Rn: Vectors - York Universityraguimov/math1025n_w16/6.1-6.5 R^n-Vectors … · Since vectors in R nare n 1 matrices, addition in R is precisely matrix addition using column matrices,

A FIRST COURSEIN LINEAR ALGEBRAAn Open Text by Ken Kuttler

Rn: Vectors

Lecture Notes by Karen Seyffarth∗

Adapted byLYRYX SERVICE COURSE SOLUTION

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Rn : Vectors Page 1/31

What is Rn?

Notation and Terminology

R denotes the set of real numbers.

R2 denotes the set of all column vectors with two entries.

R3 denotes the set of all column vectors with three entries.

In general, Rn denotes the set of all column vectors with n entries.

Rn : Vectors What is Rn? Page 2/31

Scalar quantities versus vector quantities

A scalar quantity has only magnitude; e.g. time, temperature.

A vector quantity has both magnitude and direction; e.g. displacement,force, wind velocity.

Whereas two scalar quantities are equal if they are represented by the same value,two vector quantities are equal if and only if they have the same magnitude anddirection.


R2 and R3

Vectors in R2 and R3 have convenient geometric representations as positionvectors of points in the 2-dimensional (Cartesian) plane and in 3-dimensionalspace, respectively.


R2

0 xa

y

b (a, b)

The vector

[ab

].

R3

0

x

y

z

(a, b, c)

a

b

c

The vector

abc

.


Notation

If P is a point in Rn with coordinates (p1, p2, ..., pn) we denote this byP = (p1, p2, ..., pn).

If P = (p1, p2, . . . , pn) is a point in Rn, then

−→0P =

p1p2...pn

is often used to denote the position vector of the point.

Instead of using a capital letter to denote the vector (as we generally dowith matrices), we emphasize the importance of the geometry and thedirection with an arrow over the name of the vector.


Notation and Terminology

The notation−→0P emphasizes that this vector goes from the origin 0 to the

point P. We can also use lower case letters for names of vectors. In this

case, we write−→0P = ~p.

Any vector

~x =

x1x2...xn

in Rn

is associated with the point (x1, x2, . . . , xn).

Often, there is no distinction made between the vector ~x and the point(x1, x2, . . . , xn), and we say that both (x1, x2, . . . , xn) ∈ Rn and

~x =

x1x2...xn

∈ Rn.


Geometric Vectors in R2 and R3

Let A and B be two points in R2 or R3.

0 x

y

B

A

•−→AB is the geometric vector from A to B.

• A is the tail of−→AB.

• B is the tip of−→AB.

• the magnitude of−→AB is its length, and is

denoted ||−→AB||.

Rn : Vectors Geometric Vectors Page 8/31

Equality of geometric vectors

0 x

y

A

B

C

D

•−→AB is the vector from A = (1, 0).

to B = (2, 2).

•−→CD is the vector from C = (−1,−1)

to D = (0, 1).

•−→AB =

−→CD because the vectors have

the same length and direction.

The fact that the points A and B are different from the points C and D is notimportant. For geometric vectors, the location of the vector in the plane (or in3-dimensional space) is not important; the important properties are its length anddirection.


Coordinatizing Vectors – Part 1

0 x

y

A

BP

−→0P is the position vector for P = (1, 2),

and−→0P =

[12

].

Since−→AB =

−→0P, it should be the case that

−→AB =

[12

]. This can be seen by

moving−→AB so that its tail is at the origin.

A geometric vector is coordinatized by putting it in standard position, meaningwith its tail at the origin, and then identifying the vector with its tip.


Algebra in Rn

Addition in Rn

Since vectors in Rn are n × 1 matrices, addition in Rn is precisely matrix additionusing column matrices, i.e.,

If ~u and ~v are in Rn, then ~u + ~v is obtained by adding togethercorresponding entries of the vectors.

The zero vector in Rn is the n × 1 zero matrix, and is denoted ~0.

Example

Let ~u =

123

and ~v =

456

. Then,

~u + ~v =

123

+

456

=

579

Rn : Vectors Algebra in Rn Page 11/31

Properties of Vector Addition

Let ~u, ~v , and ~w be vectors in Rn. Then the following properties hold.

1 ~u + ~v = ~v + ~u (vector addition is commutative).

2 (~u + ~v) + ~w = ~u + (~v + ~w) (vector addition is associative).

3 ~u +~0 = ~u (existence of an additive identity).

4 ~u + (−~u) = ~0 (existence of an additive inverse).


Scalar Multiplication

Since vectors in Rn are n × 1 matrices, scalar multiplication in Rn is preciselymatrix scalar multiplication using column matrices, i.e., If ~u is a vector in Rn andk ∈ R is a scalar, then k~u is obtained by multiplying every entry of ~u by k.

