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The three-dimensional coordinate
system is used in chemistry to help
understand the structure of crystals.
For example, isometric crystals are
shaped like cubes
S E L E C T E D A P P L I C AT I O N S
Three-dimensional analytic geometry concepts have many real-life applications. The applications
listed below represent a small sample of the applications in this chapter.
Exercise 63, page 818
Exercise 66, page 819
Exercise 43, page 826
Exercise 43, page 833
- Data Analysis: Milk Consumption,
Exercise 47, page 842
Exercise 48, page 842
Exercise 12, page 849
11.1 The Three-Dimensional Coordinate System
11.2 Vectors in Space
11.3 The Cross Product of Two Vectors
11.4 Lines and Planes in Space
Analytic Geometry
in Three Dimensions
Arnold Fisher/Photo Researchers, Inc.
811
The Three-Dimensional Coordinate System
Recall that the Cartesian plane is determined by two perpendicular number lines
called the x -axis and the y -axis. These axes, together with their point of intersec-
tion (the origin), allow you to develop a two-dimensional coordinate system for
identifying points in a plane. To identify a point in space, you must introduce a
third dimension to the model. The geometry of this three-dimensional model is
called solid analytic geometry.
You can construct a three-dimensional coordinate system by passing a
z -axis perpendicular to both the x - and y -axes at the origin. Figure 11.1 shows the
positive portion of each coordinate axis. Taken as pairs, the axes determine three
coordinate planes: the xy -plane, the xz -plane, and the yz -plane. These three
coordinate planes separate the three-dimensional coordinate system into eight
octants. The first octant is the one in which all three coordinates are positive. In
this three-dimensional system, a point P in space is determined by an ordered
triple where x , y , and z are as follows.
directed distance from yz -plane to P
directed distance from xz -plane to P
directed distance from xy -plane to P
FIGURE 11.1 FIGURE 11.
A three-dimensional coordinate system can have either a left-handed or a
right-handed orientation. In this text, you will work exclusively with right-
handed systems, as illustrated in Figure 11.2. In a right-handed system, Octants
II, III, and IV are found by rotating counterclockwise around the positive z -axis.
Octant V is vertically below Octant I. Octants VI, VII, and VIII are then found by
rotating counterclockwise around the negative z -axis. See Figure 11.3.
x
y
z
x
y
yz -plane
xy -plane
xz -plane
z
( x , y , z )
z
y
x
x , y , z ,
812 Chapter 11 Analytic Geometry in Three Dimensions
What you should learn
- Plot points in the three- dimensional coordinate system.
- Find distances between points in space and find midpoints of line segments joining points in space.
- Write equations of spheres in standard form and find traces of surfaces in space.
Why you should learn it The three-dimensional coordinate system can be used to graph equations that model surfaces in space, such as the spherical shape of Earth, as shown in Exercise 66 on page 819.
The Three-Dimensional Coordinate System
NASA
11.
Octant I Octant II Octant III Octant IV Octant V Octant VI Octant VII Octant VIII
x
z
y x
z
y x
z
y x
z
y x
z
y x
z
y x
z
y x
z
y
FIGURE 11.
Using the Midpoint Formula in Space
Find the midpoint of the line segment joining and
Solution
Using the Midpoint Formula in Space, the midpoint is
as shown in Figure 11.6.
FIGURE 11.
Now try Exercise 31.
The Equation of a Sphere
A sphere with center and radius is defined as the set of all points
such that the distance between and is as shown in
Figure 11.7. Using the Distance Formula, this condition can be written as
By squaring each side of this equation, you obtain the standard equation of a
sphere.
Notice the similarity of this formula to the equation of a circle in the plane.
Equation of sphere in space
Equation of circle in the plane
As is true with the equation of a circle, the equation of a sphere is simplified
when the center lies at the origin. In this case, the equation is
x^2 y^2 z^2 r^2. Sphere with center at origin
x h ^2 y k ^2 r^2
x h ^2 y k ^2 z j ^2 r^2
x h ^2 y k ^2 z j ^2 r.
x , y , z x , y , z h , k , j r ,
h , k , j r
Midpoint: ( 52 , 1,^72 )
x
y
2 1
4 3
− 1
− (^4) − 3 − 2
− 3
4
(^3 2 ) 4
2
z
2 ^
(^)
2
5, 2, 3 0, 4, 4.
