14.1vectors in three-dimensional rectangular coordinate system 14.2vector product and scalar triple...
TRANSCRIPT
14.1 Vectors in Three-dimensional Rectangular Coordinate
System14.2 Vector Product and Scalar Triple
ProductChapter Summary
Case Study
Vectors in Three-dimensional Space14
P. 2
A plane is approaching Hong Kong International Airport. The flight-control operator of the control toweris trying to give instructions to thepilot to provide a safe route for landing.
Case StudyCase Study
To describe the position and the route ofthe plane, we can introduce the three-dimensional rectangular coordinate system as shown in the figure.
Let the airport be the origin of the coordinate system. Then we use the triplet (x, y, z) to describe the horizontal position (x, y as in the two-dimensional case) and the height of the plane (z).
The plane is located at the point A(2, 1, 4),B(−1, −2, 3), C(2, 0, 2), D(1, 1, 1) are points in space, such that the plane follows the route A B C D O.
We are now ready for landing. Please indicate a flight route.
Captain, please follow the following flight route.
P. 3
1144 .1 .1 Vector in Three-dimensional Vector in Three-dimensional Rectangular Coordinate Rectangular Coordinate SystemSystem
A. A. Vectors in Three-dimensional SpaceVectors in Three-dimensional Space
The addition, subtraction, scalar multiplication, negative, parallelism of vectors, and the rules of operations of vectors are also defined in the same way as in the case of plane vectors.
For any two points A and B in space, the directed line segment from A to B is called the vector from A to B, and is denoted by . The magnitude of is denoted by , which is the same as the vectors on the plane defined before.
AB ABAB
For example, for the cube ABCDEFGH, we have1. (equal vectors) HGDCEFAB 2. (negative vectors) BAAB 3. (parallel vectors) HGDCEFAB //////
4. (addition of vectors ) DFADBFABAF 5. (subtraction of vectors) DBDEABAEBE
P. 4
Example 14.1T
Solution:(a)
(b)
(c)
1144 .1 .1 Vector in Three-dimensional Vector in Three-dimensional Rectangular Coordinate Rectangular Coordinate SystemSystem
A. A. Vectors in Three-dimensional SpaceVectors in Three-dimensional Space
AEDADE AEAD
cbbc
CGBCABAG AEADAB
cba
EBBD )()( ABEAADBA ABAEADAB
AEAD bc
The figure shows a cube ABCDEFGH. Let = a, = b and = c. Express the following in terms of a, b and c. (a) (b) (c) EBBD AGDE
AE ADAB
P. 5
Example 14.2T
Solution:
1144 .1 .1 Vector in Three-dimensional Vector in Three-dimensional Rectangular Coordinate Rectangular Coordinate SystemSystem
A. A. Vectors in Three-dimensional SpaceVectors in Three-dimensional Space
The figure shows a cube ABCDEFGH. Prove that .2GCEGDBFDEC
EGDBFDEC DBFDEGEC
FBGC GCGC
GC2
P. 6
1144 .1 .1 Vector in Three-dimensional Vector in Three-dimensional Rectangular Coordinate Rectangular Coordinate SystemSystem
BB. . Representation of Representation of Vectors in Three-dimensional Vectors in Three-dimensional Rectangular Coordinate SystemRectangular Coordinate System
The directions of these axes are aligned in such a way that they obey the right-hand rule.
The three-dimensional Cartesian coordinate system R3 consists of three mutually perpendicular axes: x, y and z.
If the x- and y-axes are represented by the index finger and the middle finger respectively, then the thumb represents the z-axis.
P. 7
1144 .1 .1 Vector in Three-dimensional Vector in Three-dimensional Rectangular Coordinate Rectangular Coordinate SystemSystem
BB. . Representation of Representation of Vectors in Three-dimensional Vectors in Three-dimensional Rectangular Coordinate SystemRectangular Coordinate System
These three values are called x-, y- and z-coordinates of the point respectively.
