4.5 Lines and hyperplanes in R n
4.5.2 Hyperplanes
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Overview
In this chapter we have defined and looked at vectors and Euclidean n-space, Rn, together with the definition and properties of the inner product in Rn. We have worked with lines in R2 and lines and planes in R3 in order to gain geometric insight into the possibilities that arise from linear combinations of vectors, so that we may be able to apply this intuition to Rn. Vectors are the fundamental building blocks of linear algebra, as we shall see in the next chapters.
At the end of this chapter and the relevant reading you should be able to:
explain what is meant by a vector and by Euclidean n-space,Rn.
define the inner product of two vectors, understand and use the properties of the inner product
show that if A= (c1 c2 . . . cn), and if x= (α1, α2, . . . , αn)T ∈Rn, then Ax=α1c1+α2c2+· · ·+αncn
state what is meant by the length and direction of a vector, what is meant by a unit vector
state and apply the relationship between the inner product and the length and angle between two vectors in R2 and R3
explain what is meant by two vectors being orthogonal and determine if two vectors are orthogonal
find the equations, vector and Cartesian, of lines in R2, lines and planes in R3, and work problems involving lines and planes
understand what is meant by a line and by a hyperplane in Rn.
Test your knowledge and understanding
Work Exercises 1.8–1.12 in the text A-H. The solutions to all exercises in the text can be found at the end of the textbook.
Work Problems 1.10 – 1.12 in the text A-H. You will find the solutions on the VLE.
Comments on selected activities
Feedback to activity 4.1
aTb= ( 1 2 3 )
4
−2 1
= (3)
abT =
1 2 3
( 4 −2 1 ) =
4 −2 1
8 −4 2
12 −6 3
.
Feedback to activity 4.2
To prove properties (ii) and (iii), apply the definition to the LHS (left-hand side) of the equation and rearrange the terms to obtain the RHS (right-hand side). For example, for
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4.5. Comments on selected activities
x, y∈Rn, using the properties of real numbers,
αhx,yi = α(x1y1+x2y2+· · ·+xnyn)
= αx1y1+αx2y2+· · ·+αxnyn
= (αx1)y1+ (αx2)y2+· · ·+ (αxn)yn =hαx,yi.
Do the same for property (iii).
The single property hαx+βy,zi=αhx,zi+βhy,zi implies property (ii) by letting β = 0 for the first equality and then letting α= 0 for the second, and property (iii) by letting α=β= 1. On the other hand, if properties (ii) and (iii) hold, then
hαx+βy,zi = hαx,zi+hβy,zi by property (iii)
= αhx,zi+βhy,zi by property (ii) . Feedback to activity 4.6
kak= 5, so
u= 1 5
4 3
and w=−2 5
4 3
. Feedback to activity 4.7
In the figure below,
-y
6
z
x
(a1, a2, a3)
(a1, a2,0)
1 PP
PP PP
the line from the origin to the point (a1, a2,0) lies in thexy-plane, and by Pythagoras’
theorem, it has length p
a21+a22. Applying Pythagoras’ theorem again to the right triangle shown, we have
kak= s
q
a21+a22 2
+ (a3)2 = q
a12+a22+a32 Feedback to activity 4.8
We have
a=
1 2 2
b=
−1 1 4
c=b−a=
−2
−1 2
The cosines of the three angles are given by
ha,bi
kakkbk = −1 + 2 + 8
√9√
18 = 1
√2
4
ha,ci
kakkck = 2 + 2−4
√9√
9 = 0;
hb,ci
kbkkck = 2−1 + 8
√18√
9 = 1
√2 Thus the triangle has a right-angle, and two angles of π/4.
Alternatively, as the vectors a and c are orthogonal, and have the same length, it follows immediately that the triangle is right-angled and isosceles.
Feedback to activity 4.9
If t= 3, then q= (3,7)T. You are asked to sketch the position vector qas this sum to illustrate that the vector q does locate a point on the line, but the vector qdoes notlie on the line.
Feedback to activity 4.10 Heres =−1.
Feedback to activity 4.11
We will work through this for the second equation and leave the first for you. We have, for s∈R,
x y
= 1
3
+s −2
−4
⇒
(x= 1−2s
y= 3−4s ⇒ 1−x
2 =s = 3−y 4 , which yields 2(1−x) = 3−y or y= 2x+ 1.