814 Chapter 11 Analytic Geometry in Three Dimensions
Example 3
Standard Equation of a Sphere
The standard equation of a sphere with center and radius is
given by
x h ^2 y k ^2 z j ^2 r^2.
h , k , j r
( , x y z , )
( , h k j , )
r
Sphere: radius r ; center ( , h k j , )
x
y
z
FIGURE 11.
Finding the Equation of a Sphere
Find the standard equation of the sphere with center and radius 3. Does
this sphere intersect the -plane?
Solution
Standard equation
Substitute.
From the graph shown in Figure 11.8, you can see that the center of the sphere
lies three units above the -plane. Because the sphere has a radius of 3, you can
conclude that it does intersect the -plane—at the point
FIGURE 11.
Now try Exercise 39.
Finding the Center and Radius of a Sphere
Find the center and radius of the sphere given by
Solution
To obtain the standard equation of this sphere, complete the square as follows.
So, the center of the sphere is and its radius is See Figure 11.9.
Now try Exercise 49.
Note in Example 5 that the points satisfying the equation of the sphere are
“surface points,” not “interior points.” In general, the collection of points
satisfying an equation involving x , y ,and z is called a surface in space.
1, 2, 3,^ 6.
x^2 y^2 z^2 2 x 4 y 6 z 8 0.
1
3 2
2
1 3 4 5
1
5 4
3 6 7
r = 3
x
y
z
xy 2, 4, 0.
xy
x 2 ^2 y 4 ^2 z 3 ^2 32
x h ^2 y k ^2 z j ^2 r^2
xy
2, 4, 3
Section 11.1 The Three-Dimensional Coordinate System 815
Example 4
Example 5
x 1 ^2 y 2 ^2 z 3 ^2 6
2
x^2 2 x 1 y^2 4 y 4 z^2 6 z 9 8 1 4 9
x^2 2 x (^) y^2 4 y (^) z^2 6 z (^) 8
x^2 y^2 z^2 2 x 4 y 6 z 8 0
Find the equation of the sphere
that has the points
and as endpoints of
a diameter. Explain how this
problem gives you a chance
to use all three formulas
discussed so far in this section:
the Distance Formula in Space,
the Midpoint Formula in Space,
and the standard equation of a
sphere.
1, 4, 2
3, 2, 6
Exploration
r = 6
x
1 2
5 4 3 3
1
5
6
y
z
FIGURE 11.
Section 11.1 The Three-Dimensional Coordinate System 817
In Exercises 1 and 2, approximate the coordinates of the points.
1. 2.
In Exercises 3–6, plot each point in the same three-dimen- sional coordinate system.
3. (a) 4. (a) (b) (b) 5. (a) 6. (a) (b) (b)
In Exercises 7–10, find the coordinates of the point.
7. The point is located three units behind the -plane, three units to the right of the -plane, and four units above the -plane. 8. The point is located six units in front of the -plane, one unit to the left of the -plane, and one unit below the -plane. 9. The point is located on the -axis, 10 units in front of the -plane. 10. The point is located in the -plane, two units to the right of the -plane, and eight units above the -plane.
In Exercises 11–16, determine the octant(s) in which is located so that the condition(s) is (are) satisfied.
**11.
16.**
In Exercises 17–24, find the distance between the points.
**17.
23.**
24. 2, 4, 0,0, 6, 3
yz > 0
xy < 0
y < 0
z > 0
x < 0, y > 0, z < 0
x > 0, y < 0, z > 0
x, y, z
xz xy
yz
yz
x
xy
xz
yz
xy
xz
yz
x y
B
A
C
543
2
2
4
3 − 5 − 6
− (^4) − 3 − 2
z
x^ y B
A
C
4
3 2
4 3 2
−^ −^344
− (^4) − (^3) − 2 − 4
z
11.1 Exercises
VOCABULARY CHECK: Fill in the blanks.
