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# 10 VECTORS AND THEGEOMETRY OF SPACE

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THREE-DIMENSIONAL COORDINATE SYSTEMS

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 of real numbers, where is the -coordinate and is the -coordinate. For this reason, a plane is called two- dimensional. To locate a point in space, three numbers are required. We represent any point in space by an ordered triple of real numbers.

In order to represent points in space, we ﬁrst choose a ﬁxed point (the origin) and three directed lines through that are perpendicular to each other, called the coordinate axes and labeled the -axis, -axis, and -axis. Usually we think of the

- and -axes as being horizontal and the -axis as being vertical, and we draw the ori- entation of the axes as in Figure 1. The direction of the -axis is determined by the right-hand rule as illustrated in Figure 2: If you curl the ﬁngers of your right hand around the -axis in the direction of a counterclockwise rotation from the positive -axis to the positive -axis, then your thumb points in the positive direction of the -axis.

The three coordinate axes determine the three coordinate planes illustrated in Fig- ure 3(a). The -plane is the plane that contains the - and -axes; the -plane con- tains the - and -axes; the -plane contains the - and -axes. These three coordinate planes divide space into eight parts, called octants. The ﬁrst octant, in the fore- ground, is determined by the positive axes.

Because many people have some difﬁculty visualizing diagrams of three-dimen- sional ﬁgures, you may ﬁnd it helpful to do the following [see Figure 3(b)]. Look at any bottom corner of a room and call the corner the origin. The wall on your left is in

FIGURE 3 (a) Coordinate planes y z

x

O

yz-plane

xz-plane

(b) z

O

right wall

left wall y

x

z x xz

z y

yz y

x xy

zx y

90 z

z z y

x

z y

x O

O

a, b, c

y b

x

a, b a

### VECTORS AND THEGEOMETRY OF SPACE

In this chapter we introduce vectors and coordinate systems for three-dimensional space.This will be the setting for the study of functions of two variables in Chapter 11 because the graph of such a function is a surface in space. In this chapter we will see that vectors provide particularly simple descriptions of lines, planes, and curves.We will also use vector-valued functions to describe the motion of objects through space. In particular, we will use them to derive Kepler’s laws of plane- tary motion.

## 10

517 FIGURE 2

Right-hand rule O

z

y x

FIGURE 1 Coordinate axes

x

z

y

(2)

the -plane, the wall on your right is in the -plane, and the ﬂoor is in the -plane.

The -axis runs along the intersection of the ﬂoor and the left wall. The -axis runs along the intersection of the ﬂoor and the right wall. The -axis runs up from the ﬂoor toward the ceiling along the intersection of the two walls. You are situated in the ﬁrst octant, and you can now imagine seven other rooms situated in the other seven octants (three on the same ﬂoor and four on the ﬂoor below), all connected by the common corner point .

Now if is any point in space, let be the (directed) distance from the -plane to let be the distance from the -plane to and let be the distance from the -plane to . We represent the point by the ordered triple of real numbers and we call , , and the coordinates of ; is the -coordinate, is the -coordi- nate, and is the -coordinate. Thus to locate the point we can start at the ori- gin and move units along the -axis, then units parallel to the -axis, and then

units parallel to the -axis as in Figure 4.

The point determines a rectangular box as in Figure 5. If we drop a per- pendicular from to the -plane, we get a point with coordinates called the projection of on the -plane. Similarly, and are the projec- tions of on the -plane and -plane, respectively.

As numerical illustrations, the points and are plotted in Figure 6.

The Cartesian product is the set of all or-

dered triples of real numbers and is denoted by . We have given a one-to-one cor- respondence between points in space and ordered triples in . It is called a three-dimensional rectangular coordinate system. Notice that, in terms of coor- dinates, the ﬁrst octant can be described as the set of points whose coordinates are all positive.

In two-dimensional analytic geometry, the graph of an equation involving and is a curve in . In three-dimensional analytic geometry, an equation in , , and rep- resents a surface in .

EXAMPLE 1 What surfaces in are represented by the following equations?

