Date post: | 06-Jan-2016 |
Category: |
Documents |
Upload: | atenhyunae |
View: | 29 times |
Download: | 0 times |
of 22
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
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 68 Abdurahim Okhunov [email protected] S1, 2015/2016
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
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 69 Abdurahim Okhunov [email protected] S1, 2015/2016
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
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 70 Abdurahim Okhunov [email protected] S1, 2015/2016
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
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 71 Abdurahim Okhunov [email protected] S1, 2015/2016
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
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 72 Abdurahim Okhunov [email protected] S1, 2015/2016
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
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 73 Abdurahim Okhunov [email protected] S1, 2015/2016
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
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 74 Abdurahim Okhunov [email protected] S1, 2015/2016
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
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 75 Abdurahim Okhunov [email protected] S1, 2015/2016
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
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 76 Abdurahim Okhunov [email protected] S1, 2015/2016
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
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 77 Abdurahim Okhunov [email protected] S1, 2015/2016
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 .
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 78 Abdurahim Okhunov [email protected] S1, 2015/2016
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
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 79 Abdurahim Okhunov [email protected] S1, 2015/2016
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
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 80 Abdurahim Okhunov [email protected] S1, 2015/2016
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
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 81 Abdurahim Okhunov [email protected] S1, 2015/2016
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
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 82 Abdurahim Okhunov [email protected] S1, 2015/2016
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
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 83 Abdurahim Okhunov [email protected] S1, 2015/2016
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
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 84 Abdurahim Okhunov [email protected] S1, 2015/2016
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.
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 85 Abdurahim Okhunov [email protected] S1, 2015/2016
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
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 86 Abdurahim Okhunov [email protected] S1, 2015/2016
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
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 87 Abdurahim Okhunov [email protected] S1, 2015/2016
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
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 88 Abdurahim Okhunov [email protected] S1, 2015/2016
2 5 2 4 0 3 5 2 1 5
8 0 6 10 5 25
, , , ,
, , , , , 5,2 31
=
=
a b