1. A _______ coordinate system can be formed by passing a -axis perpendicular to both the -axis and the -axis at the origin. 2. The three coordinate planes of a three-dimensional coordinate system are the _______ , the _______ , and the _______. 3. The coordinate planes of a three-dimensional coordinate system separate the coordinate system into eight _______. 4. The distance between the points and can be found using the _______ _______ in Space. 5. The midpoint of the line segment joining the points and given by the Midpoint Formula in Space is _______. 6. A _______ is the set of all points such that the distance between and a fixed point is 7. A _______ in _______ is the collection of points satisfying an equation involving and 8. The intersection of a surface with one of the three coordinate planes is called a _______ of the surface.
x , y , z.
h , k , j r.
x , y , z x , y , z
x 1 , y 1 , z 1 x 2 , y 2 , z 2
x 1 , y 1 , z 1 x 2 , y 2 , z 2
x y
z
In Exercises 25–28, find the lengths of the sides of the right triangle. Show that these lengths satisfy the Pythagorean Theorem.
25. 26.
In Exercises 29 and 30, find the lengths of the sides of the triangle with the indicated vertices, and determine whether the triangle is a right triangle, an isosceles triangle, or neither.
29. 30.
In Exercises 31–38, find the midpoint of the line segment joining the points.
**31. 32.
38.**
In Exercises 39–46, find the standard form of the equation of the sphere with the given characteristics.
39. 40. 41. Center: radius: 3 42. Center: radius: 6 43. Center: diameter: 10 44. Center: diameter: 8 45. Endpoints of a diameter: 46. Endpoints of a diameter:
In Exercises 47–56, find the center and radius of the sphere.
**47.
56.**
In Exercises 57– 60, sketch the graph of the equation and sketch the specified trace.
57. -trace 58. -trace 59. -trace 60. -trace
In Exercises 61 and 62, use a three-dimensional graphing utility to graph the sphere.
61. 62.
63. Crystals Crystals are classified according to their symmetry. Crystals shaped like cubes are classified as isometric. The vertices of an isometric crystal mapped onto a three-dimensional coordinate system are shown in the figure. Determine
FIGURE FOR 63 FIGURE FOR 64
64. Crystals Crystals shaped like rectangular prisms are classified as tetragonal. The vertices of a tetragonal crystal mapped onto a three-dimensional coordinate system are shown in the figure. Determine 65. Architecture A spherical building has a diameter of 165 feet. The center of the building is placed at the origin of a three-dimensional coordinate system. What is the equation of the sphere?
x , y , z .
x
y
z
( x , y , z )
x
y
z
( x , y , z )
x , y , z .
x^2 y^2 z^2 6 y 8 z 21 0
x^2 y^2 z^2 6 x 8 y 10 z 46 0
x^2 y 1 ^2 z 1 ^2 4; xy
x 2 ^2 y 3 ^2 z^2 9; yz
x^2 y 3 ^2 z^2 25; yz
x 1 ^2 y^2 z^2 36; xz
4 x^2 4 y^2 4 z^2 4 x 32 y 8 z 33 0
9 x^2 9 y^2 9 z^2 6 x 18 y 1 0
2 x^2 2 y^2 2 z^2 2 x 6 y 4 z 5 0
9 x^2 9 y^2 9 z^2 18 x 6 y 72 z 73 0
x^2 y^2 z^2 8 y 6 z 13 0
x^2 y^2 z^2 4 x 8 z 19 0
x^2 y^2 z^2 6 x 4 y 9 0
x^2 y^2 z^2 4 x 2 y 6 z 10 0
x^2 y^2 z^2 8 y 0
x^2 y^2 z^2 5 x 0
2
2
3
3 (^3 ) (^5 )
6 4 2
r = 2
y
x
z
65 4 5 6
r = 4
x^ y
z
2 3
2
4
− 3 ( 2, 5, 0)− (^) y
x
z
3
(^2 )
4
− 3 − 4
− (^3) − (^2)
− 4
(0, 4, 0) (^) y
x
z
818 Chapter 11 Analytic Geometry in Three Dimensions
What you should learn
- Find the component forms of the unit vectors in the same direction of, the magnitudes of, the dot products of, and the angles between vectors in space.
