1 ANALYSIS OF VECTORS GEOMETRICALLY

Some of the quantities we measure in our daily lives are completely determined by their

magnitudes, for example, length, mass, area, temperature, and energy. When we speak of a

length of 3 cm or an area of 5 cm2, we only need one number to describe each of these

quantities. We call such quantities scalar qquantities.

On the other hand, to describe a force, we need to record its direction as well as its size. For

example, to describe the velocity of a moving object, we must specify both the speed and the

direction of travel. Quantities such as displacement, velocity, acceleration, and other forces

that have magnitude as well as direction are called vector qquantities. We usually show a

vector quantity as an arrow that points in the direction of the action, with length that shows

the magnitude of the action in terms of a suitable unit. The way to represent such quantities

mathematically is through the use of vectors.

1. Directed Line SegmentWhen we move from Antalya to Berlin

by bus, we have two quantities: the

direction from Antalya to Berlin, and

the length of the displacement between

these cities.

We can sketch a line segment AB as shown in the figure with starting

point A and finishing point B to represent the movement from Antalya to

Berlin. The line segment AB with an arrow has direction and length. The

arrow head specifies the direction, and the length of the arrow specifies

the magnitude, at a suitable scale. A and B are the endpoints of the segment.

Point A is called the initial ppoint and point B is called the terminal ppoint

of the line segment. The resulting segment AB is called a directed lline ssegment.

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Definition directed lline ssegment

A line segment with direction is called a directed lline ssegment.

We write−→AB to denote a directed line segment from point A to point B.

Directed line segments are used in daily life. For example, some

traffic signs for drivers use directed line segments.

In technology we also use directed line segments.

�� ANALYSIS OF VECTORS GEOMETRICALLY

2 Vectors in the Plane

A. BASIC VECTOR CONCEPTS

3Analysis of Vectors Geometrically

Points M, N, P and K on line d are given. Write all the

directed line segments with endpoints M, N, P, or K.

The directed line segments with endpoints M, N, P, or K are −→MN,

−→MP,

−→MK,

−→NP,

−→NK,

−→NM,

−→PK,

−→PN,

−→PM,

−→KP,

−→KN, and

−→KM.

Notice that −→MN is not the same as

−→NM, and

−→MP is not the same as

−→PM. This is because the

line segments have direction. Pairs such as −→MN and

−→NM have the same magnitude but

opposite direction.

Solution

EXAMPLE 1

Definition vector

A directed line segment in the plane is called a vector.

The length of the directed line segment is the length of thevector.

The direction of the directed line segment is the direction ofthe vector.

We write−→AB to mean a vector with initial point A and terminal point B. Alternatively, we can

name a vector with a lower-case letter such as →u or

→p.

3. Equal Vectors

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Definition equal vvectors

Two vectors that have the same direction and length are

called equal vvectors. We show that two vectors →u and

→v are

equal by writing →u =

→v .

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In the figure, D, E, and F are the midpoints of AB, AC and

BC respectively, and DE || BC, EF || AB, DF || AC.

Name all the equal vectors.

In triangle ABC,|DE| = |BF| = |FC|

|EF| = |AD| = |DB|

|DF| = |AE| = |EC|.

So−→DE =

−→BF =

−→FC

−→EF =

−→AD =

−→DB

−→DF =

−→AE =

−→EC

−→ED =

−→FB =

−→CF

−→FE =

−→DA =

−→BD

−→FD =

−→EA =

−→CE.

Solution

EXAMPLE 2

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2. Definition of a Vector

and

For example, consider a line segment −→AB with length 2 cm.

We can say the length of vector−→AB is 2 cm, and write |

−→AB| = 2 cm.


Check Yourself 1ABCD is a parallelogram in the figure.1. How many pairs of equivalent directed line segments are there?2. How many pairs of equal vectors are there?

Answers1. 4 2. 4

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1. Addition of VectorsLet

−→PQ and

−→QR be two vectors in a plane.

−→PQ +

−→QR denotes the sum of the vectors

−→PQ and

−→QR. There are two ways to find the sum of two or more vectors.

a. The Polygon MethodImagine we want to add n vectors together. Using the polygon method, we draw the firstvector. Then we place the initial point of the second vector at the terminal point of the firstvector, the initial point of the third vector at the terminal point of the second vector, and soon until we place the initial point of the nth vector at the terminal point of the (n – 1)th

vector. The sum is the vector whose initial point is the initial point of the first vector andwhose terminal point is the terminal point of the last vector.

Let us look at an example.

Let −→AB and

−→CD be two vectors in a plane, as in the

diagram. We place the initial point of −→AB at the

terminal point of −→CD to make

−→DE (

−→AB =

−→DE).

Using the polygon method,−→CD +

−→AB =

−→CD +

−→DE =

−→CE.

B. VECTOR OPERATIONS

Definition opposite vvectors

Two vectors are called opposite vectors if and only if their

magnitudes (lengths) are the same but their directions are opposite.

Definition zero vvector

A vector whose initial and terminal points are the same is called a zero vvector.

We write a zero vector as →0.

A zero vector has no direction and no size.

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For example, in the figure, −→AB and

−→BA are opposite vectors.

−→CD and

−→DC are also opposite vectors. We can write

−→AB = –

−→BA and

−→CD = –

−→DC.


b. The Parallelogram Method

To add one vector to another using the parallelogram method, we draw the first vector, and

then we draw the second vector with its initial point at the initial point of the first vector. We

make a parallelogram by drawing two additional sides, each passing through the terminal

point of one of the vectors and parallel to the other vector. We find the sum by drawing a

vector along the diagonal from the common initial point to the intersection of the two lines.

Look at the example of adding →u and

→v using the parallelogram method:

Now look at an example of adding more than two vectors using the polygon method.

As shown in the figure,→u +

→v +

→w +

→x =

−→AE.

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Find →u +

→v +

→w in the figure on the right.EXAMPLE 3 �

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Let us choose A as a fixed point. We can use the polygon method or the parallelogram

method to add the given vectors →u +

→v +

→w .

Solution

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The velocity of a boat is 25 m/min north and the velocity of a river current is 3 m/min east.

Draw a scale diagram to show the velocities as vectors and find the sum.

First we choose a starting point A and

write −→AN = velocity of the boat due north.

−→AE = velocity of the current due east.−→AN and

−→AE are perpendicular, and

−→AK is

the sum of −→AN and

−→AE:

|−→AK| =

This is the sum of the vectors.

2 225 +3 = 634.

Solution

EXAMPLE 4

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c. Properties of Vector AdditionLet

→u,

→v, and

→w be three vectors in a plane P.

1. The sum of any two vectors in P is

also a vector in P (closure property).��

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2. The sum of any two vectors in P is

commutative (commutative property).

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4. The sum of the zero vector and a vector

in P is the vector itself (identity

element).