(a) (b)

SOLUTION

(a) The equation represents the set , which is the set of all points in whose -coordinate is . This is the horizontal plane that is parallel to the xy-plane and three units above it as in Figure 7(a).

z 3

3 z  3 x, y, z

z  3

y 5

z  33

V

3 x y z

2 x y

3

a, b, c

P

3

### 

x, y,z  ⺢

FIGURE 6

(3, _2, _6)

y z

x

0

_6 _2 3 _5

y z

x 0

(_4, 3, _5) 3

_4 (0, 0, c)

R(0, b, c) P(a, b, c)

(0, b, 0) z

y x

0 S(a, 0, c)

Q(a, b, 0) (a, 0, 0)

FIGURE 5

3, 2, 6

4, 3, 5

xz yz

P

Sa, 0, c

R0, b, c

xy P

a, b, 0

Q xy

P Pa, b, cz c

y b

x a

O

a, b, c

z c

y b x

a P c

b a

a, b, c

P P

xy

c P,

xz b

P,

yz a

P O

z y

x

xy yz

xz

FIGURE 4 z

y x

O

b

a c

P(a, b, c)

(3)

(b) The equation represents the set of all points in whose -coordinate is 5. This is the vertical plane that is parallel to the -plane and ﬁve units to the

right of it as in Figure 7(b).

NOTE When an equation is given, we must understand from the context whether it represents a curve in or a surface in . In Example 1, represents a plane in , but of course can also represent a line in if we are dealing with two- dimensional analytic geometry. See Figure 7, parts (b) and (c).

In general, if is a constant, then represents a plane parallel to the -plane, is a plane parallel to the -plane, and is a plane parallel to the -plane.

In Figure 5, the faces of the rectangular box are formed by the three coordinate planes (the -plane), (the -plane), and (the -plane), and the

planes , , and .

EXAMPLE 2 Describe and sketch the surface in represented by the equation .

SOLUTION The equation represents the set of all points in whose - and -coor- dinates are equal, that is, . This is a vertical plane that intersects the -plane in the line , . The portion of this plane that lies in

the ﬁrst octant is sketched in Figure 8.

The familiar formula for the distance between two points in a plane is easily extended to the following three-dimensional formula.

DISTANCE FORMULA IN THREE DIMENSIONS The distance between the points and is

To see why this formula is true, we construct a rectangular box as in Figure 9, where and are opposite vertices and the faces of the box are parallel to the coor- dinate planes. If and are the vertices of the box indicated in the ﬁgure, then

BP2



z2 z1

AB



y2 y1

P1A



x2 x1

### 

Bx2, y2,z1 Ax2, y1,z1

P2

P1

P1P2

### 

 sx2 x12 y2 y12 z2 z12

P2x2, y2,z2

P1x1, y1,z1

P1P2

z  0 y x xy

x, x, z

### 

x ⺢, z  ⺢

y

3 x

y x3

V

z  c y b

xx a 0 yz y 0 xz z  0 xy z  k xy

xz

y k k x k yz

2 y 5

323 y 5

xz3 y

y 5

FIGURE 7 (c) y=5, a line in R@

0 y

5

x

(b) y=5, a plane in R#

(a) z=3, a plane in R#

y 0

z

x 5

0 z

x y 3

SECTION 10.1 THREE-DIMENSIONAL COORDINATE SYSTEMS 519

0

y z

x FIGURE 8

The plane y=x

FIGURE 9 0 z

y x

P¡(⁄, ›, z¡)

A(¤, ›, z¡)

P™(¤, ﬁ, z™)

B(¤, ﬁ, z¡)

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Because triangles and are both right-angled, two applications of the Pythag- orean Theorem give

and

Combining these equations, we get

Therefore

EXAMPLE 3 The distance from the point to the point is

EXAMPLE 4 Find an equation of a sphere with radius and center . SOLUTION By deﬁnition, a sphere is the set of all points whose distance from is . (See Figure 10.) Thus is on the sphere if and only if . Squaring both sides, we have or

The result of Example 4 is worth remembering.