- Determine whether vectors in space are parallel or orthogonal.
- Use vectors in space to solve real-life problems.
Why you should learn it Vectors in space can be used to represent many physical forces, such as tension in the cables used to support auditorium lights, as shown in Exercise 44 on page 826.
Vectors in Space
Physical forces and velocities are not confined to the plane, so it is natural to
extend the concept of vectors from two-dimensional space to three-dimensional
space. In space, vectors are denoted by ordered triples
Component form
The zero vector is denoted by Using the unit vectors
and in the direction of the positive -axis, the
standard unit vector notation for is
Unit vector form
as shown in Figure 11.14. If v is represented by the directed line segment from
to as shown in Figure 11.15, the component form of
is produced by subtracting the coordinates of the initial point from the
corresponding coordinates of the terminal point
FIGURE 11.14 FIGURE 11.
Q q ( (^) 1 , q (^) 2 , q 3 )
P p ( (^) 1 , p (^) 2 , p 3 ) v
x
y
z
〈 v^ 1 ,^ v^ 2 , v 3 〉
i
k j
x
y
z
v v 1 , v 2 , v 3 q 1 p 1 , q 2 p 2 , q 3 p 3 .
v
P p 1 , p 2 , p 3 Q q 1 , q 2 , q 3 ,
v v 1 i v 2 j v 3 k
v
j 0, 1, 0, k 0, 0, 1 z
0 0, 0, 0. i 1, 0, 0,
v v 1 , v 2 , v 3 .
820 Chapter 11 Analytic Geometry in Three Dimensions
Vectors in Space
SuperStock
11.
Vectors in Space
1. Two vectors are equal if and only if their corresponding components are
equal.
2. The magnitude (or length ) of is
3. A unit vector in the direction of is
4. The sum of and is
Vector addition
5. The scalar multiple of the real number and is
Scalar multiplication
6. The dot product of and is
u (^) v u 1 v 1 u 2 v 2 u 3 v 3. Dot product
u u 1 , u 2 , u 3 v v 1 , v 2 , v 3
c u cu 1 , cu 2 , cu 3 .
c u u 1 , u 2 , u 3
u v u 1 v 1 , u 2 v 2 , u 3 v 3 .
u u 1 , u 2 , u 3 v v 1 , v 2 , v 3
u v 0.
v
v
u v ,
u u 1 , u 2 , u 3 u u 12 u 22 u 32.
Finding the Component Form of a Vector
Find the component form and magnitude of the vector having initial point
and terminal point Then find a unit vector in the direction of
Solution
The component form of is
which implies that its magnitude is
The unit vector in the direction of is
Now try Exercise 3.
Finding the Dot Product of Two Vectors
Find the dot product of and
Solution
Note that the dot product of two vectors is a real number, not a vector.
Now try Exercise 19.
As was discussed in Section 6.4, the angle between two nonzero vectors is
the angle between their respective standard position vectors, as
shown in Figure 11.16. This angle can be found using the dot product. (Note that
the angle between the zero vector and another vector is not defined.)
FIGURE 11.
If the dot product of two nonzero vectors is zero, the angle between the
vectors is (recall that Such vectors are called orthogonal. For
instance, the standard unit vectors , i j , and k are orthogonal to each other.
90 cos 90 0).
v − u
v
u
Origin
θ
0, 3, 2 (^) 4, 2, 3 0 4 3 2 2 3
u
v
v
0, 2, 2 (^) 0,
2 ^
(^) 0,
2
v
v 02 22 22 8 2 2.
v 3 3, 6 4, 4 2 0, 2, 2
v
3, 4, 2 3, 6, 4. v.
v
Section 11.2 Vectors in Space 821
Some graphing utilities have the capability to perform vector operations, such as the dot product. Consult the user’s guide for your graphing utility for specific instructions.
Te c h n o l o g y
Example 1
Example 2
Angle Between Two Vectors
If is the angle between two nonzero vectors and , then cos
u v
u v
u v.