Every point in space can be represented in the three-dimensional coordinate system by the triplet (x, y, z), where x, y and z represent the directed distances from the yz-, zx- and xy-planes respectively.
The point of intersection of the three axes is called the origin O and its coordinates are (0, 0, 0). In the figure, i, j and k are the unit vectors in the positive directions of x-, y- and z-axes respectively. They have a common starting point at the origin, and their terminal points are (1, 0, 0), (0, 1, 0) and (0, 0, 1) respectively. P(x, y, z) is a point in R3, so we can express the position vector asOP
By Pythagoras’ theorem, we have
. kji zyxOP
222 zyxOP
P. 8
1144 .1 .1 Vector in Three-dimensional Vector in Three-dimensional Rectangular Coordinate Rectangular Coordinate SystemSystem
BB. . Representation of Representation of Vectors in Three-dimensional Vectors in Three-dimensional Rectangular Coordinate SystemRectangular Coordinate System
kji )()()( 121212 zzyyxxAB 2
122
122
12 )()()( zzyyxxAB
Now if two points A(x1, y1, z1) and B(x2, y2, z2) in R3 are given, the vector from A to B can be found by subtracting the position vector from which is the same as we did in the case of R2, then we have
OA OB
P. 9
Example 14.3T
Solution:
1144 .1 .1 Vector in Three-dimensional Vector in Three-dimensional Rectangular Coordinate Rectangular Coordinate SystemSystem
BB. . Representation of Representation of Vectors in Three-dimensional Vectors in Three-dimensional Rectangular Coordinate SystemRectangular Coordinate System
Given two points P(–3, –2, 8) and Q(0, –5, 4). Find the unit vector in the direction of . PQ
kji 823 OPkj 45 OQ
OPOQPQ
kjikjikj
433)823()45(
222 )4()3(3 PQ
34
Unit vector 34
433 kji
PQ
kji34
4
34
3
34
3
P. 10
1144 .1 .1 Vector in Three-dimensional Vector in Three-dimensional Rectangular Coordinate Rectangular Coordinate SystemSystem
BB. . Representation of Representation of Vectors in Three-dimensional Vectors in Three-dimensional Rectangular Coordinate SystemRectangular Coordinate System
Property 14.1 (a) pi + qj + rk = si + tj + uk if and only if p = s, q = t and
r = u, and (b) pi + qj + rk = 0 if and only if p = q = r = 0.
Consider two vectors pi + qj + rk and si + tj + uk in R3. As i, j and k are non parallel vectors, we have
P. 11
Example 14.4T
Solution:
Given three points A(–3, 1, 5), B(2, 5, –1) and C(–6, 4, 3). Find the coordinates of a point D if (a) ABCD forms a parallelogram, (b) ABDC forms a parallelogram.
1144 .1 .1 Vector in Three-dimensional Vector in Three-dimensional Rectangular Coordinate Rectangular Coordinate SystemSystem
BB. . Representation of Representation of Vectors in Three-dimensional Vectors in Three-dimensional Rectangular Coordinate SystemRectangular Coordinate System
Let the coordinates of D be (x, y, z). (a) If ABCD forms a parallelogram, .DCAB
kji )51()15()]3(2[ AB kji 645 kji )3()4()6( zyxDC
kjikji 645)3()4()6( zyx
We have
963
044
1156
zz
yy
xx
The coordinates of D are (11, 0, 9).
P. 12
Example 14.4T
Solution:
Given three points A(–3, 1, 5), B(2, 5, –1) and C(–6, 4, 3). Find the coordinates of a point D if (a) ABCD forms a parallelogram. (b) ABDC forms a parallelogram.
1144 .1 .1 Vector in Three-dimensional Vector in Three-dimensional Rectangular Coordinate Rectangular Coordinate SystemSystem
BB. . Representation of Representation of Vectors in Three-dimensional Vectors in Three-dimensional Rectangular Coordinate SystemRectangular Coordinate System
(b) If ABDC forms a parallelogram, .CDAB kji 645 AB
kji )3()4()6( zyxCD
kjikji 645)3()4()6( zyx
We have
363
844
156
zz
yy
xx
The coordinates of D are (1, 8, 3).