Feedback to activity 4.12 A vector equation of the line is
x= −1
1
+t 4
1
=p+tv, t∈R,
where we have used p to locate a point on the line, and the direction vector, v=q−p.
The point (7,3) is on the line (t= 2), and this is the only point of this form on the line, since once 7 is chosen for the xcoordinate, the y coordinate is determined.
Feedback to activity 4.13
Once given, for example, that thex coordinate is x= 3, the parameter t of the vector equation is determined, therefore, so are the other two coordinates. We saw in Example 4.7 that t = 2 satisfies the first two equations and it certainly does not satisfy the third equation, 1 = 0−t.
Feedback to activity 4.14
This is similar to the earlier activity in R2. A vector equation of the line is x=
−1 1 2
+t
4 1
−1
=p+tv, t∈R.
The point (7,1,3) is not on this line, but the point (−5,0,3) is on the line. The value t=−1 will then satisfy all three component equations. There is, of course, only one possible choice for the values of cand d.
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4.5. Comments on selected activities
Feedback to activity 4.15
The lines are not parallel because their direction vectors are not parallel.
Feedback to activity 4.30
If vand w are parallel, then this equation represents a line in the direction v. If w=λv, then this line can be written as
x=p+ (s+λt)v, where r=s+λt∈R. Feedback to activity 4.21
Using the properties of the inner product, we have for any s, t ∈R, hn, sv+twi=shn,vi+thn,wi=s·0 +t·0 = 0.
Feedback to activity 4.22
Equating components in the vector equation, we have 3 =s and 7 =t from the first two equations, and these values do satisfy the third equation, 2 = 4−3s+t.
Feedback to activity 4.23
The parallel planes must each contain the direction vectors of each of the lines as displacement vectors, so the vector equations of the planes are, respectively
x y z
=
1 3 4
+s
1 2
−1
+t
−2 1 7
and
x y z
=
5 6 1
+s
1 2
−1
+t
−2 1 7
, where s, t∈R.
The parallel planes have the same normal vector, which we need for the Cartesian equations. Recall that in Example 4.21 on page 69 we found a Cartesian equation and a normal vector to the first plane, the plane which contains L1,
3x−y+z = 4 with n=
3
−1 1
.
Note that the point (1,3,4) is on this plane because it satisfies the equation, but the point (5,6,1) does not. Substituting (5,6,1) into the equation 3x−y+z =d, we find the Cartesian equation of the parallel plane which contains L2 is
3x−y+z = 10.
Feedback to activity 4.24
As stated, to verify that the line is in both planes, show that its direction vector is perpendicular to the normal vector of each plane, and that the point (2,−1,0) satisfies both equations.
Feedback to activity 4.25
To describe a line in Rn you need n−1 Cartesian equations. A vector parametric equation of a hyperplane in Rn would require n−1 parameters.
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Chapter 5
Linear systems I: Gaussian elimination
Introduction
Being able to solve systems of many linear equations in many unknowns is a vital part of linear algebra. This is where we begin to use matrices and vectors as essential
elements of obtaining and expressing the solutions. In this chapter we investigate linear systems and present a useful method known as Gaussian elimination.
Aims
The aims of this chapter are to:
Define linear systems
Learn how to solve linear systems by using the method of Gaussian elimination.
Essential reading
R Anthony, M. and M. Harvey. Linear Algebra: Concepts and Methods. Chapter 2.
Sections 2.1-2.3.
This chapter of the guide closely follows the first half of Chapter 2 of the textbook. You should read the corresponding sections of the textbook and work through all the
activities there while working through the sections of this subject guide.
Further reading
The material of this chapter is also discussed in the following book:
R Anthony, M. and N. Biggs. Mathematics for Economics and Finance: Methods and Modelling. Chapters 16, 17.
If you would like to have another source containing the material covered in this chapter, you can look up the key concepts and definitions (as listed in the synopsis which
follows) in any elementary linear algebra textbook, such as the two listed in Chapter 1, using either the table of contents or the index.
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Synopsis
We begin by expressing a system in matrix form and defining elementary row
operations on an augmented matrix. These operations mimic standard operations on systems of equations. We then learn a precise algorithm to apply these operations in order to put the matrix in a form called reduced echelon form, from which the general solution to the system is readily obtained. The method of manipulating matrices in this way to obtain the solution is known as Gaussian elimination.