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3. The sum of any three vectors in P is associative (associative property).→u +(

→v +

→w ) = (

→u +

→v ) +

→w

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5. The additive inverse of any vector →u is –

→u:

→u + (–

→u) =

→0 (additive inverse).

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In a triangle ABC, P is the midpoint of −→AB. Express

−→CP in terms of

−→CA and

−→CB.

−→CP =

−→CA +

−→AP

+−→CP =

−→CB +

−→BP

–––––––––––––––––––––––––––––––––––––––2⋅

−→CP =

−→CA +

−→CB +

−→AP +

−→BP

−→CP = ⋅(

−→CA +

−→CB)1

2

Solution

EXAMPLE 5

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Check Yourself 2

1. Find the following using the vectors in the figure.

a. →v –

→u b. →

u + →w c. →

w + →v –

→u

2. In a triangle ABC, D ∈ [BC] and |BD| = 2 ⋅ |DC|.

Express −→AD in terms of

−→AB and

−→AC.

Answers

1. use the polygon method 2.−→AD =

−→AC +

−→AB

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2. Subtraction of VectorsSince subtraction is the inverse of addition, we can find the difference of two vectors

→u and

→v by adding the vectors

→u and –

→v (opposite of

→v) using either the parallelogram method

(→u –

→v =

→u + (–

→v )) or the polygon method.

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In a triangle ABC, G is the centroid. Find −→GA +

−→GB +

−→GC.

Let us label a point G′ on the extension of CG which

satisfies |CG| = |GG′|. Since G is the centroid of

A¿BC, |CG| = 2⋅|GK|. Therefore |GG′| = 2⋅|GK|,

which means that K is the midpoint of GG′. We con-

clude that AG′BG is a parallelogram because K is the

midpoint of both diagonals AB and GG′. So we have−→AG′ =

−→GB which gives us

−→GA +

−→GB =

−→GG′.

On the other hand, we have −→CG =

−→GG′ = –

−→GC. Using

this result in −→GA +

−→GB =

−→GG′, we get

−→GA +

−→GB = –

−→GC which gives us

−→GA +

−→GB +

−→CG =

→0.

Solution

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3. Multiplication of a Vector by a ScalarMultiplying a vector by a scalar makes the vector longer or shorter depending on the value of

the scalar. If the scalar is greater than 1 or less than –1, multiplying makes a longer vector. If

the scalar is between –1 and 1 and non-zero, it makes a shorter vector.

If the scalar is positive, multiplying does not change the direction.

If the scalar is negative, multiplying will make the vector’s direction opposite.

For a real number a and a vector →u,

1. if a > 0 then vector a⋅→u has the same direction as →u and the length |a⋅→u| = a⋅|→

u|.

2. if a < 0 then vector a⋅→u has the opposite direction to →u and the length |a⋅→u| = |a|⋅|→

u|.

3. if a = 0 then a⋅→u = →0.

Using −→AB as shown in the figure, draw vector

diagrams to show 2⋅−→AB, – 4⋅

−→AB, and ⋅

−→AB.1

2

Since 2 and are positive, 2⋅−→AB and ⋅

−→AB have the

same direction as −→AB. However, 2⋅

−→AB is twice as long

as −→AB and ⋅

−→AB is half as long.

On the other hand, –4⋅−→AB has opposite direction to

−→AB (since –4 is a negative scalar) and it is four times

as long as −→AB.

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Solution

EXAMPLE 7

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a. Properties of the Multiplication of a Vector by a ScalarFor any vectors

→u,

→v, and

→w and real numbers a and b, the following properties are satisfied.

1. a⋅→u is a vector in the plane

2. (a⋅b)⋅→u = a⋅(b⋅→u )

3. (a + b)⋅→u = a⋅→u + b⋅→u

4. a⋅(→u +

→v ) = a⋅→u + b⋅→v

5. 1⋅→u = →u

6. a⋅→0 =

→0


Check Yourself 3

1. Multiply the vector →u by the scalars –2, 3, 0.5 and

and draw a vector diagram to show them.

2. Points A, B, C, and M are on the same line. M is between

A and C. −→AB = 2⋅

−→AC. Express the vector

−→MC in terms of

the vectors −→MA and

−→MB.

Answers

2.−→MC = ⋅(

−→MA +

−→MB)1

2

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1. Parallel Vectors

C. PARALLEL VECTORS

Definition parallel vvectors

Let →a and

→b be two vectors.

→a and

→b are called parallel vvectors if and only if

→a = k⋅→b where

k ≠ 0 and k ∈ . We write→a||

→b to show that two vectors are parallel.

For example, in the diagram, |→a| = 2 cm,

|→b| = 1 cm and |

→c| = 4 cm.

We can express vector →a as

→a = ⋅→c and

→a = –2⋅→b.

Therefore the vectors →a,

→b, and

→c are parallel, i.e.

→a||

→b||

→c.

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Points A, B, C, and M are on the same line. M is between C and B. −→AB = 3⋅

−→AC. Express the

vector −→MC in terms of vectors

−→MA and

−→MB.

−→AB = 3⋅

−→AC so

−→CB = 2⋅

−→AC (1)

−→MA +

−→AC =

−→MC (2)

−→CM +

−→MB = 2⋅

−→AC (3)

−→AC = ⋅

−→MC + ⋅

−→MB (4)

−→MA – ⋅

−→MC + ⋅

−→MB =

−→MC by (2) and (4).

−→MA + ⋅

−→MB = ⋅

−→MC +

−→MC

−→MA + ⋅

−→MB = ⋅

−→MC

So −→MC = ⋅

−→MA + ⋅

−→MB.1

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Solution

EXAMPLE 8

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In a triangle ABC, D and E are the midpoints of sides AB and AC respectively.

Show that −→DE ||

−→BC.

EXAMPLE 9−→BA +

−→AC =

−→BC and

−→DA +

−→AE =

−→DE by the addition of vectors.

−→DA = ⋅

−→BA,

−→AE = ⋅

−→AC

−→DE = ⋅

−→BA + ⋅

−→AC = ⋅(

−→BA +

−→AC) = ⋅

−→BC

Now −→DE = ⋅

−→BC, so

−→DE ||

−→BC by the definition of parallel vectors.1

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Solution

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−→BA +

−→AC =

−→BC (1) and

−→EA +

−→AF =

−→EF (2).

−→EA = ⋅

−→BA

+ −→AF = ⋅

−→AC

–––––––––––––––––––––––––––––––−→EA +

−→AF = ⋅(

−→BA +

−→AC)

−→EF = ⋅

−→BC by (1) and (2). Therefore,

−→EF ||

−→BC.1

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Solution

EXAMPLE 10 �

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In a quadrilateral ABCD, points E and F are the midpoints of side AB and diagonal AC,

respectively. Show that EF || BC.

2. Non-Parallel Vectors

By the definiton of parallel vectors we can conclude that if →a and

→b are non-zero,

non-parallel vectors, then h⋅→a = k⋅→b when h = k = 0. Look at the proof:

Suppose that h ≠ k ≠ 0, →a = ⋅→b.