P. 13
Example 14.5T
Solution:
Consider the three vectors a = 3i – 4j + 2k, b = i – 3j – k and c = 5i + 2j + k. If m = –16i – 8j – 7k, express m in terms of a, b and c.
1144 .1 .1 Vector in Three-dimensional Vector in Three-dimensional Rectangular Coordinate Rectangular Coordinate SystemSystem
BB. . Representation of Representation of Vectors in Three-dimensional Vectors in Three-dimensional Rectangular Coordinate SystemRectangular Coordinate System
Let m = a + b + c. )25()3()243( kjikjikji kji 7816
kji )2()234()53( kji 7816
728234
1653
Consider the determinant of the coefficient matrix: 112234513
55
By Cramer’s rule, 165 and 110,55
355
165,2
55
110,1
55
55 cbam 32
P. 14
Example 14.6T
Solution:
1144 .1 .1 Vector in Three-dimensional Vector in Three-dimensional Rectangular Coordinate Rectangular Coordinate SystemSystem
BB. . Representation of Representation of Vectors in Three-dimensional Vectors in Three-dimensional Rectangular Coordinate SystemRectangular Coordinate System
(a)
2
OBOA
OC
)]726()583[(2
1kjikji
kji 32
9
(b)
21
21
OBOA
OC
)]726(2)583[(3
1kjikji
kji 33
45
Given two points A and B with = 3i – 8j + 5k and = 6i + 2j – 7k. C is a point on the line segment AB.Find if (a) C is the mid-point of AB, (b) C divides AB in the ratio 2 : 1.
OBOA
OC
P. 15
1144 .1 .1 Vector in Three-dimensional Vector in Three-dimensional Rectangular Coordinate Rectangular Coordinate SystemSystem
CC. . Scalar ProductScalar Product
In the three-dimensional rectangular coordinate system R3, as the unit base vectors i, j and k are mutually perpendicular, we have i i = j j = k k = 1
i j = j i = 0j k = k j = 0i k = k i = 0
If a = x1i + y1j + z1k and b = x2i + y2j + z2k are two non-zero vectors, then a b = x1x2 + y1y2 + z1z2,
where is the angle between a and b.
,cos2
22
22
22
12
12
1
212121
zyxzyx
zzyyxx
We also have the following properties of scalar product:a b = b a a (b + c) = a b + a c (ka) b = k(a b) = a (kb)
P. 16
Example 14.7T
Solution:
Two vectors r = 2i + 3j – k and s = i + 2k are given. (a) Find the value of r s. (b) Hence find the angle between r and s.
1144 .1 .1 Vector in Three-dimensional Vector in Three-dimensional Rectangular Coordinate Rectangular Coordinate SystemSystem
CC. . Scalar ProductScalar Product
)2()32( kikjisr (a)
0)2)(1()0)(3()1)(2(
(b) 0srsr
The angle between r and s is 90.
P. 17
Example 14.8T
Solution:
If the vectors ci + 5j – 3k and 2ci + cj + k are perpendicular to each other, find the value(s) of c.
ci + 5j – 3k and 2ci + cj + k are perpendicular to each other. 0)2()35( kjikji ccc0)1)(3())(5()2)(( ccc
0352 2 cc0)12)(3( cc
2
1or 3c
1144 .1 .1 Vector in Three-dimensional Vector in Three-dimensional Rectangular Coordinate Rectangular Coordinate SystemSystem
CC. . Scalar ProductScalar Product
P. 18
Example 14.9T
Solution:
Given three points A(2, 1, 6), B(– 5, 3, 5) and C(0, –6, 5) are vertices of ABC. Solve ABC. (Give the answers in surd form or correct to the nearest degree.)