EQUATION OF A SPHERE An equation of a sphere with center and radius is

In particular, if the center is the origin , then an equation of the sphere is

EXAMPLE 5 Show that is the equation of a sphere, and ﬁnd its center and radius.

SOLUTION We can rewrite the given equation in the form of an equation of a sphere if we complete squares:

Comparing this equation with the standard form, we see that it is the equation of a sphere with center 2, 3, 1and radius s8  2s2.

x  22 y  32 z  12 8

x2 4x  4  y2 6y  9  z2 2z  1  6  4  9  1 x2 y2 z2 4x  6y  2z  6  0

x2 y2 z2 r2 O

x  h2 y  k2 z  l2 r2 r

Ch, k, l

x  h2 y  k2 z  l2 r2

PC

P2 r2

PC

 r

r C

Px, y, z

Ch, k, l

r

V

 s1  4  4  3

PQ

### 

 s1  22 3  12 5  72

Q1, 3, 5

P2, 1, 7

P1P2

### 

 sx2 x12 y2 y12 z2 z12

 x2 x12 y2 y12 z2 z12



x2 x1

2

y2 y1

2

z2 z1

2

P1P2

2

P1A

2

AB

2

BP2

2

P1B

2

P1A

2

AB

2

P1P2

2

P1B

2

BP2

### 

2

P1AB P1BP2

FIGURE 10 0 z

x

y r

P(x, y, z)

C(h, k, l)

(5)

EXAMPLE 6 What region in is represented by the following inequalities?

SOLUTION The inequalities

can be rewritten as

so they represent the points whose distance from the origin is at least 1 and at most 2. But we are also given that , so the points lie on or below the

-plane. Thus the given inequalities represent the region that lies between (or on)

the spheres and and beneath (or on) the

-plane. It is sketched in Figure 11.

xy

x2 y2 z2 4 x2 y2 z2 1

xy

z 0

x, y, z

1 sx2 y2 z2 2 1 x2 y2 z2 4

z 0 1 x2 y2 z2 4

3

SECTION 10.1 THREE-DIMENSIONAL COORDINATE SYSTEMS 521

8. Find the distance from to each of the following.

(a) The -plane (b) The -plane (c) The -plane (d) The -axis (e) The -axis (f ) The -axis 9. Determine whether the points lie on straight line.

(a) , ,

(b) , ,

10. Find an equation of the sphere with center and radius 5. Describe its intersection with each of the coordi- nate planes.

Find an equation of the sphere that passes through the point and has center .

12. Find an equation of the sphere that passes through the ori- gin and whose center is .

13–16 Show that the equation represents a sphere, and ﬁnd its center and radius.

13.

14.

15.

16.

17. (a) Prove that the midpoint of the line segment from

to is

### 

x1 x2 2, y1 y2 2,z1 z2 2

### 

P2x2, y2,z2 P1x1, y1,z1

4x2 4y2 4z2 8x  16y  1 x2 y2 z2 x  y  z x2 y2 z2 4x  2y

x2 y2 z2 6x  4y  2z  11

1, 2, 3

3, 8, 1

4, 3, 1

11.

2, 6, 4

F3, 4, 2

E1, 2, 4

D0, 5, 5

C1, 3, 3

B3, 7, 2

A2, 4, 2

z y

x xz

yz xy

3, 7, 5

1. Suppose you start at the origin, move along the -axis a dis- tance of 4 units in the positive direction, and then move downward a distance of 3 units. What are the coordinates of your position?

2. Sketch the points , , , and

on a single set of coordinate axes.

3. Which of the points , , and

is closest to the -plane? Which point lies in the -plane?

4. What are the projections of the point (2, 3, 5) on the -, -, and -planes? Draw a rectangular box with the origin and

as opposite vertices and with its faces parallel to the coordinate planes. Label all vertices of the box. Find the length of the diagonal of the box.

Describe and sketch the surface in represented by the

equation .

6. (a) What does the equation represent in ? What does it represent in ? Illustrate with sketches.

(b) What does the equation represent in ? What does represent? What does the pair of equations

, represent? In other words, describe the set of points such that and . Illustrate with a sketch.

