LECTURE 5 Vectors in Plane and Space S1 2015-2016

IIUM, Faculty of Engineering, Engineering Calculus II, MTH 1212

Department Engineering in Science Chapter II.1: Vectors & Geometry in Space

Sections 2, Venue: E1416 Sections 3, Venue: E1310 67 Abdurahim Okhunov [email protected] S1, 2015/2016

2 VECTORS and GEOMETRY in SPACE

2.1 Vectors in plane and space. Definition. Application; 2.2 Dot ant Cross product ; 2.3 Line & planes in space. Quadric surfaces. 2.43 Geometric and Algebraic vectors.

In this chapter we introduce vectors and coordinate systems for two and three

dimensional space. From the basic mathematics we know that, many measurable

quantities, such as length, area, volume, mass and temperature can be completely

described by specifying their magnitude (). Such as quantities called scalar.

Other quantities, such as velocity, force, and acceleration, require () both a

magnitude () and a direction () for their description. Such as quantities

called vector quantities, or simply vectors.

To locate a point in a plane, two numbers are necessary. We know that any point

in the plane can be represented as an ordered pair ,a b of real numbers, where a is

the x coordinate and b is the y coordinate. For this reason, a plane is called two-

dimensional. To locate a point in a space, three numbers are required. We represent any

point in space by an ordered triple , ,a b c of real numbers.

In order to represent points in space, we first choose a fixed point 0 (the origin)

and three directed lines through 0 that are perpendicular to each other, called the

coordinate axes and labelled the x axis, y axis, and z axis. Usually we think of

the x and y axes as being horizontal and the z axis as being the vertical, and we

draw the orientation of the axes as in the Figure 2.1, below. The three coordinate axes

determine the three coordinate planes illustrated in the Figure 2.2:

z

x

y

Figure 2.1 Figure 2.2

xy

plane

yz

plane

xz

plane

y

x

z




The xy plane is the plane that contains the x and y axes; the yz plane

contains the y and z axes, and xz plane contains the x and z axes. These

three coordinate planes divide space into eight parts, called octant, in the foreground, is

determined by the positive axes.

Vectors in the plane:

The term vector is used by scientists to indicate a quantity (such as displacement

or velocity or force) that has both magnitude and direction. A vector is often represented

by an arrow or a directed line segment. The length of the arrow represents the

magnitude of the vector and the arrow points in the direction of the vector. We denote a

vector by printing a letter in boldface v or putting an arrow above the letter v .

Definition and Application We begin by considering the Cartesian plane with the familiar x and y axes.

A vector is a directed line segment that corresponds to a displacement from one point

A to another point B , see Figure 2.3.

The vector from A to B is denoted by AB

, the point A is called its initial point,

or tail, and the point B is called its terminal point, or head. Often, a vector is simply

denoted by a single boldface, lowercase letter such v .

The set of all vectors with two components is denoted by 2 (where denotes

the set of real numbers from which the components of vectors in 2 are chosen). Thus,

3 5, 3. , 2, , and 5

3, 4

are all in 2 .

x

y

A

B

0 Figure 2.3




The set of all points in the plane corresponds to the set of all vectors whose tails

are at the origin 0 . To each point A , there corresponds the vector 0A

a , to each vector

a with tail at 0 , there corresponds its head A .

It is natural to represent such vectors using coordinates. For example, in Figure

2.4, ,3 2A and we write the vector 0 3 2,B

a using square brackets. Similarly,

the other vectors in Figure 2.4 are 1,0 3B

b and ,0 2 1C

c .

Example 1 (Vector in Plane):

If 1, 2A and ,3 4B are two pointes in plane, find AB

and draw it:

(i) In standard position and;

(ii) Draw with its tail at the point ,2 1C .

Solution:

We compute 3 1 4, 2 4 2,AB

a .

If AB

is then translated to CD

, where

,2 1C , then we must have

2 4 1 2 6, 1,D .