Then →a||

→b. This is a contradiction, since

→a and

→b are non-parallel. As a result, h = k = 0.

kh

EXAMPLE 11 Prove that the diagonals of a parallelogram intersect at their midpoints by using vectors.

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Look at the diagram. Let −→AB =

→a and

−→BC =

→b, so

−→AC =

→a +

→b

−→DB =

→a –

→b

−→AE = m⋅(→

a + →b )

−→EB = n⋅(→

a – →b )

Solution


Check Yourself 41. Name all the pairs of parallel

vectors in the figure.

2. Add the vector pairs →u1 and

→u5,→

u2 and →u6, and

→u3 and

→u7.

3. Find the additive inverse of →u7

and →u1.

4. Subtract →u8 from

→u4.

5. In a quadrilateral ABCD, P and

K are the midpoints of sides AB

and DC respectively. Express −→PK

in terms of −→DA and

−→CB.

Answers

1. look at the directions and lengths 2. use the polygon method 3. use the polygon method

4. use the polygon method 5.−→PK = – ⋅(

−→DA +

−→CB)

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Let us draw the triangle ABC as in the figure.

If D, E and F are midpoints then

−→AD = ⋅(

−→AB +

−→AC) by the result of Example 5.

−→BF = ⋅(

−→BA +

−→BC)

+ −→CE = ⋅(

−→CA +

−→CB)

––––––––––––––––––––––––––––––– −→AD +

−→BF +

−→CE = ⋅(

−→AB +

−→BA +

−→AC +

−→CA +

−→BC +

−→CB )

Therefore, −→AD +

−→BF +

−→CE =

→0.

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Solution

EXAMPLE 12 In a triangle ABC, D, E and F are the midpoints of sides BC, BA and AC respectively. Find the

sum −→AD +

−→BF +

−→CE.

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→0

→0

−→AE +

−→EB = m⋅(→

a + →b ) + n⋅(→

a – →b )

→a = m⋅(→

a + →b ) + n⋅(→

a – →b)

(m + n – 1)⋅→a =

→b⋅(n – m).

Since →a and

→b are non-zero and non-parallel, we have (m + n – 1) = (n – m) = 0. Therefore

m + n = 1 and m = n, and so m = n = .12


Traffic signs are important fordrivers and pedestrians. If peopledo not know the meaning ofthese signs, they can have somedifficulties in traffic.

For example, if a driver drives inthe opposite direction to a ‘oneway’ sign, he or she might havean accident.

Project: Describe some other areas in which we use directed line segments(for example: flowcharts, keyboards, ...).

A. Basic Vector Concepts

11.. Draw any two parallel directed line segments with

the same length but opposite direction. Express

one of the line segments in terms of the other.

EXERCISES 1

22.. How many equal directed line segments can we

find on two parallel lines?

33.. Make a scale diagram showing the vectors in each

statement and find their sum.

a. A 6 km trip east is followed by a 3 km trip

southeast.

b. The velocity of a swimmer is 5 m/min west

and the velocity of a river current is 2 m/min

north.

44.. Using the vectors given on the right,

sketch the following vectors.

a. →u +

→v b. →

w – (→u +

→v)

c. –→w –

→v –

→u d. →

u + (→w +

→v)

e. →u – 3⋅→v + 2⋅→w

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B. Vector Operations

55.. In a plane, [AB] is given. Point K is the midpoint

of [AB] and point O is any point in the same

plane. Express −→OK in terms of

−→OA and

−→OB.

66.. In a triangle ABC, points D and E lie on [BC] and

|BD| = |DE| = |EC|. Express the vector −→AD +

−→AE

in terms of −→AB and

−→AC.


1122.. In a six-sided polygon ABCDEF, −→AB =

−→ED,

−→BC =

−→FE,

and −→CD =

−→AF. Show that FBCE is a parallelogram.

In a parallelogram ABCD, K is the midpoint of

side DC, and the intersection point of diagonals

AC and BK is T. Show that −→AC = 3⋅

−→TC.

1188..

In a quadrilateral ABCD, E and F are the midpoints

of the diagonals AC and BD respectively.

Show that−→AB +

−→AD +

−→CB +

−→CD = 4⋅

−→EF.

1155..

1144.. In a triangle ABC, |BD| = |DE| = |EC|, and

E, D ∈ [BC]. If |−→AD +

−→AE| = 9 cm, find |

−→AB +

−→AC|.

Show that the centroid of a triangle divides amedian in the ratio 1:2 using vectors.

1177..

1133.. A trapezoid is a four-sided figure with only twoparallel sides. A line segment which joins themidpoints of the non-parallel sides is called themedian of the trapezoid. Prove that the median ofa trapezoid is parallel to the two parallel sides,and has magnitude equal to half of their sum.

In the figure,

T is the midpoint of BC,

ABC is a triangle,

2⋅|AK| = |KB|, and

2⋅|AM| = |MC|.

Use vectors to show that

|−→AL| = ⋅|

−→LT|.1

2

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1111.. ABCD is a quadrilateral and M, N, P, Q are themidpoints of AB, BC, CD, and DA respectively.Show that MNPQ is a parallelogram.

C. Parallel Vectors

1100.. In a triangle ABC, G is the point of intersection of the

medians and P is the midpoint of −→BG. Show that

−→PA +

−→PC = 4⋅

−→PG.

Consider any two points A and B in a plane.For

any point P in the same plane, the symmetry of

point P with respect to point A is Q and the

symmetry of point Q with respect to point B is

point R. Show that −→PR is always 2⋅

−→AB. (Hint: let A,

B, and C be collinear. If |AB| = |BC| then A is

the symmetry of C with respect to B.)

99..

88.. Point O is in the plane of a triangle ABC. Point G

is the centroid of triangle ABC. Show that−→OA +

−→OB +

−→OC = 3⋅

−→OG.

77.. Show that (−→AB +

−→BC) +

−→CD =

−→AB + (

−→BC +

−→CD)

by using the parallelogram method.

We have studied vectors geometrically. Now let us look at a method for describing vectors

analytically.

We will begin this section by looking at some important axioms.

1. Axioms

1. For each pair of points P and Q there exists a

unique vector →v such that

−→PQ =

→v.

�� ANALYSIS OF VECTORS ANALYTICALLY

2. For each point P and vector →v, there is a

unique point Q such that →v =

−→PQ.

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15Analysis of Vectors Analytically

A. BASIC CONCEPTS OF VECTORS IN THE ANALYTICPLANE

Conclusion

1. Two points in a plane determine two opposite vectors.

2. In a plane, if one point is fixed as an initial point then all the other points in the planecan be chosen as the terminal point of any vector.

2. The Rectangular Coordinate System

The rectangular ccoordinate ssystem is formed bytwo perpendicular intersecting number lines, asshown in the diagram opposite.