7. Find the lengths of the sides of the triangle . Is it a right triangle? Is it an isosceles triangle?

(a) , ,

(b)P2, 1, 0, Q4, 1, 1, R4, 5, 4

R1, 2, 1

Q7, 0, 1

P3, 2, 3

PQR z  5 y 3

x, y, z

z  5

y 3z  5 3 yx 3 4 32

x y  2 3

5.

2, 3, 5

xz

yz xy yz

xz R0, 3, 8

Q5, 1, 4

P6, 2, 3

1, 1, 2 0, 5, 2 4, 0, 1 2, 4, 6

x EXERCISES

### 10.1

FIGURE 11 z

x y

0 1 2

(6)

(b) Find the lengths of the medians of the triangle with ver-

tices , , and .

18. Find an equation of a sphere if one of its diameters has end- points and .

Find equations of the spheres with center that touch (a) the -plane, (b) the -plane, (c) the -plane.

20. Find an equation of the largest sphere with center (5, 4, 9) that is contained in the ﬁrst octant.

21–30 Describe in words the region of represented by the equation or inequality.

21. 22.

23. 24.

26.

27. 28.

30.

x2 y2 z2 2z x2 z2 9

29.

x z x2 y2 z2 3

z2 1 0 z 6

25.

y 0 x 3

x 10 y 4

3 xz yz

xy

2, 3, 6

19.

4, 3, 10

2, 1, 4

C4, 1, 5

B2, 0, 5

A1, 2, 3

31–34 Write inequalities to describe the region.

31. The half-space consisting of all points to the left of the -plane

32. The solid rectangular box in the ﬁrst octant bounded by the

planes , , and

The region consisting of all points between (but not on) the spheres of radius and centered at the origin, where

34. The solid upper hemisphere of the sphere of radius 2 centered at the origin

Find an equation of the set of all points equidistant from the points and . Describe the set.

36. Find the volume of the solid that lies inside both of the spheres

and x2 y2 z2 4

x2 y2 z2 4x  2y  4z  5  0 B6, 2, 2

A1, 5, 3

35.

r R

R r 33.

z  3 y 2

x 1 xz

VECTORS

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 magni- tude of the vector and the arrow points in the direction of the vector. We denote a vector by printing a letter in boldface or by putting an arrow above the letter

For instance, suppose a particle moves along a line segment from point to point . The corresponding displacement vector , shown in Figure 1, has initial point (the tail) and terminal point (the tip) and we indicate this by writing ABl. Notice that the vector CDlhas the same length and the same direction as even though it is in a different position. We say that and are equivalent (or equal) and we write . The zero vector, denoted by 0, has length . It is the only vector with no speciﬁc direction.

COMBINING VECTORS

Suppose a particle moves from , so its displacement vector is ABl

. Then the par- ticle changes direction and moves from , with displacement vector BCl

as in Figure 2. The combined effect of these displacements is that the particle has moved from . The resulting displacement vector ACl

is called the sum of ABl and BCl

and we write

ACl ABl

BCl

In general, if we start with vectors and , we ﬁrst move so that its tail coincides with the tip of and deﬁne the sum of and as follows.u u v

v v

u



 A to C

B to C A to B

0

u v u v

v

u B v

A v

B

A

vl.

v

### 10.2

FIGURE 1 Equivalent vectors A

B v

C

D u

FIGURE 2 C

B

A

(7)

DEFINITION OF VECTOR ADDITION If and are vectors positioned so the initial point of is at the terminal point of , then the sum is the vector from the initial point of to the terminal point of .

The deﬁnition of vector addition is illustrated in Figure 3. You can see why this deﬁ- nition is sometimes called the Triangle Law.

In Figure 4 we start with the same vectors and as in Figure 3 and draw another copy of with the same initial point as . Completing the parallelogram, we see that

. This also gives another way to construct the sum: If we place and so they start at the same point, then lies along the diagonal of the parallelo- gram with and as sides. (This is called the Parallelogram Law.)