(-1, 2)

(6, 1)

y

x

B

A

(2,-1)

[4, 2]

C

D

(3, 4)

Figure 2.5

0 C

y

x

(-1, 3)

(3, 2)

(2, -1)

A

B

a b

c

Figure 2.4




We will divide vectors to three positions:

1. Vectors on the one dimension space or on the line; 2. Vectors on the plane; 3. Vectors on the space.

Vectors on the line:

Vectors on the plane: Vectors on the space:

Same operations with vectors:

Let given vectors 4,3, 1a and 1 2 3, , b

Addition: 3 1 1 2 4, , , ,3 4 1 7 a b

Distract: 3 1 1 2 4, , , ,3 2 3 1 a b Product:

a) Dot product 3 1 1 2 4 3 3 2 12 13 a b

b) Cross product 1 4 3 4 3 1

3 1 42 3 1 3 1 2

1 2 3

i j k

i j k

a b

Multiplication to number: , , ,5 5 3 1 4 5 3 5 1 5 4 15 5, , , 20 a

0 1 2 3 -1 -2 -3

(a, b, c)

(a, 0, c)

(a, 0, 0)

(a, b, 0)

(0, b, 0)

(0, b,c)

(0,0,c)

0 j

k

i

z

y x

D

C

B

A

a

b

y

x




Combining Vectors

Suppose a particle moves from A to B , so its displacement vector is AB

.

Then the particle changes direction and moves from B to C , with displacement vector

BC

as in Figure 2.6.

The combined effect of these displacements is that the particle has moved from A

to C . The resulting displacement vector AC

is called the sum of AB

and BC

and we

write AC AB BC

.

In general, if we start with vectors u and v , we first move v so that its tail

coincides with the tip of u and define the sum of u and v as follows:

Definition of Vector Addition:

If u and v are vectors positioned so the initial point of v is at the terminal point

of u , then the sum u + v is the vector from the initial point of u to the terminal point of

v .

This definition is sometimes called the Triangle Law, why? Its you can see from

Figure 2.7.

In Figure 2.8, below we start with the same vectors u and v as in Figure

2.7 and draw another copy of v with the same initial point as u . Completing the

C

B

A

AC

BC

AB

Figure 2.6

C

B

A

u v v

u

Figure 2.7

u C




parallelogram, we see that u + v = v +u . This also gives another way to construct the

sum. If we place u and v so they start at the same point, then u + v lies along the

diagonal of the parallelogram with u and v as sides. This is called the Parallelogram

Law.

Example 2 (Vectors sum in Plane):

Draw the sum of the vectors a and b as shown in Figure 2.9.

Solution:

The first we translate vector b and place its tail at the tip a , being careful to draw

a copy of b that has the same length and direction, see Figure 2.10.

Then we draw the vector a b , starting at the initial point of a and ending at the

terminal point of copy of b . Alternatively, we could place b so it starts where a starts

and construct a b by the Parallelogram Law.

u

v u + v v

u

B

C

Figure 2.8

a

b

Figure 2.9

a+b b

a

b

a

Figure 2.10

b

a

b




Example 3 (Vectors sum in Plane):

If 3, 1A and 1, 4B , compute and draw A B .

Solution:

We compute 3 1, 4 4 3,1A B . This vector is drawn using the headto

tail rule, see Figure 2.11.

Subtraction:

By the difference u v of two vectors we mean u v u v .

So we can construct u v by first drawing the negative of v , v , and then

adding it to u by the Parallelogram Law as in Figure 2.12(a). Alternatively,

since v u v u , the vector u v , when added to v , gives u . So we could

construct u v as in Figure 2.12(b) by means of the Triangle Law.

Figure 2.12 a) Figure 2.12 b)

Figure 2.11

y

A+B

)

B

A

x

u-v

u

-v

v u-v

u

v




Example 4 (Vectors in Plane):

If 2,1a and 3 1, b , then what are distract this vectors?

Solution:

,1 3 2 1 4 1, a b .