1. The horizontal number line is called the x-aaxis.

2. The vertical number line is called the y-aaxis.

The origin is the point of intersection. At thisintersection, both number lines are 0. Therectangular coordinate system is split into fourquadrants, which are marked in the diagramwith roman numerals.

Each point in the coordinate system is associated with a pair oof rreal nnumbers. In an x, ysystem, the x-ccoordinate always comes first and the y-ccoordinate always comes second in thepair (x, y). The first coordinate is called the abscissa of the point and the second coordinateis called the ordinate of the point.

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A(2, 3) lies in quadrant I.

B(–1, 2) lies in quadrant II.

C(–3, –4) lies in quadrant III.

D(2, 0) lies on the x-axis.

E(0, 5) lies on the y-axis.

Solution

EXAMPLE 13 Plot each pair of coordinates and name the quadrant or axis in which the point lies.

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Check Yourself 5

Plot the pairs of real numbers and name the quadrant or axis in which the point lies.

A(3, 2), B(–2, 1), C(–4, –3), D(0, 2), E(5, 0)

From this definition we can conclude the following:

1. For every vector in the plane there exists a position

vector −→OP which is determined by a pair (x, y), that is,

−→OP = (x, y).

2. If −→OP =

−→AB, then OPBA is a parallelogram.

Look at the diagram. We can calculate that x = x2 – x1 and

y = y2 – y1. Therefore the vector −→AB determined by the

points A(x1, y1) and B(x2, y2) has position vector−→OP =

−→AB =

−→OB –

−→OA = (x, y) = (x2 – x1, y2 – y1).

−→OP =

−→KL =

−→OL –

−→OK = (4 – 2,5 – 1)

−→OP = (2, 4)

Solution

EXAMPLE 14 Find the position vector of −→KL with endpoints

K(2, 1) and L(4, 5).

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3. Position Vector

Definition position vvector

A vector −→OP whose initial point is at the origin of the rectangular coordinate plane and which

is parallel to a vector −→AB is called the position vvector of

−→AB in the plane. In other words, if

−→OP is the position vector of

−→AB, then

−→OP ||

−→AB, |

−→OP| = |

−→AB|, and

−→OP =

−→AB.

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4. Components of a VectorWe have seen how to describe a vector in the plane withreference to its unique position vector.

We can also express a vector in the plane as the sum oftwo vectors, one of which is parallel to the x-axis and theother parallel to the y-axis. These two vectors are calledthe components of the original vector. The componentparallel to the x-axis is called the horizontal ccomponentof the vector and the component parallel to y-axis iscalled the vertical ccomponent. Expressing a vector as thesum of its components like this is called resolving the vector. For example, in the figureopposite, the vector

→u is the sum of the two components

→ux and

→uy.

We can represent the vector →u as an ordered pair of real numbers:

→u = (u1, u2) or

→u = ,

where u1 is the horizontal scalar component of →u, and

u2 is the vertical scalar component of →u.

Now, →ux = (u1, 0) and

→uy = (0, u2).

Look at the diagram opposite. By applying thePythagorean theorem for triangle PQR in the figure, wecan see that the length of the vector

→u = (u1, u2) is

|→u| = 2 2

1 2+ .u u

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1

2

u

u

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a. |→u| =

b. |→v| =

c. |→w| = ++2 23 4 9 16 9 16 25

( ) +( ) = = = =15 5 25 25 25 25

2 23 + 0 = 9 = 3

2 22 +(–3) = 4+9 = 13Solution

EXAMPLE 15 Find the length of each vector.

a. →u = (2, –3) b. →v = (3, 0) c. →w = 3 4

( , )5 5

→u = (3 – 1, 6 – 2) = (2, 4)

|→u| = 2 22 + 4 = 4+16 = 20

Solution

EXAMPLE 16 Find the length of the vector →u with initial point (1, 2) and terminal point (3, 6).

|−→AB| =

(1 – a)2 + 4 = 20(1 – a)2 = 16

1 – a = 4 or 1 – a = –4 a = –3 or a = 5

2 2 2(1– ) +(4 – 2) = (1– ) + 4 = 2 5a aSolution

5. Equal Vectors

Definition equal vvectors

Two vectors are equal if and only if their corresponding scalar components are equal.

In other words, the vectors →u = (u1, u2) and

→v = (v1, v2) are equal if and only if u1 = v1 and

u2 = v2.

→u =

→v , so a + b = –1 and 3 = b – a.Solution

EXAMPLE 18 →u and →v are equal vectors with

→u = (a + b, 3),

→v = (–1, b – a). Find a and b.

Let the terminal point of →v be (x, y), so

(x – 2, y – 4) = (3, 7)x – 2 = 3 and y – 4 = 7 by the equality of vectors.

So x = 5 and y = 11.

Solution

EXAMPLE 19 The vector →v = (3, 7) has initial point (2, 4). What is its terminal point?

Check Yourself 6

1. Find the position vector of −→ML with endpoints M(3, 2) and L(2, 3).

2. Find the length of each vector.

a. →u = (3, 1) b. →v = (0, 3) c.−→AB with endpoints A(2, 1) and B(5, 4)

Answers

1.−→ML = (–1, 1) 2. a. ò10 b. 3 c. 3ñ2

a + b = –1

b – a = 3 , so b = 1 and a = –2.

⎫⎬⎭


EXAMPLE 17 Find the possible values of a given |−→AB| = 2ñ5 and the endpoints A(a, 2) and B(1, 4).


−→MN = (p – 2, 1 – k) = (2, 3)

p – 2 = 2 and 1 – k = 3p = 4 and k = –2

So p + k = 4 – 2 = 2.

Solution

EXAMPLE 20 The initial point and terminal point of −→MN = (2, 3) are M(2, k) and N(p, 1) respectively. Find p + k.

Check Yourself 7

1. A(2, 1), B(1, –3), C(3, 2), and D(a, b) are given. If −→AC =

−→BD, find |

−→AD|.

2. Let →u = (a, a + 1) and

→v = (3, b) such that

→u =

→v. Find a and b.

Answers

1. 3 2. a = 3, b = 4

B. VECTOR OPERATIONS

→u +

→v = (1 – 2, –2 + 1) = (–1, –1).Solution

EXAMPLE 21 →u = (1, –2) and →v = (–2, 1). Find

→u +

→v .

EXAMPLE 22 A(–3, 2) and −→AB = (5, –2), are given. Find the coordinates of B.

Two identical pictures are hung on a wallby pieces of string as shown in the figure.Each piece of string exerts a force uponthe picture to support its weight. The sumof the forces of the first picture is equal tothe force of the second picture. We canrepresent the forces as vectors. The boxshows that the sum of the two vectors ofthe first picture equals the vector of the second picture.

1. Addition of VectorsIf →u = (u1, u2) and

→v = (v1, v2), then→u +

→v = (u1 + v1, u2 + v2).

The figure opposite shows how the analytic definition of

addition corresponds to the geometric one.

