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
left wall y
z x xz
z z y
a, b, c
a, b a
VECTORS AND THE GEOMETRY 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.
517 FIGURE 2
Right-hand rule O
FIGURE 1 Coordinate axes
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) 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).
⺢3 z 3 x, y, zz 3
z 3 ⺢3
⺢3 x y z
⺢2 x y
a, b, c
⺢3x, y,z ⺢
(3, _2, _6)
_6 _2 3 _5
(_4, 3, _5) 3
_4 (0, 0, c)
R(0, b, c) P(a, b, c)
(0, b, 0) z
0 S(a, 0, c)
Q(a, b, 0) (a, 0, 0)
3, 2, 6
4, 3, 5
Sa, 0, c
R0, b, c
a, b, 0
P Pa, b, cz c
a, b, c
y b x
a P c
a, b, c
FIGURE 4 z
P(a, b, c)
(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, thenBP2z2 z1 ABy2 y1 P1Ax2 x1
Bx2, y2,z1 Ax2, y1,z1
P1P1P2 sx2 x12 y2 y12 z2 z12
z 0 y x xy
x, x, zx ⺢, z ⺢
y x ⺢3
z c y b
xx a 0 yz y 0 xz z 0 xy z k xy
y k k x k yz
⺢2 y 5
⺢3 ⺢2 ⺢3 y 5
xz ⺢3 y
FIGURE 7 (c) y=5, a line in R@
(b) y=5, a plane in R#
(a) z=3, a plane in R#
x y 3
SECTION 10.1 THREE-DIMENSIONAL COORDINATE SYSTEMS ■ 519
x FIGURE 8
The plane y=x
FIGURE 9 0 z
P¡(⁄, ›, z¡)
A(¤, ›, z¡)
P™(¤, ﬁ, z™)
B(¤, ﬁ, z¡)
Because triangles and are both right-angled, two applications of the Pythag- orean Theorem give
Combining these equations, we get
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, 1and radius s8 2s2. ■
x 22 y 32 z 12 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 h2 y k2 z l2 r2 r
Ch, k, l
x h2 y k2 z l2 r2PCP2 r2 PC r
Px, y, z
Ch, k, l
s1 4 4 3PQ s1 22 3 12 5 72
Q1, 3, 5
P2, 1, 7P1P2 sx2 x12 y2 y12 z2 z12
x2 x12 y2 y12 z2 z12x2 x12y2 y12z2 z12 P1P22P1A2AB2BP22 P1B2P1A2AB2 P1P22P1B2BP22
FIGURE 10 0 z
P(x, y, z)
C(h, k, l)
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. ■
x2 y2 z2 4 x2 y2 z2 1
x, y, z
1 sx2 y2 z2 2 1 x2 y2 z2 4
z 0 1 x2 y2 z2 4
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.
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
17. (a) Prove that the midpoint of the line segment from
to isx1 x2 2, y1 y2 2,z1 z2 2
P2x2, y2,z2 P1x1, 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
2, 6, 4
F3, 4, 2
E1, 2, 4
D0, 5, 5
C1, 3, 3
B3, 7, 2
A2, 4, 2
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
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)P2, 1, 0, Q4, 1, 1, R4, 5, 4
R1, 2, 1
Q7, 0, 1
P3, 2, 3
PQR z 5 y 3
x, y, z
y 3z 5 ⺢3 yx 3 4 ⺢⺢32
x y 2 ⺢3
2, 3, 5
yz xy yz
xz R0, 3, 8
Q5, 1, 4
P6, 2, 3
1, 1, 2 0, 5, 2 4, 0, 1 2, 4, 6
FIGURE 11 z
0 1 2
(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.
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
x2 y2 z2 2z x2 z2 9
x z x2 y2 z2 3
z2 1 0 z 6
y 0 x 3
x 10 y 4
⺢3 xz yz
2, 3, 6
4, 3, 10
2, 1, 4
C4, 1, 5
B2, 0, 5
A1, 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 B6, 2, 2
A1, 5, 3
R r 33.
z 3 y 2
x 1 xz
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.
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
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
A to C
B to C A to B
u v u v
u B v
FIGURE 1 Equivalent vectors A
FIGURE 2 C
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 cv cc v
v v 2v
a b a
a b b
a and b
u v v
u v v uv u
FIGURE 3 The Triangle Law u+v v
FIGURE 4 The Parallelogram Law v
u 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
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.