1144 .1 .1 Vector in Three-dimensional Vector in Three-dimensional Rectangular Coordinate Rectangular Coordinate SystemSystem
CC. . Scalar ProductScalar Product
kji )65()13()25( AB kji 27kji )55()36()5(0[ BC ji 95
kji )65()16()20( AC kji 72222 )1(2)7( AB 63
222 0)9(5 BC 106222 )1()7()2( AC
)63)(63(
)1)(1()7)(2()2)(7(cos
BAC54
1
9389.88BAC89 (cor. to the nearest degree)
63
P. 19
Example 14.9T
Solution:
Given three points A(2, 1, 6), B(– 5, 3, 5) and C(0, –6, 5) are vertices of ABC. Solve ABC. (Give the answers in surd form or correct to the nearest degree.)
1144 .1 .1 Vector in Three-dimensional Vector in Three-dimensional Rectangular Coordinate Rectangular Coordinate SystemSystem
CC. . Scalar ProductScalar Product
)106)(63(
)0)(1()9)(2()5)(7(cos
ABC6363
53
5305.45ABC46 (cor. to the nearest degree)
ABCBACACB 180 5305.459389.88180
46 (cor. to the nearest degree)
P. 20
Example 14.10T
1144 .1 .1 Vector in Three-dimensional Vector in Three-dimensional Rectangular Coordinate Rectangular Coordinate SystemSystem
CC. . Scalar ProductScalar Product
Solution:
Given two vectors x = 6j – 5k and y = 3i – 4j. (a) Find the angle between x and y, correct to the
nearest degree. (b) Find the length of the projection of y on x.
)43()56( jikjyx 24)5)(0()4)(6()3)(0( Let be the angle between x and y.
yx
yxcos
(a)
222222 0)4(3)5(60
24
615
24
128 (cor. to the nearest degree)
The angle between x and y is 128.
(b) The length of the projection of y on x
cosyx
yx
222 )5(60
24
61
24
P. 21
Suppose we have two non-zero vectors a and b in the three-dimensional space. The vector product of a and b, denoted by a b, is the vector which is perpendicular to both a and b, with the magnitude equal to
|a b| = |a||b|sin,where is the angle between a and b (with 0° ≤ ≤ 180°).
1144 ..22 Vector Product and Scalar Vector Product and Scalar Triple ProductTriple Product
AA. . Definition of Vector ProductDefinition of Vector Product
Note:1. a b is read as ‘a cross b’. Therefore the vector product is also called the cross product. 2. The vector product is only defined in the three-dimensional space.3. In contrast to the scalar product of two vectors, the vector product
is a vector while the scalar product is a scalar.
a b = |a||b|sin , where is the angle between a and b, and is a unit vector whose direction is defined by the right-hand rule.
n̂n̂
Particularly, the direction of a b is defined in such a way that a, b and a b always obey the right-hand rule. In conclusion,
P. 22
1144 ..22 Vector Product and Scalar Vector Product and Scalar Triple ProductTriple Product
AA. . Definition of Vector ProductDefinition of Vector Product
In particular, if b = a, we have
For the unit vectors i, j and k:
For and two non-zero vectors a and b, a b = 0 if and only if a and b are parallel to each other.
a a = 0.
i i = j j = k k = 0 i j = k j i = k j k = i k j = i k i = j i k = j
P. 23
1144 ..22 Vector Product and Scalar Vector Product and Scalar Triple ProductTriple Product
BB. . Properties of Vector ProductProperties of Vector Product
Proof of (a):
b a = (a b)
Property 14.2 Properties of Vector Product(a) b a = (a b)(b) (a + b) c = a c + b c(c) a (b + c) = a b + a c(d) (ka) b = a (kb) = k(a b)(e) |a b|2 = |a|2|b|2 – (a b)2
n baba ˆsin)ˆ(sin nabab n ba ˆsin
P. 24
1144 ..22 Vector Product and Scalar Vector Product and Scalar Triple ProductTriple Product
BB. . Properties of Vector ProductProperties of Vector Product
Proof of (d):If k = 0 or a = 0 or b = 0, then
(ka) × b = a × (kb) = (k a × b) = 0.