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Let us write B(a, b). Now, −→AB = (a – (–3), b – 2) = (a + 3, b – 2)

(a + 3, b – 2) = (5, –2).

a + 3 = 5 and b – 2 = –2 by the equality of vectors, so a = 2 and b = 0.

Therefore, the coordinates of B are (2, 0).

Solution

first picture second picture

EXAMPLE 24 Given →w = (1, –3), K(3, 2) and P(–1, 4), find

→w +

−→KP and

→w –

−→KP.


2. Subtraction of VectorsIf →u = (u1, u2) and

→v = (v1, v2) then

→u –

→v = (u1 – v1, u2 – v2).

The figure opposite shows how the analytic definition

of vector subtraction corresponds to the geometric

one.

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→u –

→v = (2 – 3, 4 – 2) = (–1, 2)Solution

EXAMPLE 23 Subtract →v = (3, 2) from

→u = (2, 4).

a. Properties of Vector SubtractionLet

→u = (u1, u2),

→v = (v1, v2), and

→w = (w1, w2) be vectors in a plane. Then the following

properties hold.

1. The difference of any two vectors in a plane is a vector.

2. →u – →v ≠ →v –

→u. Therefore, vector subtraction is not commutative.

3. →u – (→v – →w ) ≠ (

→u –

→v ) –

→w . Therefore, vector subtraction is not associative.

4. →u – →0 ≠

→0 –

→u. Therefore there is no identity element for subtraction.

a. Properties of Vector Addition

Let →u = (u1, u2),

→v = (v1, v2), and

→w = (w1, w2) be vectors in a plane. Then the following

properties hold.

1. The sum of any two vectors in a plane is a vector. (closure property)

2. →u + →v =

→v +

→u (commutative property)

3. →u + (→v +

→w ) = (

→u +

→v ) +

→w (associative property)

4.→0 is the identity element:→u +

→0 = (u1 + 0, u2 + 0) = (u1, u2) =

→u.

5. –→u is additive inverse of

→u:

→u + (–

→u) = (u1 – u1, u2 – u2).


3. Multiplication of a Vector by a ScalarLet

→v = (v1, v2) and c ∈ , then c⋅→v = (c⋅v1, c⋅v2).

–2⋅→u = (–2⋅(–3), (–2)⋅2) = (6, –4)

3⋅→v = (3⋅2, 3⋅(–1)) = (6, –3)

3⋅→u + 2⋅→v = 3⋅(–3, 2) + 2⋅(2, –1) = (–9, 6) + (4, –2) = (–5, 4)

Solution

EXAMPLE 26 →u = (–3, 2) and →v = (2, –1) are given. Find –2⋅→u, 3⋅→v , and 3⋅→u + 2⋅→v .

a. Properties of the Multiplication of a Vector by a Scalar Let

→u = (u1, u2),

→v = (v1, v2) and c, d ∈ . Then the following properties hold:

1. c⋅(→u + →v) = c⋅→u + c⋅→v. Look at the proof:

c⋅(→u +→v ) = c⋅(u1 + v1, u2 + v2)

= (c⋅u1 + c⋅v1, c⋅u2 + c⋅v2)

= c⋅(u1, u2) + c⋅(v1, v2)

= c⋅→u + c⋅→v.

2. (c + d)⋅→u = c⋅→u + d⋅→u

3. (c⋅d)⋅→u = c⋅(d⋅→u ) = d⋅(c⋅→u )

4. 1⋅→u = →u

5. 0⋅→u = →0

6. c⋅→0 =

→0

7. |c⋅→u| = |c|⋅|→u|.

−→KP = (–1 – 3, 4 – 2) = (–4, 2)

→w +

−→KP = (1, –3) + (–4, 2) = (1 – 4, –3 + 2) = (–3, –1)

→w –

−→KP = (1, –3) – (–4, 2) = (1 + 4, –3 – 2) = (5, –5)

Solution

−→MN +

−→NK =

−→MK

|−→MN +

−→NK| = |

−→MK| = 13

(m – 1)2 + 25 = 169

(m – 1)2 = 144

m – 1 = 12 or m – 1 = –12

m = 13 or m = –11

− − −2 2( 1) +( 1 4) =13m

Solution

EXAMPLE 25 M(1, 4), N(3, 2m), K(m, –1), and |−→MN +

−→NK| = 13 are given. Find m.


4. Standard Base Vectors

a. u – 3⋅→v = (–3, 1) – 3⋅(2, –2) = (–3, 1) + (–6, 6) = (–3 – 6, 1 + 6) = (–9, 7)

b. 3⋅→u + 2⋅→v = 3⋅(–3, 1) + 2⋅(2, –2) = (–9, 3) + (4, –4) = (–9 + 4, 3 –4) = (–5, –1)

c. 4⋅→u + →v = 4⋅(–3, 1) + (2, –2) = (–12, 4) + (2, –2) = (–12 + 2, 4 – 2) = (–10, 2)

Solution

EXAMPLE 28 Find each vector, given →u = (–3, 1) and

→v = (2, –2).

a. →u – 3⋅→v b. 3⋅→u + 2⋅→v c. 4⋅→u + →v

2⋅→u – 3⋅→v = 2⋅(–2, 3) – 3⋅(1, –1) = (–4, 6) – (3, –3) = (–4 – 3, 6 + 3) = (–7, 9)

Therefore, |2⋅→u – 3⋅→v | = − 2 2( 7) +9 = 49+81= 130.

Solution

EXAMPLE 27 Find |2⋅→u – 3⋅→v| given →u = (–2, 3) and

→v = (1, –1).

Check Yourself 8

1. Find |→w| given

→u = (3, –2),

→v = (–1, 4),

→w = (a, b) and

→v – →w = 3⋅→u.

2. Find →x given 2⋅→x +

→y = (1, 2), and

→x – →y = (–4, 4).

Answers

1. |→w| = 10ñ2 2. →x = (–1, 2)

→u

|→u|

1

|→u|

= ⋅→u 21

2 2 2 21 2 1 2

= , .+ +

uu

u u u u

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

→u

|→u|

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

22 2 2 222 2 1 21 1

2 2 2 2 2 22 2 2 21 2 1 2 1 21 2 1 2

+= + = + = =1.

+ + ++ +

u u u uu uu u u u u uu u u u

So

We sometimes use to find the direction of →u.

→u

|→u|

There are two important unit vectors, →i and

→j, defined as

→i = (1, 0) and

→j = (0, 1).

These vectors are special because we can use them to express any vector.

We call these vectors standard bbase vvectors.

Definition unit vvector

A vector of length 1 is called a unit vvector.