EXAMPLE 1 Draw the sum of the vectors shown in Figure 5.

SOLUTION First we translate and place its tail at the tip of , being careful to draw a copy of that has the same length and direction. Then we draw the vector

[see Figure 6(a)] starting at the initial point of and ending at the terminal point of the copy of .

Alternatively, we could place so it starts where starts and construct by the Parallelogram Law as in Figure 6(b).

It is possible to multiply a vector by a real number . (In this context we call the real number a scalar to distinguish it from a vector.) For instance, we want to be the same vector as , which has the same direction as but is twice as long. In general, we multiply a vector by a scalar as follows.

DEFINITION OF SCALAR MULTIPLICATION If is a scalar and is a vector, then the scalar multiple is the vector whose length is times the length of and whose direction is the same as if and is opposite to if

. If c 0or v 0, then cv 0. c 0

v c 0

v

v cv c

c

### 

v

v

v v 2v

c

c

FIGURE 6

a b

a+b

(a)

a

a+b b

(b)

a b a

b b

a

a b b

a b

a and b

V

v u

u v v

u

u v  v  uv u

v u

FIGURE 3 The Triangle Law u+v v

u

FIGURE 4 The Parallelogram Law v

v+

u

u u

v u+

v

v u

u v u

v

v u

SECTION 10.2 VECTORS 523

Visual 10.2 shows how the Triangle and Parallelogram Laws work for various vectors .u and v FIGURE 5

a b

(8)

This deﬁnition is illustrated in Figure 7. We see that real numbers work like scal- ing factors here; that’s why we call them scalars. Notice that two nonzero vectors are parallel if they are scalar multiples of one another. In particular, the vector

has the same length as but points in the opposite direction. We call it the negative of .

By the difference of two vectors we mean

So we can construct by ﬁrst drawing the negative of , , and then adding it to by the Parallelogram Law as in Figure 8(a). Alternatively, since

the vector , when added to , gives . So we could construct as in Fig- ure 8(b) by means of the Triangle Law.

EXAMPLE 2 If are the vectors shown in Figure 9, draw .

SOLUTION We ﬁrst draw the vector pointing in the direction opposite to and twice as long. We place it with its tail at the tip of and then use the Triangle Law to draw as in Figure 10.

COMPONENTS

For some purposes it’s best to introduce a coordinate system and treat vectors alge- braically. If we place the initial point of a vector at the origin of a rectangular coor- dinate system, then the terminal point of has coordinates of the form or , depending on whether our coordinate system is two- or three-dimensional (see Figure 11). These coordinates are called the components of and we write

or

We use the notation for the ordered pair that refers to a vector so as not to confuse it with the ordered pair that refers to a point in the plane.

For instance, the vectors shown in Figure 12 are all equivalent to the vector OPl

whose terminal point is . What they have in common is that the terminal point is reached from the initial point by a displacement of three units to the right and two upward. We can think of all these geometric vectors as representations of the algebraic vector . The particular representation OPl

from the origin to the point P3, 2is called the position vector of the point .a 3, 2 P

P3, 2

 3, 2

a1, a2

a1, a2

a a1, a2, a3 a a1, a2

a

a1, a2, a3 a a1, a2

a

FIGURE 9 a

b

FIGURE 10 _2b a

a-2b

a 2b a

2b b

a 2b a and b

FIGURE 8

Drawing u-v (a)

v u

u-v _v

(b) v

u-v

u

u v u

v

u v v u  v  u,

u

v v u v

u v  u  v

u v v

v  1v v

_1.5v

v 2v

_v

1v

2

FIGURE 7

Scalar multiples of v

FIGURE 11

a=ka¡, a™l

a=ka¡, a™, a£l (a¡, a™)

O y

x a

z

x y

a O

(a¡, a™, a£)

(9)

In three dimensions, the vector OPl is the position vector of the point . (See Figure 13.) Let’s consider any other representation ABl of , where the initial point is and the terminal point is . Then we

must have , , and and so ,

, and . Thus we have the following result.