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
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 P3, 2is called the position vector of the point .a 3, 2 P
a a1, a2, a3 a a1, a2
a1, a2, a3 a a1, a2
FIGURE 9 a
FIGURE 10 _2b a
a 2b a
a 2b a and b
Drawing u-v (a)
u v u
u v v u v u,
v v u v
u v u v
u v v
v 1v v
Scalar multiples of v
a=ka¡, a™, a£l (a¡, a™)
(a¡, a™, a£)
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.
ca a b a1 b1, a2 b2
b b1, b2 a a1, a2a sa12 a22 a32
a a1, a2, a3a sa12 a22
a a1, a2 OP
a 2 2, 1 3, 1 4 4, 4, 3
B2, 1, 1
A2, 3, 4
a x2 x1, y2 y1,z2 z1 a Bx2, y2,z2
a3 z2 z1
a2 y2 y1x1 a1 x2 Ayx1 a1, y1,2z y1 2 z1 a3 z2 Bx2, y2a,1z2 x 2 x1 a
Pa1, a2, a3 a a1, a2, a3
Representations of the vector a=k3, 2l (1, 3)
Representations of a=ka¡, a™, a£l O
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
b¡ x a¡
If and , then
Similarly, for three-dimensional vectors,
EXAMPLE 4 If and , ﬁnd and the vectors
, , , and .
■ 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
PROPERTIES OF VECTORS If , , and are vectors in and and are scalars, then
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
c da ca da ca 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
a a1, a2, . . . , an
a a1, a2, . . . , an
8, 0, 6 10, 5, 25 2, 5, 31
2a 5b 24, 0, 3 52, 1, 5
3b 32, 1, 5 32, 31, 35 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, 5a s42 02 32 s25 5
2a 5b 3b
aV b a 4, 0, 3 b 2, 1, 5a
ca1, 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
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 2i 2j 3k 34i 7k
k j i
2a 3b b 4i 7 k
a i 2j 3k
a a1, a2 a1i a2j
1, 2, 6 i 2j 6k k j i V3
a a1i a2j a3k
a11, 0, 0 a20, 1, 0 a30, 0, 1
a a1, a2, a3 a1, 0, 0 0, a2, 0 0, 0, a3 a a1, a2, a3
Standard basis vectors in V™ and V£ (a) 0 y
(1, 0) i (0, 1)
j i k
j 0, 1
i 1, 0 x y z
k 0, 0, 1
j 0, 1, 0
i 1, 0, 0
b1 a1, b2 a2 b1, b2 a1, a2 a b a1, a2 b1, b2 a1 b1, a2 b2
SECTION 10.2 VECTORS ■ 527
(b) a=a¡i+a™ j+a£k (a) a=a¡i+a™ j 0
a™ j (a¡, a™)
a£k (a¡, a™, a£)
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
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
Equating components, we get
Solving the ﬁrst of these equations for and substituting into the second, we getT1sin 50 T1cos 50
cos 32 sin 32 100T2 T1sin 50 T2sin 32 100 T1cos 50 T2cos 32 0
(T1cos 50 T2cos 32
(T1sin 50 T2sin 32
)j 100 j
T1 T2 w 100 j
w T1 T2
T2T2cos 32 i T2sin 32 j
T1T1cos 50 i T1sin 50 j
32i j 2k 23i13 j23k2 i j 2k s22 12 22 s9 3
2 i j 2kucaca 1a
u ca c 1a
u 1a a aa
a 0 i j k
FIGURE 20 50°
FIGURE 19 100 T¡
So the magnitudes of the tensions are
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 jT2 T1cos 50
cos 32 64.91 lbT1 sin 50 tan 32 cos 50100 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 .
■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■
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
38v 4 v x
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, 12a
0, 4, 0
1, 0, 2
0, 0, 3
0, 1, 2
B4, 2, 1
A4, 0, 2
B2, 3, 1
A0, 3, 1
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.
4. Copy the vectors in the ﬁgure and use them to draw the following vectors.
(e) (f )
2a b 12b
a b a b
v w u
w v u v w
u v u v