For any non-zero vector →u = (u1, u2), is a unit vector because and

1

|→u|

2 21 2

1=

+u u

→u

|→u|

For example, the vector →w = is a unit vector.3 4

( , )5 5


b. The properties of addition and scalar multiplication of vectors show that we can

manipulate vectors in the same way we manipulate algebraic expressions. Therefore,

2⋅→u + 3⋅→v = 2⋅(3⋅

→i + 2⋅→j ) + 3⋅(–

→i + 6⋅→j )

= (6⋅→i + 4⋅→j ) + (–3⋅

→i + 18⋅

→j )

= (6 – 3)⋅→i + (4 + 18)⋅

→j

= 3⋅→i + 22⋅

→j.

a. →u = 5 ⋅→i + (–8) ⋅

→j = 5

→i – 8

→j.Solution

EXAMPLE 29 a. Write the vector →u = (5, –8) in terms of

→i and

→j.

b. If →u = 3 ⋅

→i + 2 ⋅

→j and

→v = –

→i + 6 ⋅

→j , find 2 ⋅→u + 3 ⋅→v.

For example, let us express the vector →v = (v1, v2) in terms of

→i and

→j:

→v = (v1, v2) = v1⋅

→i + v2⋅

→j.

In this expression,

v1⋅→i is the horizontal component,

v2⋅→j is the vertical component,

v1 is the horizontal scalar component, and

v2 is the vertical scalar component.

We can prove the proposition above by using algebraic operations on vectors and theproperties of real numbers:→v = (v1, v2) = (1⋅v1 + 0, 0 + 1⋅v2) = (1⋅v1, 0) + (0, 1⋅v2)

= v1⋅(1, 0) + v2⋅(0, 1)

= v1⋅→i + v2⋅

→j.

Let →v be a vector in the plane with its initial point at the

origin. Let θ be the positive angle between the positive

x-axis and →v (see the figure). If we know the length and

direction of →v , then we can resolve the vector into

horizontal and vertical components in terms of θ:→v has length |

→v|, and

→v = (v1, v2) = v1⋅

→i + v2⋅

→j.

So v1 = |→v |⋅ cos θ and v2 = |

→v |⋅ sin θ.

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cos 60° = sin 30° =

sin 60° = cos 30° =

sin 45° = cos 45° =

sin 90° = cos 0° = 1

sin 0° = cos 90° = 0

cos (180 – α) = –cos αsin (180 – α) = sin αcos (90 – α) = sin αsin (90 – α) = cos α

22

32

12

EXAMPLE 30 a. |→v| = 4 cm and the angle between

→v and the positive x-axis is 60°. Find the horizontal and

vertical components of →v and express

→v in terms of

→i and

→j.

b. Find the angle θ between the vector →u = –ñ3⋅

→i +

→j and the positive x-axis.


As we have seen, vectors describe quantities that have both magnitude and direction. They

have a wide range of applications, such as in navigation, mechanics, and engineering.

In navigation, the direction of movement of an object such as a ship is usually given as abearing, that is, as an acute angle measured from due north or due south. For example, the

bearing N 60° E stands for a direction that points 60° to the east of due north. Look at some

more examples of bearings in the figure:

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Check Yourself 91. Express

→v = (3, 2) in terms of standard base vectors.

2. Resolve the vector →u with length 3 cm into its components if the angle between the

positive x-axis and →u is 120°.

3. Find the angle θ between the vector →v = –3⋅

→i – 3ñ3⋅→j and the positive x-axis.

Answers

1. →v = 3 ⋅→i + 2 ⋅

→j 2. →ux = ,

→uy = 3. θ = 240° 3 3

(0, )2

− 3( , 0)

2

C. VECTOR APPLICATIONS

The speed of a moving object along a bearing is called the velocity of the object. We can see

that velocity has direction (a bearing) and magnitude (speed). Therefore we can use a

vector to represent a velocity.

Note that N 60° E is notthe same as E 60° N.

a. We have →v = (v1, v2), where the scalar components are given by v1 = 4⋅cos 60° = 2 and

v2 = 4⋅sin60° = 2ñ3. Therefore, the horizontal component is 2⋅→i and the vertical

component is 2ñ3⋅→j . Therefore,

→v = 2⋅

→i + 2ñ3⋅

→j.

b. From the figure we see that θ has the property that

tan (180° – θ) = .

Thus 180 – θ = 30°, and so θ = 150°.

1 3=

33

Solution

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Let us choose a coordinate system with the origin at

the initial position of the boat as shown in the figure.

Let →u and

→v represent the velocities of the river and

the boat, respectively.

Then →u = 6⋅

→i and |

→v | = 12,

→v = (12⋅cos θ)⋅

→i + (12⋅sin θ)⋅

→j

where the angle θ is as shown in the figure.

Solution

EXAMPLE 32 A woman wants to move by boat from one shore of a straight river to the point directlyopposite on the other shore. The speed of the boat in still water is 12 km/h and the river isflowing east at the rate of 6 km/h. In what direction should the woman head the boat in orderto arrive at the point directly opposite?

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In the figure, the vector →u represents the velocity of

wind blowing in the direction N 60° E, and the

vector →v represents the velocity of an airplane flying

through this wind at point A. It is obvious that the

wind affects both the speed and the direction of the

airplane.

From the figure we can see that actual velocity of the

plane (relative to the ground) is given by the vector→w =

→u +

→v.

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We first construct a rectangular coordinate system to

indicate the four directions east, west, north and

south.

Look at the figure. We can draw a vector −→LP to

represent the direction and magnitude of the air

velocity of the plane, and a vector −→OP to represent

the direction and magnitude of the wind velocity. We

can now expect the resultant velocity to be in the

direction of −→OP with a magnitude proportional to the

length of −→OP. Therefore, we have |

−→OP| =

So the velocity of the airplane is approximately 412 km/h along −→OP.

≅2 2400 +100 412.

Solution

EXAMPLE 31 An airplane is headed due west at an air speed of 400 km/h and the wind is blowing from the

north at 100 km/h. Find the resultant velocity of the airplane relative to the ground.

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, so they are parallel.2 1= = 2

112

Solution

EXAMPLE 33 Show that →u = (2, 1) and

→v = (1, ) are parallel.

12

EXAMPLE 34 Find the relation between x and y given A(3, –1), B(2, 3), C(5, –4), D(x, y), and −→CD ||

−→AB.

Check Yourself 10

1. A jet plane is flying in the direction N 20° E with a speed of 500 km/h. Find the east and

north components of the velocity.

2. A vector is 4 cm long and makes an angle of 60° with the positive x-axis. Resolve the

vector into its components.

Answers

1. 500 ⋅ cos70° ⋅→i , 500 ⋅ sin70° ⋅→j 2.→u = 2⋅→i + 2ñ3⋅→j

We know from the geometrical analysis of vectors that two non-zero vectors are parallel if and

only if multiplying one of them by a suitable scalar equals the other, that is,

for any c ≠ 0, →u ≠ 0, and

→v ≠ 0,

→u ||

→v if and only if

→u = c ⋅ →v.

It follows that if →u = (u1, u2) and

→v = (v1, v2), then (u1, u2) = (c ⋅ v1, c ⋅ v2).