The definition of subtraction in Example 4 also agree with the way we calculate a

vector such as AB

. If the point A and B correspond to the vectors a and b in

standard position, then AB

b a , as shown in Figure 2.13.

[Observe that the headtotail rule applied to this

diagram gives the equation a b a b . If we had

accidentally drawn b a with its head at A instead

of at B , the diagram would have read

b a b a , which is clearly wrong.]

Example 5: If a and b are the vectors shown in Figure below draw 2a b .

Solution:

We first draw the vector 2 b pointing in the

direction opposite to b and twice as long. We place it

with its tail at the tip of a and then use the Triangle

Law to draw 2 a b as in Figure.

a

b

b

a -2b

a

-2b

A

B

y

b

a

x

Figure 2.13




Scalar Multiplication

It is possible to multiply a vector by a real number c . (In this context we call the real

number c a scalar to distinguish it from a vector). For instance, we want 2v to be the

same vector as v v , which has the same direction as v but is twice as long. In general,

we multiply a vector by a scalar as follows.

Definition of Scalar Multiplication:

If c is a scalar and v is a vector, then the scalar multiple cv is the vector whose

length is c times the length of v and whose direction is the same as v if c 0> and

is opposite to v if c 0< . If c = 0 or 0v then c 0 v .

This definition illustrated in Figure 2.14.

We see from the Figure 2.14, that real numbers work like scaling factors

here, thats why we call them scalars . Notice that two nonzero vectors are

parallel is they are scalar multiples of one another. In particular, the vector

( 1) v v= has the same length as v but points in the opposite direction. We

call it negative of v .

Example 6 (Vectors in plane):

If 2 4, v , compute and draw 2v , 12

v , and 2 v .

-v

1

2v

2v

v

0c > 0c <

Figure 2.14

- 1

2v

-2v




Solution:

We calculate as follows:

1 1 1 1

2 2 2 2

2 2 2 4 2 2 2 4 4 8

2 4 2 4 1 2

2 2 2 4 2 2 2 4 4 8

, , ,

, , ,

, , ,

v

v

v

.

These vectors are shown in Figure 2.15.

Position vector

Since the location of the initial point is irrelevant, we typically draw vectors with

their initial point located at the origin. Such a vector is called a position vector.

For the position vector a with initial point at the origin and terminal point at the

point 1 2,a aA , we denote the vector by 1 2,0a A a a (see the Figure 2.16).

We call 1a and 2a the components of the vector a ; 1a is the first component and

2a is the second component. Be careful to distinguish between the point 1 2,a a and the

position vector 1 2,a a .

y

x 0

1 2,A a a

a

2a

1a

2a

1a

a

Figure 2.16

(1/2)v

y

x

Figure 2.15

2v

-2v




Not that, from the Figure 2.16, that the magnitude of the position vector we a follows

directly from the Pythagorean Theorem.

So, we have

2 21 2a a a Magnitude of a vector (2.1)

Observe that if we multiply any vector (with any direction) by the scalar c 0 , we get

a vector with zero length, the zero vector.

0,00 .

Finally, we define the additive inverse a of a vector a in the expected way:

1 2 1 2 1 2, 1 , ,a a a a a a a .

Not that, this says that the vector a is a vector with the opposite direction as a and since

1 2 1,1 a a a a a .

a has the same length as a .

Example 7 (Finding Position vector):

Find the vector with:

a) Initial point at 2 3,A and terminal point at 3 1,B ;

b) Initial point at B and terminal point at A .

c) If 2 4, v , compute and draw 2 v , 1

2 v , and 2 v .




Solution:

a) We show this graphically in Figure 2.17.

,2 4,3 1 3 1AB .

b) Similarly, the vector initial point at

3 1,B and terminal point at 2 3,A is

given by

,3 4,2 3 1 1BA .

Theorem (Unit vector):

For any nonzero position vector 1 2,a aa , a unit vector having the same

direction as a is given by 1

u aa .

The process of dividing a nonzero vector by its magnitude is sometimes

called normalization.