Assume that k 0 and a and b are non-zero.
When k < 0, n baba ˆsin kk
n ba ˆsink bak
n ba ˆsin k
When k > 0, n baba ˆsinkk
bakn ba ˆsink
Similarly, it can be proved that a × (kb) = k(a × b).
(ka) × b = a × (kb) = k(a × b)
Let be the angle between a and b, and be the unit vector in the direction of a × b.
n̂
P. 25
1144 ..22 Vector Product and Scalar Vector Product and Scalar Triple ProductTriple Product
BB. . Properties of Vector ProductProperties of Vector Product
Proof of (e):
|a × b|2 = (|a||b|sin )2
Since |a × b| = |a||b|sin
= |a|2|b|2sin2 = |a|2|b|2 − |a|2|b|2cos2 = |a|2|b|2 – (a b)2
sin2 = 1 – cos2a – b = |a||b|cos
P. 26
1144 ..22 Vector Product and Scalar Vector Product and Scalar Triple ProductTriple Product
CC. . Calculation of Vector ProductCalculation of Vector Product
We can use the determinant to represent the vector product:
If a = x1i + y1j + z1k and b = x2i + y2j + z2k,
then .
222
111
zyxzyxkji
ba
P. 27
Example 14.11T
Solution:
For the following pairs of vectors m and n, find the vector products m × n. (a) m = 3i + 8j, n = 6k(b) m = –4i + 2j + 6k,
(a)
(b)
1144 ..22 Vector Product and Scalar Vector Product and Scalar Triple ProductTriple Product
CC. . Calculation of Vector ProductCalculation of Vector Product
kjin4
3
4
1
2
1
600
083
kji
nm kji00
83
60
03
60
08 ji 1848
4
3
4
1
2
1624
kji
nm kji4
1
2
124
4
3
2
164
4
3
4
162
kji 000 0
P. 28
Example 14.12T
Solution:
1144 ..22 Vector Product and Scalar Vector Product and Scalar Triple ProductTriple Product
CC. . Calculation of Vector ProductCalculation of Vector Product
P, Q and R are three points with position vectors i + j + k, –2j and –i + 3j – k respectively. Find the unit vectors which are perpendicular to and . PRPQ
kjikjij 3)(2PQ
kjikjikji 222)()3( PR
PRPQ222
131
kji
kji22
31
22
11
22
13
ki 88
28)8(08 222 PRPQ
Unit vectors which are perpendicular to and PQ PR
PRPQ
PRPQ
28
88 ki
ki
2
1
2
1
P. 29
1144 ..22 Vector Product and Scalar Vector Product and Scalar Triple ProductTriple Product
DD. . Applications of Vector ProductApplications of Vector Product
Consider a parallelogram ABCD.
Area of the parallelogram ABCD = ADAB
Since the area of ABD is half that of parallelogram ABCD, we can obtain a formula for the area of triangle:
Area of ABD = ADAB2
1
The above formula can be further rewritten as
Area of ABD = 222
)(2
1ADABADAB
P. 30
Example 14.13T
Solution:
Find the area of the triangle formed by vertices X(2, 1, 1), Y(0, –1, 0) and Z(–2, 1, –1).
1144 ..22 Vector Product and Scalar Vector Product and Scalar Triple ProductTriple Product
DD. . Applications of Vector ProductApplications of Vector Product
kjikjij 22)2(XY
kikjikji 24)2()2( XZ
204
122
kji
XZXY kji04
22
24
12
20
12
ki 84
Area of XYZ XZXY 2
1
52
)8(042
1 222
P. 31
Since the scalar product of two vectors is a scalar, thus a (b c) is a scalar as the name suggests.