Given the points and , the vector with represen- tation ABlis

EXAMPLE 3 Find the vector represented by the directed line segment with initial

point ) and terminal point .

SOLUTION By (1), the vector corresponding to ABl is

The magnitude or length of the vector is the length of any of its representations and is denoted by the symbol or . By using the distance formula to compute the length of a segment , we obtain the following formulas.

The length of the two-dimensional vector is

The length of the three-dimensional vector is

How do we add vectors algebraically? Figure 14 shows that if and , then the sum is , 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 15 we see that the components of are and . So to multi- ply a vector by a scalar we multiply each component by that scalar.

ca2

ca1

ca a b  a1 b1, a2 b2

b b1, b2 a a1, a2

a

 sa12 a22 a32

a a1, a2, a3

a

 sa12 a22

a a1, a2 OP

 v 

v

### 

v

a 2  2, 1  3, 1  4  4, 4, 3

B2, 1, 1

A2, 3, 4

V

a x2 x1, y2 y1,z2 z1 a Bx2, y2,z2

Ax1, y1,z1

1

a3 z2 z1

a2 y2 y1x1 a1 x2 Ayx1 a1, y1,2z y1 2 z1 a3 z2 Bx2, y2a,1z2 x 2 x1 a

Pa1, a2, a3 a  a1, a2, a3

FIGURE 12

Representations of the vector a=k3, 2l (1, 3)

(4, 5)

x y

O

P(3, 2)

FIGURE 13

Representations of a=ka¡, a™, a£l O

z

y x

position vector of P

P(a¡, a™, a£)

A(x, y, z)

B(x+a¡, y+a™, z+a£) SECTION 10.2 VECTORS 525

FIGURE 14 0 y

x

b b™

a+b

a

(a¡+b¡, a™+b™)

a™ a™

FIGURE 15

ca™

ca¡

ca a™

a

(10)

If and , then

Similarly, for three-dimensional vectors,

EXAMPLE 4 If and , ﬁnd and the vectors

, , , and .

SOLUTION

We denote by the set of all two-dimensional vectors and by the set of all three-dimensional vectors. More generally, we will later need to consider the set of all -dimensional vectors. An -dimensional vector is an ordered -tuple:

where are real numbers that are called the components of . Addition and scalar multiplication are deﬁned in terms of components just as for the cases

and .

PROPERTIES OF VECTORS If , , and are vectors in and and are scalars, then

1. 2.

3. 4.

5. 6.

7. 8.

These eight properties of vectors can be readily veriﬁed either geometrically or algebraically. For instance, Property 1 can be seen from Figure 4 (it’s equivalent to the

1a a

cda  cda

c  da  ca  da ca  b  ca  cb

a a  0 a 0  a

a b  c  a  b  c a b  b  a

d c Vn

c b a n 3

n 2

a a1, a2, . . . , an

a a1, a2, . . . , an

n n

n

Vn

V3

V2

 8, 0, 6  10, 5, 25  2, 5, 31

2a 5b  24, 0, 3  52, 1, 5

3b 32, 1, 5  32, 31, 35  6, 3, 15

 4  2, 0  1, 3  5  6, 1, 2

a b  4, 0, 3  2, 1, 5

 4  2, 0  1, 3  5  2, 1, 8

a b  4, 0, 3  2, 1, 5

a

### 

 s42 02 32  s25  5

2a 5b 3b

a b

aV b a 4, 0, 3 b 2, 1, 5

a

### 

ca1, a2, a3  ca1, ca2, ca3

a1, a2, a3  b1, b2, b3  a1 b1, a2 b2, a3 b3

a1, a2, a3  b1, b2, b3  a1 b1, a2 b2, a3 b3 ca ca1, ca2

a b  a1 b1, a2 b2 a b  a1 b1, a2 b2

b b1, b2 a a1, a2

Vectors in dimensions are used to list various quantities in an organized way. For instance, the components of a six-dimensional vector

might represent the prices of six differ- ent ingredients required to make a partic- ular product. Four-dimensional vectors

are used in relativity theory, where the ﬁrst three components specify a position in space and the fourth repre- sents time.

 x, y, z, t

p  p1, p2, p3, p4, p5, p6 n

(11)

Parallelogram Law) or as follows for the case :

We can see why Property 2 (the associative law) is true by looking at Figure 16 and applying the Triangle Law several times: The vector PQl

is obtained either by ﬁrst con- structing a b and then adding c or by adding a to the vector b  c.