Figure 2.17

AB

3 1,B

2 3,A

1, 4

x

-4

-1

y




Vectors in Space:

We now extend several ideas from the two dimensional Euclidean space, 2 , to

the three dimensional Euclidean space, 3. We specify each points in three

dimensions by an ordered triple , ,a b c , where the coordinates ,a b and c represent

the distance from the orign along each of three coordinates axes , andx y z , as

indicated in Figure 2.18.

Everything we have just done extends easily to three dimensions. The set of all

ordered triples of real numbers is denoted by 3 . Points and vectors are located using

three mutually perpendicular coordinate axes that meet at the origin O .

To locate the point , ,a b c in 3 , where ,a b and c are all positive, first move

along the x axis a distance of a units from the origin. This will put you at the point

, ,a 0 0 . Continuing from this point, move parallel to the y axis a distance of b

units from , ,a 0 0 . This leaves you at the point , ,a b 0 . Finally, continuing from this

point and moving c units parallel to the z axis leaves you at the point , ,a b c , (see

Figure 2.18, above).

For example a point such as ,1 3, 2A can be located as follows: First travel 1

unit along the x axis, then move 2 units parallel to the y axis, and finally move 3

units parallel to the z axis. The corresponding vector 3,1, 2a is then OA , as

shown in Figure 2.19.

(a, b, c)

(a, 0, c)

(a, 0, 0)

(a, b, 0)

(0, b, 0)

(0, b,c)

(0,0,c)

0 j

k

i

z

y x

Figure 2.18

0

y x

z




Recall that in 2 , the coordinate axes divide the xy plane into four quadrants. In

a similar fashion, the three coordinate planes in 3 (the xy plane, the yz plane

and xz plane) divide space into eight octants, (see Figure 2.2). The first octant is the

one with ,0 0x y and 0z . We do not usually distinguish among the other seven

octants.

Magnitude (Length) of a Vector:

The magnitude of a vector is simply its length. By the Pythagorean

Theorem and Figure 2.20 we have the following definition.

Figure 2.19

z

0

A(1,2,3

)

a

3

1 y

x

2

y

x

0

1y

2y

1v

2v

1x

2

x

1 1,x yP

2 2,x yQ

2 1x x

2 1y y

1 2,v vv

2 1 2 1

2 22 1 2 1

,Q

Q

P x x y y

P x x y y

Figure 2.20




Definition: Given the point 1 1,P x y and 2 2,Q x y the magnitude or length

(or norm) of 2 1 2 1

,P x x y yQ , denoted PQ is the distance

between P and Q : 2 2

2 1 2 1P x x yQ y .

The magnitude of the position vector 1 2,v vv is 2 2

1 2v v v .

By using the distance formula to compute the length of a segment OP , we

obtain the following formulas:

The length of the twodimensional vector 1 2,a aa is 2 2

1 2a a a

The length of the threedimensional vector 1 2 3, ,a a aa is

2 2 2

1 2 3a a a a

Example (Vectors length, LA,DP, p.17):

For example given the vector 2 22 3 2 3 1, 3 .

Magnitude of position vector a is: 2 2

1 2a a a . If a 0 then a 0

this is called a zero vector. Any vector with magnitude of 1 i.e. u 1 is

called a unit vector.

We will occasionally () find it convenient to write vectors in terms

of some standard vectors,

we define the standard

basis vectors i and j by

, 0= 1i and , 1= 0j and

this can be extended to

three dimensional case.

i

0

j

(0,1)

x

y

(1,0)

Figure 2.21

j

k

i

z

y

x




Notice that 1i j (see the Figure 2.21)

Magnitude of 1 2 3, ,a a a a is 2 2 21 2 3a a a a

Vector addition:

1 2 3 1 2 3

1 1 2 2 3 3

a b

, ,

a a a b b b

a b a b a b

Standard basis vectors: , , , , ,1 0 0 0 1 0 0 1, , ,0i j k

Any vector 1 2 3 1 2 3, , ,a ,i ja a a a a a k

Finding the equation of a sphere of radius r centred at (a, b, c): the sphere consists of all

points (x, y, z) whose distance from (a, b, c) is r:

