧e
0
∧ e
2
+ n
3
m
′
3
e
1
∧ e
2
∧e
0
∧ e
3
= (m
1
n
′
1
+ m
2
n
′
2
+ m
3
n
′
3
+ n
1
m
′
1
+ n
2
m
′
2
+ n
3
m
′
3
)e
0
∧ e
1
∧ e
2
∧ e
3
.
(2) We can write L as L = p
1
∧ p
2
if the line L is defined by two points p
1
and p
2
.
Similarly, we can write L
′
as L
′
= p
3
∧ p
4
if the line L
′
is defined by two points
p
3
and p
4
. Then, lines L and L
′
are coplanar if and only if s uch four points p
1
,
p
2
, p
3
, and p
4
are coplanar. This condition is written as p
1
∧p
2
∧p
3
∧p
4
= 0, or
L ∧ L
′
= 0. From the above (1), this is equivalent to hm, n
′
i+ hn, m
′
i = 0.
7.6. (1) If we let L = p
2
∪p
3
, Eq. (7.79) is written as Π = p
1
∪L. From Proposition 7.4,
its dual is Π
∗
= p
∗
1
∩ L
∗
. Since L
∗
= p
∗
2
∩ p
∗
3
by Propos itio n 7.5, we can write it
as Π
∗
= p
∗
1
∩ p
∗
2
∩ p
∗
3
. Hence, Eq. (7.81) holds.
(2) If we let L = Π
1
∩Π
2
, Eq. (7.80) is written as p = L ∩Π
3
. From Proposition 7.4,
its dual is p
∗
= L
∗
∪Π
∗
3
. Since L
∗
= Π
∗
1
∪Π
∗
2
by Proposition 7.5, we can write it
as p
∗
= Π
∗
1
∪ Π
∗
2
∪Π
∗
3
. Hence, Eq. (7.82) holds.
Chapter 8
8.1. As shown by Eq. (8.9), all points p satisfy
kpk
2
= x
2
1
+ x
2
2
+ x
2
3
− 2x
∞
= 0,
and x
0
= 1. This means that all points p are embedded in a hypersurface
x
∞
=
1
2
(x
2
1
+ x
2
2
+ x
2
3
),
and in a hyperplane x
0
= 1 in 5D, i.e., in a 3D parabolic surface defined as their
intersection.
8.2. (1) An element x has the norm
kxk
2
= hx
1
e
1
+ x
2
e
2
+ x
3
e
3
+ x
4
e
4
+ x
5
e
5
, x
1
e
1
+ x
2
e
2
+ x
3
e
3
+ x
4
e
4
+ x
5
e
5
i
= x
2
1
+ x
2
2
+ x
2
3
+ x
2
4
− x
2
5
.
Hence, all elements x such that kxk
2
= 0 form a hypercone around axis e
5
given
by
x
5
= ±
q
x
2
1
+ x
2
2
+ x
2
3
+ x
2
4
.
180 Answers
(2) From the definition of e
0
and e
∞
, we can write
e
4
= e
0
−
1
2
e
∞
, e
5
= e
0
+
1
2
e
∞
.
Hence, all elements x in R
4,1
have the following form:
x = x
1
e
1
+ x
2
e
2
+ x
3
e
3
+ x
4
(e
0
−
1
2
e
∞
) + x
5
(e
0
+
1
2
e
∞
)
= (x
4
+ x
5
)e
0
+ x
1
e
1
+ x
2
e
2
+ x
3
e
3
+
1
2
(x
5
− x
4
)e
0
.
If we let
x
0
= x
4
+ x
5
, x
∞
=
1
2
(x
5
− x
4
),
all elements x are written in the form of E q. (8.1). Since x
4
and x
5
are expressed
in terms of x
0
and x
∞
as
x
4
=
1
2
x
0
− x
∞
, x
5
=
1
2
x
0
+ x
∞
,
the norm kxk
2
can be written as follows:
kxk
2
= x
2
1
+ x
2
2
+ x
2
3
+ x
2
4
− x
2
5
= x
2
1
+ x
2
2
+ x
2
3
+
1
2
x
0
− x
∞
2
−
1
2
x
0
+ x
∞
2
= x
2
1
+ x
2
2
+ x
2
3
− 2x
0
x
∞
.