So→u ||

→v if aand oonly iif 2211

22 22

.uu

cv v

== ==

The actual velocity of the boat is now→w =

→u +

→v = 6⋅→i + (12⋅cos θ)⋅→i + (12⋅sin θ)⋅

→j

= (6 + 12⋅cos θ)⋅→i + (12⋅sin θ)⋅→j.

Since the woman wants to land at a point directly opposite, her direction should have

horizontal component 0.

6 + 12⋅cos θ = 0

cos θ =

θ = 120°

Therefore, the woman should head the boat in the direction N 30° W.

1–

2

D. PARALLEL VECTORS


Check Yourself 11

1. Show that →u = (–2, –3) and

→v = (4, 6) are parallel.

2. Find the value of k if →u = (1, k) and

→v = (3, 6) are parallel.

Answers

1. check the scalar components 2. k = 2

E. LINEAR COMBINATION OF VECTORS1. Linear Combination of Vectors

−→AB = (2 – 3, 3 – (–1)) = (–1, 4)−→CD = (x – 5, y + 4)−→CD ||

−→AB so

4x – 20 = –y – 4, so 4x + y – 16 = 0.

−−

5 +4= .

1 4x y

Solution

Let us write E(x, y) and F(m, n).

−→BE =

−→EA

−→BE = (x + 3, y + 2)−→EA = (3 – x, 2 – y)

x + 3 = 3 – x

2x = 0

x = 0

y + 2 = 2 – y

2y = 0

y = 0

and−→CF =

−→FA

−→CF = (m – 2, n + 3)−→FA = (3 – m, 2 – n)

m – 2 = 3 – m

2m = 5

m =

n + 3 = 2 – n

2n = –1

n =1

–2

52

Solution

EXAMPLE 35 A triangle ABC has vertices A(3, 2), B(–3, –2), and C(2, –3). E and F are the midpoints of

sides AB and AC respectively. Find the coordinates of E and F.

Definition

Let →u1, →u2, ...,

→uk be vectors in the plane and let c1, c2, ..., ck be scalars.

The expression c1⋅→u1 + c2⋅

→u2 + ... + ck⋅

→uk is called a linear ccombination of the vectors.

Remember!If two parallel vectors

→a

and →b have at least one

point in common, then→a and

→b lie on the same

straight line (they arecollinear).For instance, if−→AB = k⋅

−→BC for some

k ∈ \ {0}, then A, B,

and C are collinear.

For example, w = 2⋅→u – 3⋅→v is a linear combination of vectors.

linear ccombination oof vvectors

Therefore the coordinates are E(0, 0) and 5 1

( ,– ).2 2

F


Let c1, c2, and c3 ∈ . Then →x = c1⋅

→u + c2⋅

→v + c3⋅

→w.

(19, 13) = c1⋅(2, –3) + c2⋅(1, 2) + c3⋅(5, 4)

(19, 13) = (2⋅c1, –3⋅c1) + (c2, 2⋅c2) + (5⋅c3, 4⋅c3)

(19, 13) = (2c1 + c2 + 5c3,– 3c1 + 2c2 + 4c3)

19 = 2c1 + c2 + 5c3 and 13 = –3c1 + 2c2 + 4c3

c2 = 19 – 2c1 – 5c3 and 13 = –3c1 + 2⋅19 – 2⋅2c1 – 2⋅5c3 + 4c3

13 = –3c1 – 4c1 – 6c3 + 38

7c1 + 6c3 = 25

There are many solutions to this equation.

Let us choose c1 = 1, then c2 = 2 and c3 = 3, so→x =

→u + 2⋅→v + 3⋅→w. This is one solution to the problem.

Solution

EXAMPLE 38 Express the vector →x = (19, 13) as a linear combination of

→u = (2, –3),

→v = (1, 2) and

→w = (5, 4).

Let c1, c2 ∈ . Then →v = c1⋅

→u1 + c2⋅

→u2.

(12, 5) = c1⋅(2, 1) + c2⋅(3, 2)

(12, 5) = (2⋅c1, c1) + (3⋅c2, 2⋅c2)

(12, 5) = (2⋅c1 + 3⋅c2, c1 + 2⋅c2)

12 = 2⋅c1 + 3⋅c2 and 5 = c1 + 2⋅c2

12 = 2⋅(5 – 2⋅c2) + 3⋅c1 5 – 2⋅c2 = c1

12 = 10 – 4⋅c2 + 3⋅c2

2 = –c2 (1)

c2 = –2 (2)

c1 = 5 + 4 = 9

Using (1) and (2) gives →v = 9⋅→u1 – 2⋅→u2.

Solution

EXAMPLE 37 Express →v = (12, 5) as a linear combination of the vectors

→u1 = (2, 1) and

→u2 = (3, 2).

→w = 2⋅→u – 3⋅→v = 2⋅(5, –2) – 3⋅(1, 3) = (10, –4) + (–3, –9) = (7, –13)Solution

EXAMPLE 36 Find the vector →w if

→w = 2⋅→u – 3⋅→v,

→u = (5, –2), and

→v = (1, 3).

If each element of a set of

vectors V can be expressed

as a linear combination of

vectors →u and

→v, then we

say that →u and

→v span the

set V.

Note We cannot express any vector as a linear combination of two parallel vectors.


2. Linearly Dependent and Independent Vectors (OPTIONAL)

Definition linearly ddependent aand iindependent vvectors

A set of vectors S = {→v1, →v2, ...

→vk} in a vector space V is called linearly iindependent if the

vector equation c1⋅→v1 +c2⋅

→v2 +...+ck⋅

→vk =

→0 has only the trivial solution c1 =0, c2 =0, ..., ck =0.

If any of ci is different from zero, then the set S is called linearly ddependent.

c1⋅→v + c2⋅

→w =

→0

c1⋅(1, 0) + c2⋅(0, 1) = (0, 0)

(c1, 0) + (0, c2) = (0, 0)

(c1, c2) = (0, 0)

c1 = 0 and c2 = 0

So S is linearly independent by the definition.

Solution

EXAMPLE 39 Show that S = {→v, →w}, is linearly independent if

→v = (1, 0), and

→w = (0, 1).

Let c1, c2, c3 ∈ and c1⋅→u + c2⋅

→v + c3⋅

→w =

→0.

c1⋅(1, 0) + c2⋅(0, 1) + c3⋅(–2, 5) = (0, 0)

(c1, 0) + (0, c2) + (–2c3, 5c3) = (0, 0)

(c1 – 2c3, c2 + 5c3) = (0, 0)

c1 – 2c3 = 0 and c2 + 5c3 = 0

c1 = 2c3 and c2 = –5c3

There are infinitely many solutions to this set of equations. For example, if

c3 = 1, then c1 = 2 and c2 = –5.

These values are non-zero, so the vector set is linearly dependent.