2 2 2

2 2 2 2

x y z

x

a b

rz

r

y

c

a b c

Distance:

The distance o between two vectors is the direct analogue of the distance between two

points on the real number line or two points in the Cartesian plane. On the number line (see

Figure 2.23), the distance between the numbers a and b is given by a b . This distance is

also equal to 2

a b , and its two-dimensional generalization is the familiar formula for

the distance d between points 1 2,a a and 1 2,b b namely,

2 2

1 1 2 2d a b a b .

0 1 2 3 -1 -2 -3

Figure 2.23

Figure 2.22

2a

(a1+b1,

a2+b2)

2b

2a

1b

b a+b

a

0 1b

y

x




Angle:

The dot product can also be used to calculate the angle between a pair of vectors. In

2 or 3 , the angle between the nonzero vectors a and b will refer to the angle

determined by these vectors that satisfies 00 180 . (see Figure 2.24).

We can compute the distance between two points in 3 by thinking of this as

essentially a two dimensional problem. For any two points 1 1 1, ,1P x y z and

2 2 2, ,2P x y z in 3, first locate the point , ,3 3 3 3P x y z and observe that the three point

are the vertices of a right triangle, with the right angle at the point 3P , see Figure 2.25.

The Pythagorean Theorem then says that the distance between 1P and 2P , denoted

,1 2d P P , satisfies

2 22

, , ,1 2 1 3 2 3P Pd d dP P P P . (2.1)

Notice that, 2P lies directly above 3P (or below, if 2 1z z ). So that

11 1 1, ,P x y z

22 2 2, ,P x y z

z

0

y

x

13 2 2, ,P x y z Figure 2.25

b

a a

b

a

b

a b

Figure 2.24




Since 1P and 3P both lie in the plane 1z z , we can ignore the third coordinates of

these point and use the usual two dimensional distance formula:

2 2 2, , , , , ,1 2 1 1 1 2 2 2 2 1 2 1P P x y z x y z x x yd d y

From formula (2.1), we have that

2 22

222 2

2 2 2

, , ,

.

1 2 1 3 2 3

2 1 2 1 2 1

2 1 2 1 2 1

P P P P P P

x x y y z z

d d d

x x y y z z

Taking the square root of both sides gives us the distance formula for 3 .

2 2 2, , , , ,1 1 1 2 2 2 2 1 2 1 2 1x y z x y z x x y y z zd (2.2)

Example 7 (Vectors in space, computing distance in 3):

Find the distance between the point 3,1, 5A and ,5 3, 2B .

Solution:

From the formula (2.2) we have

22 2

22 2

1 3 5 5 2 3 5 1 2 3 3 5

4 5 8 1

, , , , ,

.05

d

Three dimension coordinate system (vectors in 3):

As in two dimension vectors in three dimensional space have both direction and

magnitude. We again visualize vectors as directed line segments joining two points. A vector

v is represented by any directed line segment with the appropriate magnitude and direction.

In three dimensional coordinate, the vector 1 2 3A0 , ,a a a a is the position vector of

the point 1 2 3, ,a a aA . The position vector a with terminal point at 1 2 3, ,a a aA is

denoted by 1 2 3, ,a a a and is shown in Figure 2.26.




Lets consider any other representation AB of a , where the initial point is 1 1 1, ,x y zA

and the terminal point is 2 2 2, ,x y zB . Then we must have 1 1 2 1 2 2, ,x x ya a y

and 1 3 2z za and so 1 2 1 2 2 1, ,a ax x y y and 3 2 1a z z .

Thus we have the following result:

Definition: Given the point 1 1 1, ,x y zA and 2 2 2, ,x y zB , the vector a with

representation AB is 2 1 2 1 2 1, ,x x y y z z a . (2.3)

The rules of algebra established for vectors in 2V hold verbatim in 3V , as seen in

Theorem 2.1.