This coincides with the definition of the 5D conformal space.
(3) Since x
0
= 1 is equivalently written as x
4
+ x
5
= 1, the conformal space can be
regarded as embe dding R
3
in the 3D parabolic surface defined as the intersection
of the hypercone
x
5
= ±
q
x
2
1
+ x
2
2
+ x
2
3
+ x
2
4
,
and the hyperplane x
4
+ x
5
= 1 in R
4,1
.
(4) By definition, the geometric products among e
i
, i = 1, 2, 3, are given by
e
2
i
= 1, e
i
e
j
+ e
j
e
i
= 0.
Next, the geometric products of e
i
, i = 1, 2, 3, with e
0
and e
∞
are given by
e
i
e
0
+ e
0
e
i
= e
i
e
4
+ e
5
2
+
e
4
+ e
5
2
e
i
=
1
2
(e
i
e
4
+ e
i
e
5
+ e
4
e
i
+ e
5
e
i
) = 0,
e
i
e
∞
+ e
∞
e
i
= e
i
(e
5
− e
4
) + (e
5
− e
4
)e
i
= e
i
e
5
− e
i
e
4
+ e
5
e
i
− e
4
e
i
= 0.
Finally, the geometric products invo lv ing e
0
and e
∞
are given by
e
2
0
=
e
4
+ e
5
2
2
=
e
2
4
+ e
4
e
5
+ e
5
e
4
+ e
2
5
4
=
1 − 1
4
= 0,
e
2
∞
= (e
5
− e
4
)
2
= e
2
5
− e
5
e
4
− e
4
e
5
+ e
2
4
= −1 + 1 = 0,
e
0
e
∞
+ e
∞
e
0
=
e
4
+ e
5
2
(e
5
− e
4
) + (e
5
− e
4
)
e
4
+ e
5
2
=
1
2
(e
4
e
5
− e
2
4
+ e
2
5
− e
5
e
4
) +
1
2
(e
5
e
4
+ e
2
5
− e
2
4
− e
4
e
5
)
=
1
2
(−1 − 1) +
1
2
(−1 − 1) = −2.
These agree with the rule of Eqs. (8.50) and (8.51).
Answers 181
8.3. (1) Let p = e
0
+ x + kxk
2
e
∞
/2. Recall that the inner products of e
∞
with all other
basis elements are 0 except for he
∞
, e
0
i = −1. From Eq. (8.10), the inner pr oduct
of p and σ is given by
hσ, pi = hc −
r
2
2
e
∞
, pi = hc, pi −
r
2
2
he
∞
, pi
= −
1
2
kc − xk
2
−
r
2
2
he
∞
, e
0
+ x +
kxk
2
2
e
∞
i = −
1
2
kc − xk
2
+
r
2
2
.
Since the tangential distance is
t(p, σ) =
p
kc − xk
2
− r
2
,
we see that hσ, pi = −t(p, σ)
2
/2. Hence, point p is on sphere σ if and only if hσ, pi
= 0.
(2) The inner product of σ and σ
′
is given as follows:
hσ, σ
′
i = hc −
r
2
2
e
∞
, c
′
−
r
′2
2
e
∞
i = hc, c
′
i −
r
2
2
he
∞
, c
′
i −
r
′2
2
hc, e
∞
i
= −
1
2
kc − c
′
k
2
+
r
2
2
+
r
′2
2
.
As we see from Fig. 8.11(b), the angle θ satisfies the law o f cosines
kc − c
′
k
2
= r
2
+ r
′2
− 2rr
′
cos θ.
Hence, the inner product hσ, σ
′
i is written as
hσ, σ
′
i = −
1
2
(r
2
+ r
′2
− 2rr
′
cos θ) +
r
2
+ r
′2
2
= rr
′
cos θ.
It follows that σ and σ
′
are orthogonal if and only if hσ, σ
′
i = 0.
