Graphics Reference
In-Depth Information
Definition.
Equation (1.22) is called the
point
-
normal form
of the equation for the
hyperplane defined by (1.20) or (1.21). The vector
n
is called a
normal
vector to the
hyperplane.
1.5.1. Example.
Consider the hyperplane defined by z = 0. This equation can be
rewritten in the form
(
)
+-
(
)
+-
(
)
=
0000100
x
-
y
z
.
Note that (0,0,0) is a point in the hyperplane and (0,0,1) is a normal vector for it.
The next proposition justifies the phrase “plane” in the word “hyperplane.”
1.5.2. Proposition.
(1) A hyperplane
X
in
R
n
is an (n - 1)-dimensional plane. If
X
is defined by the
equation
n
•
p
= d, then any basis for the vector subspace
{
}
n
KpRnp
=Œ
∑=
0
is a basis for
X
.
(2) Conversely, every (n - 1)-dimensional plane in
R
n
is a hyperplane.
Proof.
To prove (1) note first that
K
is a vector subspace. This can be seen either by
a direct proof or by observing that
K
is the kernel of the linear transformation
n
T
:
RR
Æ
defined by
()
=∑.
T
pnp
It follows easily from Theorem B.10.3 that
K
is an (n - 1)-dimensional vector sub-
space of
R
n
. If
p
0
is any point of
X
, then it is easy to show that
{
}
XpqqK
=+ Œ
,
0
proving the first part of the lemma. The converse, part (2), follows from Theorem
1.4.5. Exercise 1.5.2 asks the reader to fill in missing details.
R
3
1.5.3. Example.
To find a basis for the (hyper)plane
X
in
defined by
2x + y - 3z = 6.
Solution.
There will be two vectors
v
1
and
v
2
in our basis. We use Proposition
1.5.2(1). The vector
n
= (2,1,-3) is a normal vector for our plane. Therefore, to find
v
1
and
v
2
is to find a basis for the kernel
K
of the map
pnp
Æ∑.