Solution

EXAMPLE 40 Show that S = {→u, →v, →w}, is linearly dependent if

→u = (1, 0),

→v = (0, 1) and

→w = (–2, 5).

Check Yourself 12

1. Express →u = (0, 1) as a linear combination of

→v = (1, 1) and

→w = (–1, 2).

2. Show that →a = (2, 0) and

→b = (0, 2) are linearly independent.

Answers

1. →u = ⋅→v + ⋅→w 2. solve the equation →u = x⋅→v + y⋅→w1

313

Note In a plane, two non-parallel and non-zero vectors are linearly independent but two parallelvectors are linearly dependent.


A. Basic Concepts of Vectors in theAnalytic Plane

11.. Plot the points A(–1, 1), B(2, –1), C(3, 1), and

D(–3, –1) in the plane.

33.. Find the length of −→MN given M(1, 4) and N(–2, –1).

22.. Sketch the position vector of the vector with the

given endpoints.

a. b. c. −→EF

E(0, –3)

F(–4, 2)

−→CD

C(1, –5)

D(0, 2)

−→AB

A(–2, –3)

B(4, –1)

55.. Describe the vector with initial point P and

terminal point Q.

44.. The figure shows the

vectors →u and

→v. Sketch

the following vectors.

a. 2→v b. –

→u

c. →u + →v d. →u – 2 ⋅ →v

e. 2→u +

→v

EXERCISES 2

c. P(3, 2), Q(8, 9) d. P(–1, 3), Q(1, 0)

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a. b.

Project:Use The Geometer’s Sketchpad,

Cabri Geometry, or Javascript

sketchpad to sketch the vectors

2→u, –→v, →u + →v, →u – 3⋅→v and−→PQ,

−→QP if P(3, –4) and Q(4, –3).

We can use computer applications such as TheGeometer’s Sketchpad, Cabri Geometry, orJavascript sketchpad to sketch vectors and solveproblems. We can use an application to aproblem, and then change certain values to seetheir effect. We can also use a computerapplication to add and subtract vectors, and tomultiply a vector by a scalar.

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The screen opposite shows a simple problem illustrated usingThe Geometer’s Sketchpad. Suppose a current flows at acertain velocity w downstream. A boat moves at a constantspeed v. Which direction the boat take in order to reach theother side of the river in the shortest possible time?The Geometer’s Sketchpad calculates the time as we move thepoint H on the screen to set the direction of the boat. Using theSketchad we can move H to find the shortest possible time inthe problem.


66.. Find the vectors →a and

→b if 2⋅→a – 3⋅

→b = (4, 2) and

→a +

→b = (2, 1).

B. Vector Operations

77.. Find →u +

→v, →u –

→v, 2⋅→u, 2⋅→u –

→v,→u + 3⋅→v, and

–4→⋅u + 3⋅→v for the given vectors

→u and

→v.

a. →u = (2, 6),→v = (1, 3)

b. →u = (–2, 3),→v = (8, –2)

c. →u = (1, 0),→v = (0, –2)

d. →u = →i

→v =

→j

e. →u = –3→ý +→j

→v =

→i –→j

f. →u = 7

→i + 5

→j

→v =

→j – →i

88.. Find |→u|, |

→v|, |2⋅→v|, | ⋅→v|, |

→u +

→v|, and |

→u – →v|

for the given vectors →u and

→v.

a. →u = 3⋅→i + →j

→v = –

→i + 2⋅→j

b. →u = 2⋅→i – →j

→v = –

→i – →j

c. →u = (2, 3),→v = (0, 1)

d. →u = (3, 4)→v = (2, 5)

13

99.. Find the horizontal and vertical scalar components

of the vector with the given length and angle with

the positive x-axis. Express the vector in terms of

standard base vectors.

a. |→v | = 20, θ = 30°

b. |→v | = 30, θ = 120°

c. |→v | = 1, θ = 225°

d. |→u | = 80, θ = 135°

e. |→v | = 4, θ = 10°

f. |→u | = ñ3, θ = 300°

1100.. Given −→AB = –7⋅→i + 2⋅→j and B(3, 11), find the

coordinates of point A.

1111.. Given −→AB = 5⋅→i + 6⋅→j and B(–4, 8), find the

coordinates of point A.

1122.. →u = 3⋅→i + 4⋅→j and →v = 4⋅→i +

→j are given. Which

vector is the longest?

1133.. Given →u = 3⋅→i + 4⋅→j, calculate | ⋅→u –

→i|.

12

1144.. A jogger runs with a constant speed of 6 km/h in

the opposite direction to the wind. Find the

actual velocity of the jogger if the wind blows at a

speed of 3 km/h due west.

1155.. The jogger in question 14 above heads due north.

What is the actual velocity of the jogger?

C. Vector Applications

1166.. A swimmer can swim with a velocity of 1.4 m/s in

still water. The current of a river is 1 m/s due

west. The swimmer swims due west in the river.

Find the swimmer’s actual velocity.


2200.. Show that →u = (a, b) and

→v = (2a, 2b) are

parallel.

2222.. In a triangle ABC, the vertices are A(–2, 3),

B(0, 1) and C(4, 1). Points D(–1, 2) and E(1, 2)

are on the sides AB and AC respectively.

Show that −→DE ||

−→BC.

2233.. In a triangle ABC, the vertices are A(1, 3),

B(2, 1), and C(3, 2). E( , 2) is on the side AB.

Find the coordinates of F if F is on AC and−→EF ||

−→BC.

32

2211.. →u || →v, →u = (1, k – 3), and

→v = (k, k – 4) are given.

Find the value of k.

E. Linear Combination of Vectors

2244.. →a = (13, 2), →b = (11, 6) and

→c = (1, 1) are given.

If →a +

→b +

→u = 4

→c – →u, find

→u.

2255.. Show that the vectors →a = (1, 2),

→b = (3, 1) and

→c = (1, 1) are linearly dependent.

Show that any three non-parallel, non-zero

vectors in a plane are linearly dependent.

2277..

2266.. Let →u, →v and

→w be unit vectors with angles of

respectively with the positive x-axis.

Express →u as a linear combination of

→v and

→w .

π π π3, ,

4 3 4

→a = (1, –2),

→b = (3, –1), and

→c = (–1, 7) are given.

Express →a +

→b +

→c in terms of

→a and

→b.

2288..

D. Parallel Vectors

1188.. A pilot heads his airplane due west. The airplane

has a speed of 425 km/h in still air. The wind is

blowing due north with a speed of 40 km/h.

a. Find the actual velocity of the airplane.

b. Find the actual speed of the plane.

1177.. The swimmer in question 16 swims due east

(against the current). Find the swimmer’s actual

velocity.

A boat heads in the direction N 36° E along a

river. The speed of the boat in still water is

16 km/h. The river is flowing directly south. It is

observed that the actual direction of the boat is

directly east. Find the speed of the river current

and the actual speed of the boat.

1199..

1 ANALYSIS OF VECTORS GEOMETRICALLY

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