8.4. (1) From Eq. (8.58), we obtain
RT
t
R
−1
= R(1 −
1
2
te
∞
)R
−1
= RR
−1
−
1
2
Rte
∞
R
−1
= 1 −
1
2
RtR
−1
Re
∞
R
−1
= 1 −
1
2
RtR
−1
e
∞
= T
RtR
−1
,
where we have noted that the infinity e
∞
is invariant to rotation around the
origin and that Re
∞
R
−1
= e
∞
holds.
(2) If point p is rotated around the origin, it moves to RpR
−1
. To compute the same
rotation around the positio n t, we first translate p by −t, then rotate it around
the origin, and finally tra slate it by t. This composition is given by
T
t
(R(T
−t
pT
−1
−t
)R
−1
)T
−1
t
= (T
t
RT
−1
t
)p(T
t
RT
−1
t
)
−1
.
(3) We see that the following holds:
T
t
RT
−1
t
= T
t
RT
−t
R
−1
R = T
t
(RT
−t
R
−1
)R = T
t
T
−RtR
−1
R = T
t−RtR
−1
R.
182 Answers
8.5. Inversion of the infinity e
∞
with respect to sphere σ is given by σe
∞
σ
†
. If we let r be
the radius of the s phere σ, Eq. (8.90) implies that σ
†
= −σ
−1
= −σ/r
2
. Hence, the
inversion of e
∞
is given by
σe
∞
σ
†
= −σe
∞
σ
−1
= −
σe
∞
σ
r
2
.
From Eq. (8.98), we can infer that this is equal to 2/r
2
times the center c of the sphere
σ. Hence, the center c is written as
c = −
1
2
σe
∞
σ.
8.6. Equation (8.104) is rewritten as
O =
1
2
(e
0
e
∞
− e
∞
e
0
).
Noting that e
2
0
= e
2
∞
= 0 and e
0
e
∞
+ e
∞
e
0
= −2, we obtain the following:
(1) We see that
O
2
=
e
0
e
∞
− e
∞
e
0
2
2
=
1
4
(e
0
e
∞
+ e
∞
e
0
)
2
− 4e
0
e
∞
e
∞
e
0
− 4e
∞
e
0
e
0
e
∞
=
1
4
(−2)
2
− 4e
0
e
2
∞
e
0
− 4e
∞
e
2
0
e
∞
= 1.
(2) The following hold:
Oe
0
=
1
2
(e
0
e
∞
− e
∞
e
0
)e
0
=
1
2
(−2 − e
∞
e
0
− e
∞
e
0
)e
0
=
1
2
(−2e
0
) = −e
0
,
e
0
O =
1
2
e
0
(e
0
e
∞
− e
∞
e
0
) =
1
2
e
0
(e
0
e
∞
+ 2 + e
0
e
∞
) =
1
2
(2e
0
) = e
0
,
Oe
∞
=
1
2
(e
0
e
∞
− e
∞
e
0
)e
∞
=
1
2
(e
0
e
∞
+ 2 + e
0
e
∞
)e
∞
=
1
2
(2e
∞
) = e
∞
,
e
∞
O =
1
2
e
∞
(e
0
e
∞
−e
∞
e
0
) =
1
2
e
∞
(−2−e
∞
e
0
−e
∞
e
0
) =
1
2
e
∞
(−2e
∞
) = −e
∞
.
From these, we obtain Eq. (8.1 12).
Chapter 9
9.1. (1) Point (X, Y, Z) is on the plane z = Z, which is at distane 1 + Z from the south
pole (0, 0, −1). This plane is magnified by 1/(1 + Z) if projected onto the xy
plane at distance 1 from the south pole. Hence,
x =
X
1 + Z
, y =
Y
1 + Z
.
If follows that
x
2
+ y
2
=
X
2
+ Y
2
(1 + Z)
2
=
1 − Z
2
(1 + Z)
2
=
1 − Z
1 + Z
,
which can be solved in terms of Z in the form
Z =
1 − x
2
− y
2
1 + x
2
+ y
2
=
2
1 + x
2
+ y
2
− 1.
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