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University of Waterloo
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Mathematics
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MATH 136
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Robert Andre
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Lecture

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3
Friday, January 17 ▯ Lecture 6 : Planes in ▯
Concepts:
1. plane n ▯(x ▯ a) = 0 in ▯ where n is a norm to the plane.
n
2. hyperplane in ▯ .
3. scalar equation of a plane n x + 1 1 + n 2 2 d =3 3a + n a +1 1a . 2 2 3 3
4. vector equation x = sv + tw + a, of a plane
5. finding the vector and standard (scalar) equation of a plane.
6. orthogonal line to a plane
In this lecture we discuss two equations containing vectors which represent a plane in ▯ . 3
3
6.1 Intuitive description of a plane P in ▯ .
We will assume an intuitive understanding of a “plane” in 3-space. We want to formulate
a mathematical representation of a plane P in 3-space in such a way that mathematical
representation corresponds to our intuitive perception of what the set of points (vectors)
in P are.
Given a plane P we can imagine a vector n = (n , n , n 1 w2ose3directed line segment is
perpendicular (orthogonal) to all line segments which lie on P.
Say a = (a ,1a ,2a )3is any point (vector) on P.
- Note that a does not represent a directed line segment which “lies”on P and so it need
not be orthogonal to n. (We use the word point (a , a , a1) o2 P 3o avoid confusion so
as we don't think it is a vector whose directed line segment “lies” on P.)
- We want to find a way to “recognize” all points (vectors) x which are on P.
- Observe that for any x on P, since a is also on P, x▯▯ a is a vector whose directed line
segment is parallel to the line segment joining a to x and so is orthogonal to n.
- This holds true for all vectors x on the plane P.
- Hence the vectors on P are precisely those vectors x that satisfy “ x ▯ a is
orthogonal to n.”
- Thus for any point x on P,
(x▯▯ a) ▯ n.
6.1.1 Definition ▯ Let a = (a , a 1 a2) b3 a vector in ▯ and n = (n , n , n 1 be2a n3n-
3
zero vector. We define a plane, P(n, a) in ▯ , as being all vectors x such that
x ▯▯a▯▯▯n. Equivalently,
P(n, a) = {x ▯ ▯ : n3 ▯ (x ▯ a) = 0 } The vector n is referred to as the normal to the plane. The equation n ▯(x ▯ a) = 0
will be referred to as a “normal vector equation of the plane” where n and a are
given and x is a “vector variable”.
Recall that, for vectors v and w which are non-zero and linearly independent we
previously referred to the set of vectors represented by vector equation x = sv + tw +
3
a of ▯ as a plane. We will soon show that the set both P(n, a) and the equation x =
sv + tw + a can be used to describe the same plane.
6.1.2 Remarks
According to the definition n ▯(x ▯ a) = 0, any two vectors define a particular plane.
One vector, n, can be used as the normal to the plane while the other vector, a,
represents a point on the plane.
- There are many possibilities for the normal n of a plane for if n is orthogonal to
all lines on a plane then so is any scalar multiple of n. That is, for any non-zero
scalar t, P(n, a) = P(tn, a) since,
n ▯(x ▯ a) = 0 ▯▯▯tn ▯(x ▯ a) = 0 (Invoking DP3)
- Observe that, according to this definition, the vector a does indeed belong to P,
since n ▯ (a ▯ a) = n ▯0 = 0.
- Also note that if b is a point on P(n, a) then P(n, a) and P(n, b) are the same plane
since
- The definition of “plane” in ▯ generalizes to a definition of a plane in higher
n n
dimensions ▯ . For example, if n and a are two vectors in ▯ , where n > 3, a set
of vectors in
n
P(n, a) = {x▯▯ ▯ : n ▯(x ▯ a) = 0 }
is called a hyperplane. We will see that this such sets P(n, a) and be represented
in the form given in definition 2.10.
- Note that the vectors n and a need not be linearly independent. It is possible to
have as normal vector a vector n which is collinear with a. For example, the set {x
: (0,0,1)▯▯(x ▯ (0,0,1)) = 0 } is the horizontal plane containing a = n = (0, 0, 1).
3
6.2 Definition ▯ The plane in ▯ , whose normal vector is n = (n , n , n )1and2con3ains the
point a = (a ,1a 2 a 3 can be described by a scalar equation. To see this:
The plane P(n, a) is the set of all x = (x , x1, x2) 3uch that n ▯(x ▯ a) = 0, that is the set
of all x = (x , x , x ) such that
1 2 3 n 1x 1 a )1+ n (x2▯▯2 ) +2n (x ▯3a 3 = 03
Equivalently, it is the set of all vectors x = (x , x1, 2 ) 3uch that
n1 1+ n x2 2n x =3 3= n a + n1 1+ n a2 2 3 3
Expressions of the form
n1 1+ n x2 2n x 3 3
are “scalar (standard) equation of the plane”. These planes have norm (n , n , n ). S1e 2 3
the following example.
6.2.1 Observation ▯ Let d be a real number and let T = {(x, y, z) : ax + by + cz = d }. If
a ▯▯0, we see that d = (d , d , d ) = (d/a, 0, 0) ▯▯T , that is, ad + bd + cd = d. Let
1 2 3 1 2 3
n = (a, b, c).
This example show that, that given any scalar equation of the form ax + by + cz = d,
we can easily find one point d = (d , d , d ) which satisfies this equation, and then
1 2 3
setting n = (a, b, c), the set of all x = (x, y, z) that satisfies this equation is P(n , d) =
{x : n ▯(x ▯ d) = 0}. So scalar equations for the plane can easily be expressed in the
form of a “normal vector equation” n ▯(x ▯ d) = 0.
3
The following theorem shows that sets in ▯ represented by the equation x = sv + tw + a
(remember, we also called these “planes” in 2.9) can also be represented by n ▯ (x ▯ a) =
0, and so are planes as we have now defined them.
6.3 Theorem ▯ Suppose v and w are linearly independent vectors in. If x = sv + tw + a
for some s, t in ▯ then x belongs to the plane represented by (v ▯ w) ▯(x ▯ a) = 0.
Proof :
First note (from properties in 5.2) that, since n = v ▯ w, n ▯▯sv = 0 and n ▯tw = 0 for
any s and t.
Suppose x = sv + tw + a. Then
So sv + tw + a ▯{x ▯ ▯ : n 3 ▯ (x ▯ a) = 0} as required. 3
6.4 Theorem ▯ Let a ▯ ▯ and let P be the plane represented by n ▯(x ▯ a) = 0 (where n
is not 0). Then there exists linearly independent vectors v and w such that v ▯ w = kn and
3
a vector d in P such that P ▯ T = {x ▯ ▯ : x = sv + tw + d for some s, t in ▯ }.
Proof :
Let a = (a, b, c

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