Example

Let ~u =

123

and k = 4. Then,

k~u = 4

123

=

48

12


Properties of Scalar Multiplication

Let ~u, ~v ∈ Rn be vectors and k , p ∈ R be scalars. Then the following propertieshold.

1 k(~u + ~v) = k~u + k~v (scalar multiplication distributes over vector addition).

2 (k + p)~u = k~u + p~u (addition distributes over scalar multiplication).

3 k(p~u) = (kp)~u (scalar multiplication is associative).

4 1~u = ~u (existence of a multiplicative identity).


The Geometry of Vector Addition

1 Vector Equality. The vectors have the same length and direction.

2 The zero vector, ~0 has length zero and no direction.

3 Addition. Let ~u, ~v be vectors. Then ~u + ~v is the diagonal of theparallelogram defined by ~u and ~v , and having the same tail as ~u and ~v .

~u

~u + ~v

~v

Rn : Vectors The Geometry of Vector Addition Page 15/31

Tip-to-Tail Method for Vector Addition

For points A, B and C , −→AB +

−→BC =

−→AC .

A

B

C

−→AB

−→BC

−→AC

−→AB

−→BC


Example

The diagonals of any parallelogram bisect each other.

To see this, denote the parallelogram by its vertices, ABCD.

A

B C

D

M

• Let M denote the midpoint

of−→AC .

Then−−→AM =

−−→MC .

• It now suffices to show

that−−→BM =

−−→MD.

−−→BM =

−→BA +

−−→AM =

−→CD +

−−→MC =

−−→MC +

−→CD =

−−→MD.

Since−−→BM =

−−→MD, these vectors have the same magnitude and direction, implying

that M is the midpoint of−→BD.

Therefore, the diagonals of ABCD bisect each other.


The Geometry of Vector Subtraction

Let ~u and ~v be vectors in R2 or R3. The vector ~u−~v = ~u + (−~v) is obtained fromthe parallelogram defined by ~u and ~v by taking the vector from the tip of ~v to thetip of ~u, i.e., the diagonal of the parallelogram, directed towards the tip of ~u.

~v

~u

~u

−~v~u − ~v


Coordinatizing Vectors – Part 2

Let A = (x1, y1, z1) and B = (x2, y2, z2) be two points in R3.

0

x

y

z

A

B

We see from the figure that−→0A +

−→AB =

−→0B, and hence

−→AB =

−→0B −

−→0A =

x2y2z2

− x1

y1z1

=

x2 − x1y2 − y1z2 − z1

.


Length of a Vector

The Distance Between Points

For A = (x1, y1, z1) and B = (x2, y2, z2) in R3, the distance between them iswritten d(A,B) and is given by

d(A,B) =√

(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2.

This is called the distance formulaIf P = (x2 − x1, y2 − y1, z2 − z1), then the distance between the origin and P isequal to the the distance between points A and B i.e.,

d(0,P) = d(A,B) =√

(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2.

Rn : Vectors Length of a Vector Page 20/31

Properties of Distance

Let P and Q be two points in Rn, and d(P,Q) the distance between them. Thenthe following properties hold.

1 The distance between P and Q is equal to the distance between Q and P,i.e., d(P,Q) = d(Q,P).

2 d(P,Q) ≥ 0 with equality if and only if P = Q.

Example

For P = (1,−1, 3) and Q = (3, 1, 0), the distance between P and Q isd(P,Q) =

√22 + 22 + (−3)2 =

√17.


Length of a Vector

More generally, if P = (p1, p2, . . . , pn) and Q = (q1, q2, . . . , qn) are points in Rn,

then the distance between P and Q is the length of the vector−→PQ, written ‖

−→PQ‖.

d(P,Q) = ‖−→PQ‖ =

√(q1 − p1)2 + (q2 − p2)2 + · · ·+ (qn − pn)2.

If ~x =

[x1x2

]∈ R2,

0x

yX = (x1, x2)

~x

then the length of the vector ~x is the distance from the origin 0 to the pointX = (x1, x2) given by d(0,X ).The length of ~x , denoted ‖~x‖, is given by:

d(0,X ) = ‖~x‖ =√x21 + x22 .


The formula for calculating the length of a vector generalizes to Rn: if

~x =

x1x2...xn

∈ Rn,

then

‖~x‖ =√x21 + x22 + · · ·+ x2n ,

which represents the distance from the origin to the point (x1, x2, . . . , xn).


Example

Let ~p =

[−34

]and ~q =

3−1−2

. Then −2~q = (−2)~q =

−624

.