You can use vectors to determine whether three points are collinear (lie on
the same line). The points and are collinear if and only if the vectors
and are parallel.
Using Vectors to Determine Collinear Points
Determine whether the points and are
collinear.
Solution
The component forms of and are
and
Because you can conclude that they are parallel. Therefore, the
points and lie on the same line, as shown in Figure 11.19.
FIGURE 11.
Now try Exercise 31.
Finding the Terminal Point of a Vector
The initial point of the vector is What is the
terminal point of this vector?
Solution
Using the component form of the vector whose initial point is and
whose terminal point is you can write
This implies that and The solutions
of these three equations are and So, the terminal point is
Now try Exercise 35.
Q 7, 1, 5.
q 1 7, q 2 1, q 3 5.
q 1 3 4, q 2 1 2, q 3 6 1.
q 1 3, q 2 1, q 3 6 4, 2, 1 .
PQ
**
q 1 p 1 , q 2 p 2 , q 3 p 3
Q q 1 , q 2 , q 3 ,
P 3, 1, 6
v 4, 2, 1 P 3, 1, 6.
− 2 2
2
4
4
− 10 − 8 − 6 − 4
R ( − 4, −11, 0) P (2,^ −^ 1, 4)
Q (5 , 4, 6)
PR = 〈−6, −10, − 4 〉 PQ^ = 3, 5, 2〈^ 〉
x
y
z
P , Q , R
PR
**
2 PQ
**
PR
** 4 2, 11 1 , 0 4 6, 10, 4 .
PQ
** 5 2, 4 1 , 6 4 3, 5, 2
PR
**
PQ
**
P 2, 1, 4, Q 5, 4, 6, R 4, 11, 0
PR
^ PQ
**
P , Q , R
Section 11.2 Vectors in Space 823
Example 5
Example 6
Application
In Section 6.3, you saw how to use vectors to solve an equilibrium problem in a
plane. The next example shows how to use vectors to solve an equilibrium
problem in space.
Solving an Equilibrium Problem
A weight of 480 pounds is supported by three ropes. As shown in Figure 11.20,
the weight is located at The ropes are tied to the points
and Find the force (or tension) on each rope.
Solution
The (downward) force of the weight is represented by the vector
The force vectors corresponding to the ropes are as follows.
For the system to be in equilibrium, it must be true that
or
This yields the following system of linear equations.
Using the techniques demonstrated in Chapter 7, you can find the solution of the
system to be
So, the rope attached at point has 360 pounds of tension, the rope attached at
point has about 536.7 pounds of tension, and the rope attached at point has
360 pounds of tension.
Now try Exercise 43.
Q R
P
z 360.0.
v 536.
u 360.
u
v
z 480
u
v
z 000
u
z 000
u v z w 0 u v z w.
z z
SR
**
SR
**
z
2 0, 0 2, 0 1
z
3
v v
SQ
**
SQ
**
v
0 0, 4 2, 0 1
v 0,
5
u u
SP
**
SP
**
u
2 0, 0 2, 0 1
u (^)
3
w 0, 0, 480 .
Q 0, 4, 0, R 2, 0, 0.
S 0, 2, 1 . P 2, 0, 0,
824 Chapter 11 Analytic Geometry in Three Dimensions
P (2, 0, 0)
Q (0, 4, 0)
R ( 2, 0, 0)
S (0, 2, 1)
u
z
w
v
1
4
3
1 4 2
3
4
(^21) 3
3 4
1
2
3
x
y
z
FIGURE 11.
Example 7
In Exercises 27–30, determine whether u and v are orthog- onal, parallel, or neither.
27. 28.
In Exercises 31–34, use vectors to determine whether the points are collinear.
31.
32.
33.
34.
In Exercises 35–38, the vector v and its initial point are given. Find the terminal point.
35. 36.
Initial point: Initial point:
37. 38.
Initial point: Initial point:
39. Determine the values of such that where 40. Determine the values of such that where
In Exercises 41 and 42, write the component form of v.