Theorem 2.1 (Algebraic properties of Vectors in n ):

If a , b and w are vectors in 3V and let c and d are scalars in n , then:

1. a b b a Commutativity 2. a b w a b w Associativity 3. 0 a a Zero vector 4. 0 a a Additive law 5. =c c c a b a b Distributivity law 6. c d c= d a a a Distributivity law 7. =c d c d a a Distributivity law 8. 1 =a a Multiplication by 1 and 9. 0 =a a Multiplication by 0.

Figure 2.26

1 1 1, ,x yA z i

Position

vector of P

0

y

x

z

1 1 1 2 1 3, ,B a a ax y z j

a) b)

1 2 3, ,a a aA

1 2 3, ,a a aa

z

0

y

x




We leave the proof of Theorem 2.1 as an exercise.

The standard basis in 3V consists of three unit vectors, each lying along one of the

three coordinate axes. Let

[ , , ]; [1 0 0 0, ]1 0,i j and [ , ]0 0 1,k

These vectors ,i j and k are called the standard basic vectors. They have length 1 and

point in the directions of the positive , ,x y and z axis

(see Figure 2.27). Similarly, in two dimensions we define

0[ ]1,i and 1[ ]0,j .

If 1 2 3, ,a a aa , then we can write

1 2 3 1 2 3

1 2 3

1 2 3

, , , , , , , ,

1, , , 1,

0 0 0 0 0 0

0 0 0 0 ,0 0 , 1

a a a a a a

a a a

a a a

a

i j k

How do we add vectors algebraically?

In Figure 2.29 shows that if 1 2

a ,a a and 1 2

b ,b b , then the sum is

1 1 2 2,a b a b a b , at least for the case where the components are positive. In

other words, to add algebraic vectors we add their components. Similarly, to subtract

vectors we subtract components. From the similar triangles in Figure 2.28, we see that

the components of c a are 1

c a and 2

c a . So to multiply a vector by a scalar we

multiply each component by that scalar.

If 1 2

a ,a a and 1 2

b ,b b , then

1 1 2 2,a b a b a b

1 1 2 2,a b a b a b

1 2,aac c c a

Similarly, for three-dimensional vector,

a

a

1

a

2

ca

ca

1

ca2

Figure 2.28

Figure 2.27

j

k

i

z

y

x




1 2 3 1 2 3 1 1 2 1 3 3, , , , , ,a a a b b b a b a b a b

1 2 3 1 2 3 1 1 2 1 3 3, , , , , ,a a a b b b a b a b a b

1 2 3 1 2 3, , , ,a a a ac c a ac c

Everything we have just done extends easily to three dimensions. The set of all

ordered triples of real numbers is denoted by 3. Points and vectors are located using

three mutually perpendicular coordinate axes that

meet at the origin O . A point such as ,1 3, 2A

can be located as follows: First travel 1 unit along

the x axis, then move 2 units parallel to the

y axis, and finally move 3 units parallel to the

z axis. The corresponding vector 3,1, 2a

is then OA , as shown in Figure 2.29.

Example 7:

If 3,4, 0a and 1 5,2, b , find a and the vectors a b , a b , 3 b , and

2 5 a b.

Solution:

To find the length vector a we will use 2 2 2

1 2 3a a a a formula, so

2 2 24 0 25 5a 3

4 0 3 2 1 5

4

, , , ,

,2 0 1,3 5 8,1,2

=

=

a b

, , , ,4 0 3 2 1 5

4 2 0 1 3 5, 6 ,1 2, ,

=

=

a b

, ,

,

3 3 2 1 5

3 2 3 1 3 5 6 3 15, , ,

=

=

b

A(1,2,3)

a

3

y

x

2

1

z

Figure 2.29




2 5 2 4 0 3 5 2 1 5

8 0 6 10 5 25

, , , ,

, , , , , 5,2 31

=

=

a b

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LECTURE 5 Vectors in Plane and Space S1 2015-2016