Three vectors in play a special role. Let

These vectors , , and are called the standard basis vectors. They have length and point in the directions of the positive -, -, and -axes. Similarly, in two dimensions

we deﬁne and . (See Figure 17.)

If , then we can write

Thus any vector in can be expressed in terms of , , and . For instance,

Similarly, in two dimensions, we can write

See Figure 18 for the geometric interpretation of Equations 3 and 2 and compare with Figure 17.

EXAMPLE 5 If and , express the vector in terms of , , and .

SOLUTION Using Properties 1, 2, 5, 6, and 7 of vectors, we have

 2i  4j  6k  12i  21k  14i  4j  15k 2a 3b  2i  2j  3k  34i  7k

k j i

2a 3b b 4i  7 k

a i  2j  3k

a a1, a2  a1i a2j

3

1, 2, 6  i  2j  6k k j i V3

a a1i a2j a3k

2

 a11, 0, 0  a20, 1, 0  a30, 0, 1

a a1, a2, a3  a1, 0, 0  0, a2, 0  0, 0, a3 a a1, a2, a3

FIGURE 17

Standard basis vectors in V™ and V£ (a) 0 y

x j

(1, 0) i (0, 1)

(b) z

x y

j i k

j 0, 1

i 1, 0 x y z

1 k

j i

k 0, 0, 1

j 0, 1, 0

i 1, 0, 0

V3

 b  a

 b1 a1, b2 a2  b1, b2  a1, a2 a b  a1, a2  b1, b2  a1 b1, a2 b2

n 2

SECTION 10.2 VECTORS 527

FIGURE 16

b c

a (a+b)+c

P Q

=a+(b+c) a+b

b+c

FIGURE 18

(b) a=a¡i+a™ j+a£k (a) a=a¡i+a™ j 0

a

a¡i

a™ j (a¡, a™)

a™ j

a£k (a¡, a™, a£)

a¡i

a y

x

z

x y

(12)

A unit vector is a vector whose length is 1. For instance, , , and are all unit vec- tors. In general, if , then the unit vector that has the same direction as is

In order to verify this we let . Then and is a positive scalar, so has the same direction as . Also

EXAMPLE 6 Find the unit vector in the direction of the vector . SOLUTION The given vector has length

so, by Equation 4, the unit vector with the same direction is

APPLICATIONS

Vectors are useful in many aspects of physics and engineering. In Section 10.9 we will see how they describe the velocity and acceleration of objects moving in space. Here we look at forces.

A force is represented by a vector because it has both a magnitude (measured in pounds or newtons) and a direction. If several forces are acting on an object, the resul- tant force experienced by the object is the vector sum of these forces.

EXAMPLE 7 A 100-lb weight hangs from two wires as shown in Figure 19. Find the tensions (forces) and in both wires and their magnitudes.

SOLUTION We ﬁrst express and in terms of their horizontal and vertical com- ponents. From Figure 20 we see that

The resultant of the tensions counterbalances the weight and so we must have

Thus

Equating components, we get

Solving the ﬁrst of these equations for and substituting into the second, we get

T1

sin 50 

T1

### 

cos 50

cos 32 sin 32  100

T2

T1

sin 50 

T2

sin 32  100



T1

cos 50 

T2

cos 32  0



T1

cos 50 

T2

cos 32

i

T1

sin 50 

T2

sin 32

### )

j 100 j

T1 T2 w  100 j

w T1 T2

T2

T2

cos 32 i 

T2

sin 32 j

6

T1 

T1

cos 50 i 

T1

### 

sin 50 j

5

T2

T1

T2

T1 1

32i  j  2k 23i13 j23k

2 i j  2k

### 

 s22 12 22  s9  3

2 i j  2k

u



ca



c

a



1a

a

 1

a

u c

u ca c 1

a

u 1

a

a

aa

### 

4

a

a 0 i j k

FIGURE 20 50°

w

50° 32°

32°

T™

FIGURE 19 100

50° 32°

T™

(13)