41. v lies in the -plane, has magnitude 4, and makes an angle of with the positive -axis. 42. v lies in the -plane, has magnitude 10, and makes an angle of 60 with the positive -axis. 43. Tension The weight of a crate is 500 newtons. Find the tension in each of the supporting cables shown in the figure.
Synthesis
True or False? In Exercises 45 and 46, determine whether the statement is true or false. Justify your answer.
45. If the dot product of two nonzero vectors is zero, then the angle between the vectors is a right angle. 46. If and are parallel vectors, then points and are collinear. 47. What is known about the nonzero vectors and if Explain. 48. Writing Consider the two nonzero vectors and Describe the geometric figure generated by the terminal points of the vectors and where and represent real numbers.
Skills Review
In Exercises 49–52, find a set of parametric equations for the rectangular equation using (a) and (b)
51. y x^2 8 52. y 4 x^3
y
x
y 3 x 2
t x t x 1.
t
t v , u t v , s u t v , s
u v.
u (^) v < 0?
u v
AB AC A , B , C
x y A
B
C D 60 cm
45 cm 70 cm
65 cm
115 cm
z
z
xz
45 y
yz
u 2 i 2 j 4 k.
c c u 12,
u i 2 j 3 k.
c c u 3,
2, 1,^ ^32 3, 2,^ ^12
v (^) 4, 32 , ^14 v (^) 52 , ^12 , 4
v 2, 4, 7 v 4, 1, 1
v 4 i 10 j k v 8 i 4 j 8 k
u 34 i 12 j 2 k u i 12 j k
v 8, 4, 10 v 2, 1, 5
u 12, 6, 15 u 1, 3, 1
826 Chapter 11 Analytic Geometry in Three Dimensions
44. Tension The lights in an auditorium are 24-pound
disks of radius 18 inches. Each disk is supported by three equally spaced cables that are inches long (see figure).
(a) Write the tension in each cable as a function of Determine the domain of the function. (b) Use the function from part (a) to complete the table.
(c) Use a graphing utility to graph the function in part (a). What are the asymptotes of the graph? Interpret their meaning in the context of the problem. (d) Determine the minimum length of each cable if a cable can carry a maximum load of 10 pounds.
T L.
18 in.
L
L
Model It
L 20 25 30 35 40 45 50
T
Section 11.3 The Cross Product of Two Vectors 827
The Cross Product
Many applications in physics, engineering, and geometry involve finding a
vector in space that is orthogonal to two given vectors. In this section, you will
study a product that will yield such a vector. It is called the cross product , and
it is conveniently defined and calculated using the standard unit vector form.
It is important to note that this definition applies only to three-dimensional
vectors. The cross product is not defined for two-dimensional vectors.
A convenient way to calculate is to use the following determinant
form with cofactor expansion. (This determinant form is used simply to
help remember the formula for the cross product—it is technically not a deter-
minant because the entries of the corresponding matrix are not all real numbers.)
Note the minus sign in front of the -component. Recall from Section 8.4 that
each of the three determinants can be evaluated by using the following
pattern.
a 1
a 2
b 1
b 2
a 1 b 2 a 2 b 1
j
u 2 v 3 u 3 v 2 i u 1 v 3 u 3 v 1 j u 1 v 2 u 2 v 1 k
u 2
v 2
u 3
v 3
i
u 1
v 1
u 3
v 3
j
u 1
v 1
u 2
v 2
k
u v
i
u 1
v 1
j
u 2
v 2
k
u 3
v 3
u v
What you should learn
- Find cross products of vectors in space.
- Use geometric properties of cross products of vectors in space.
- Use triple scalar products to find volumes of parallelepipeds.
Why you should learn it
The cross product of two vectors in space has many applications in physics and engineering. For instance, in Exercise 43 on page 833, the cross product is used to find the torque on the crank of a bicycle’s brake.
The Cross Product of Two Vectors
Carl Schneider/Getty Images
11.
Definition of Cross Product of Two Vectors in Space
Let
and
be vectors in space. The cross product of and is the vector
u v u 2 v 3 u 3 v 2 i u 1 v 3 u 3 v 1 j u 1 v 2 u 2 v 1 k.
u v
u u 1 i u 2 j u 3 k v v 1 i v 2 j v 3 k
Put u in Row 2.