The lengths of these vectors are

‖~p‖ =√

(−3)2 + 42 =√

9 + 16 = 5,

‖~q‖ =√

(3)2 + (−1)2 + (−2)2 =√

9 + 1 + 4 =√

14,

and

‖ − 2~q‖ =√

(−6)2 + 22 + 42

=√

36 + 4 + 16

=√

56 =√

4× 14

= 2√

14 = 2‖~q‖.


Unit Vectors

DefinitionA unit vector is a vector of length one.

Example 100

,

010

,

001

,

√220√22

, are examples of unit vectors.

Example

If ~v 6= ~0, then1

‖~v‖~v

is a unit vector in the same direction as ~v .


Example

~v =

−132

is not a unit vector, since ‖~v‖ =√

14. However,

~u =1√14

~v =

−1√143√142√14

is a unit vector in the same direction as ~v , i.e.,

‖~u‖ =1√14‖~v‖ =

1√14

√14 = 1.

Example

If ~v and ~w are nonzero that have

the same direction, then ~v = ‖~v‖‖~w‖ ~w ;

opposite directions, then ~v = − ‖~v‖‖~w‖ ~w .


The Geometry of Scalar Multiplication

Scalar Multiplication. If ~v 6= ~0 and a ∈ R, a 6= 0, then a~v has length‖a~v‖ = |a| · ‖~v‖, and

I has the same direction as ~v if a > 0;I has direction opposite to ~v if a < 0.

Parallel Vectors. Two nonzero vectors are called parallel if they have thesame direction or opposite directions. It follows that nonzero vectors ~v and~w are parallel if and only if one is a scalar multiple of the other.

Rn : Vectors The Geometry of Scalar Multiplication Page 27/31

Problem

Let P = (1,−2, 1),Q = (−3, 0, 5), X = (2,−1, 5) and Y = (4,−2, 3) be points in

R3. Is−→PQ parallel to

−→XY ? Is

−→PX parallel to

−−→QY ?

Solution

−→PQ =

−424

,−→XY =

2−1−2

, and these vectors are parallel if−→PQ = k

−→XY for

some scalar k , i.e., −424

= k

2−1−2

or

−424

=

2k−k−2k

.

This gives a system of three equations in one variable, which is consistent, and

has unique solution k = −2. Therefore,−→PQ is parallel to

−→XY .

−→PX =

114

,−−→QY =

7−2−2

, and these vectors are parallel if−→PX = `

−−→QY for

some scalar `. You will find that no such ` exists, so−→PX is not parallel to

−−→QY .

Rn : Vectors The Geometry of Scalar Multiplication Page 28/31

Vector problems and examples

Problem

Find the point, M, that is midway between P1 = (−1,−4, 3) and P2 = (5, 0,−3).

Solution

0

P1 M

P2

−→0M =

−−→0P1 +

−−→P1M =

−−→0P1 +

1

2

−−−→P1P2 =

−1−4

3

+1

2

64−6

=

−1−4

3

+

32−3

=

2−2

0

.

Therefore M = (2,−2, 0).

Rn : Vectors Vector problems and examples Page 29/31

Problem

Find the two points trisecting the segment between P = (2, 3, 5) andQ = (8,−6, 2).

Solution

0

P A

B

Q

•−→0A =

−→0P + 1

3

−→PQ

•−→0B =

−→0P + 2

3

−→PQ

Since−→PQ =

6−9−3

,

−→0A =

235

+

2−3−1

=

404

and−→0B =

235

+

4−6−2

=

6−3

3

.

Therefore, the two points are A = (4, 0, 4) and B = (6,−3, 3).


Example

If ABCD is an arbitrary quadrilateral, then the the midpoints of the four sides ofABCD are the vertices of a parallelogram.

A

B

C

D

Let M1 denote the midpoint of−→AB,

M1

M2 the midpoint of−→BC ,

M2

M3 the midpoint of−→CD, and

M3

M4 the midpoint of−→DA.

M4

It suffices to prove that−−−→M1M2 =

−−−→M4M3.

−−−→M1M2 =

−−→M1B +

−−→BM2

−−−→M4M3 =

−−→M4D +

−−→DM3

= 12

−→AB + 1

2

−→BC = 1

2

−→AD + 1

2

−→DC

= 12

−→AC = 1

2

−→AC

Since−−−→M1M2 =

−−−→M4M3, the points M1, M2, M3, M4 are the vertices of a

parallelogram.


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Rn: Vectors - York Universityraguimov/math1025n_w16/6.1-6.5 R^n-Vectors … · Since vectors in R nare n 1 matrices, addition in R is precisely matrix addition using column matrices,