So the magnitudes of the tensions are

and

Substituting these values in (5) and (6), we obtain the tension vectors

T2 55.05 i  34.40 j

T1 55.05 i  65.60 j

T2



T1

cos 50

cos 32 64.91 lb

T1

### 

 sin 50  tan 32 cos 50 100 85.64 lb

SECTION 10.2 VECTORS 529

5– 8 Find a vector with representation given by the directed line segment ABl

. Draw ABl

and the equivalent representation starting at the origin.

5. , 6. ,

, 8. ,

9–12 Find the sum of the given vectors and illustrate geometrically.

9. , 10. ,

11. , 12. ,

13–16 Find a b, 2a  3b, , and .

13. ,

14. ,

15. ,

16. ,

Find a unit vector with the same direction as . 18. Find a vector that has the same direction as but

has length 6.

If lies in the ﬁrst quadrant and makes an angle with the positive -axis and , ﬁnd in component form.

20. If a child pulls a sled through the snow with a force of 50 N exerted at an angle of above the horizontal, ﬁnd the hor- izontal and vertical components of the force.

Two forces and with magnitudes 10 lb and 12 lb act on an object at a point as shown in the ﬁgure. Find the resultant force acting at as well as its magnitude and its F P

P F2

F1

21.

38

v

### 

 4 v x

3 v

19.

2, 4, 2

8 i j  4k 17.

b 2 j  k a 2 i  4 j  4 k

b 2 i  j  5k a i  2 j  3k

b i  2 j a 4 i  j

b 3, 6

a 5, 12

a

a b

### 

0, 4, 0

1, 0, 2

0, 0, 3

0, 1, 2

5, 7

2, 1

2, 4

3, 1

B4, 2, 1

A4, 0, 2

B2, 3, 1

A0, 3, 1

7.

B5, 3

A2, 2

B2, 1

A2, 3

Name all the equal vectors in the parallelogram shown. a

2. Write each combination of vectors as a single vector.

(a) PQl QRl (b) RPl PSl (c)QSl PSl (d) RSl SPl PQl

3. Copy the vectors in the ﬁgure and use them to draw the following vectors.

(a) (b)

(c) (d)

4. Copy the vectors in the ﬁgure and use them to draw the following vectors.

(a) (b)

(c) (d)

(e) (f )

a b

b 3a

2a b 12b

2a

a b a b

v w u

w v  u v w

u v u v

Q

R S

P











B E

A

D C

1.

EXERCISES

### 10.2

Animal or vegetable fats and oils and their fractiors, boiled, oxidised, dehydrated, sulphurised, blown, polymerised by heat in vacuum or in inert gas or otherwise chemically

Milk and cream, in powder, granule or other solid form, of a fat content, by weight, exceeding 1.5%, not containing added sugar or other sweetening matter.

For 5 to be the precise limit of f(x) as x approaches 3, we must not only be able to bring the difference between f(x) and 5 below each of these three numbers; we must be able

[This function is named after the electrical engineer Oliver Heaviside (1850–1925) and can be used to describe an electric current that is switched on at time t = 0.] Its graph

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You are given the wavelength and total energy of a light pulse and asked to find the number of photons it

Wang, Solving pseudomonotone variational inequalities and pseudocon- vex optimization problems using the projection neural network, IEEE Transactions on Neural Networks 17

volume suppressed mass: (TeV) 2 /M P ∼ 10 −4 eV → mm range can be experimentally tested for any number of extra dimensions - Light U(1) gauge bosons: no derivative couplings. =&gt;

We explicitly saw the dimensional reason for the occurrence of the magnetic catalysis on the basis of the scaling argument. However, the precise form of gap depends

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• Formation of massive primordial stars as origin of objects in the early universe. • Supernova explosions might be visible to the most

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