Put v in Row 3.
Find each cross product. What can you conclude?
a. i j b. i k c. j k
Exploration
Geometric Properties of the Cross Product
The first property listed on the preceding page indicates that the cross product is
not commutative. In particular, this property indicates that the vectors and
have equal lengths but opposite directions. The following list gives some
other geometric properties of the cross product of two vectors.
For proofs of the Geometric Properties of the Cross Product, see Proofs in
Mathematics on page 848.
Both and are perpendicular to the plane determined by u and v.
One way to remember the orientations of the vectors u , v , and is to
compare them with the unit vectors i , j , and as shown in Figure 11.21.
The three vectors u , v , and form a right-handed system.
Using the Cross Product
Find a unit vector that is orthogonal to both
and
Solution
The cross product as shown in Figure 11.22, is orthogonal to both and
Because
a unit vector orthogonal to both and is
Now try Exercise 15.
u v
u v
i
j
k.
u v
u v 6 ^2 3 ^2 62
6 i 3 j 6 k
u v
i
j
k
(^0)
u v , u v.
u 3 i 4 j k v 3 i 6 j.
u v
k i j ,
u v
u v v u
v u
u v
Section 11.3 The Cross Product of Two Vectors 829
2
− 2
− 4
− 6
2 x 4
y
2
4
6
8
4 6
u
v
u ⋅ v
z
FIGURE 11.
j
i
k = i × j
xy -plane
x
y
z
FIGURE 11.
Geometric Properties of the Cross Product
Let and be nonzero vectors in space, and let be the angle between
and
1. is orthogonal to both and
3. if and only if and are scalar multiples of each other.
4. u v area of parallelogram having u and v as adjacent sides.
u v 0 u v
u v u v sin
u v u v.
u v.
u v
Example 2
In Example 2, note that you could have used the cross product to form
a unit vector that is orthogonal to both u and v. With that choice, you would have
obtained the negative of the unit vector found in the example.
The fourth geometric property of the cross product states that is the
area of the parallelogram that has u and v as adjacent sides. A simple example of
this is given by the unit square with adjacent sides of i and j. Because
and it follows that the square has an area of 1. This geometric
property of the cross product is illustrated further in the next example.
Geometric Application of the Cross Product
Show that the quadrilateral with vertices at the following points is a parallelo-
gram. Then find the area of the parallelogram. Is the parallelogram a rectangle?
Solution
From Figure 11.23 you can see that the sides of the quadrilateral correspond to
the following four vectors.
Because and you can conclude that is parallel to
and is parallel to It follows that the quadrilateral is a parallelogram
with and as adjacent sides. Moreover, because
the area of the parallelogram is
You can tell whether the parallelogram is a rectangle by finding the angle
between the vectors and
Because the parallelogram is not a rectangle.
Now try Exercise 27.
arcsin 0.
sin
AB
**
AD
**
AB
**
AD
**
AD
**
AB.
**
AB
**
AD
**
AB
**
AD
**
i
j
k
(^6)
26 i 18 j 6 k
AD
**
AB
\ CB
**
AD.
**
CD
^ AB
**
CB
**
AD
**
CD ,
**
AB
**
CB
**
0 i 2 j 6 k AD
**
AD
**
0 i 2 j 6 k
CD
**
3 i 4 j k AB
**
AB
**
3 i 4 j k
A 5, 2, 0, B 2, 6, 1, C 2, 4, 7, D 5, 0, 6
k 1,
i j k
u v
v u
830 Chapter 11 Analytic Geometry in Three Dimensions
Example 3
If you connect the terminal points of two vectors u and v that have the same initial points, a triangle is formed. Is it possible to use the cross product u v to determine the area of the triangle? Explain. Verify your conclusion using two vectors from Example 3.
Exploration
x
y
4
8
(^6 )
6
8
B (2, 6, 1)
D (5, 0, 6)^ C^ (2, 4, 7)
A (5, 2, 0)
z
FIGURE 11.