The volume of a parallelepiped (a prism with all faces are parallelograms) with sides a, b and c is given by |a (b × c)|.
1144 ..22 Vector Product and Scalar Vector Product and Scalar Triple ProductTriple Product
EE. . Scalar Scalar TripleTriple Product Product
The expression a (b c) is called the scalar triple product of a, b and c.
In the three-dimensional rectangular coordinate system, suppose a = x1i + y1j + z1k, b = x2i + y2j + z2k and c = x3i + y3j + z3k, then
.)(
333
222
111
zyxzyxzyx
cba
Note:If a (b c) = 0, the volume of the parallelepiped with sides a, b and c equals zero. This only when a, b and c are coplanar.
P. 32
1144 ..22 Vector Product and Scalar Vector Product and Scalar Triple ProductTriple Product
EE. . Scalar Scalar TripleTriple Product Product
Property 14.3 Properties of Scalar Triple Product(a) (a b) c = a (b c)(b) a (b c) = b (c a) = c (a b)
Since a determinant is unchanged when interchanging the rows twice and for any non-zero vectors x and y,x y = y x, we have the following properties of the scalar triple product:
P. 33
Example 14.14T
Solution:
If p = 2i + j + 3k, q = 3i – j – 2k and r = –i + 2j – k, find (a) r × p, and (b) q (r × p).
(a)
(b)
1144 ..22 Vector Product and Scalar Vector Product and Scalar Triple ProductTriple Product
EE. . Scalar Scalar TripleTriple Product Product
312
121 kji
pr
kji12
21
32
11
31
12
kji 57
)( prq )57( kjiq )57()23( kjikji )5)(2()1)(1()7)(3(
30
P. 34
Example 14.15T
Solution:
1144 ..22 Vector Product and Scalar Vector Product and Scalar Triple ProductTriple Product
EE. . Scalar Scalar TripleTriple Product Product
Consider A(2, 1, 0), B(–3, 4, 5), C(0, –2, 4) and D(1, 2, 5). Find the volume of the parallelepiped with sides , and . AD
ACAB
kjikji
535)05()14()23(
AB
kjikji
432)04()12()20(
AC
kjikji
5)05()12()21(
AD
Volume of the parallelepiped
511
432
535
88
P. 35
14.1 Vectors in Three-dimensional Rectangular Coordinate System
Chapter Chapter SummarySummary
1. Every point in the space can be represented in the three-dimensional coordinate system by the triplet (x, y, z), where x, y and z represent the directed
distances from the yz-, zx- and xy-planes respectively.
2. The distance between two points A(x1, y1, z1) and B(x2, y2, z2) is given by
.)()()( 221
221
221 zzyyxxAB
P. 36
Chapter Chapter SummarySummary14.1 Vectors in Three-dimensional Rectangular Coordinate System
1. The rules of operations and properties of vectors in the space are the same as vectors on a plane.2. In R3, we define three mutually perpendicular unit vectors i, j and
k, which point in the positive direction of x-, y- and z-axes respectively.
3. For a point P(x, y, z) in R3, the position vector can be expressed
as , where .222 zyxOP kji zyxOP
P. 37
Chapter Chapter SummarySummary14.1 Vectors in Three-dimensional Rectangular Coordinate System
Scalar ProductIf a = x1i + y1j + z1k and b = x2i + y2j + z2k, are two non-zero vectors, then
where is the angle between a and b.
,212121 zzyyxx ba
,cos2
22
22
22
12
12
1
212121
zyxzyx
zzyyxx
P. 38
14.2 Vector Product and Scalar Triple Product
Chapter Chapter SummarySummary
Vector Product1. If a = x1i + y1j + z1k and b = x2i + y2j + z2k, are non-zero vectors and is the angle between them, then
n bab a ˆsin||||
222
111
zyx
zyx
kji
2. Area of ABC
ACAB2
1
222)(
2
1ACABACAB
P. 39
Scalar Triple Product
14.2 Vector Product and Scalar Triple Product
Chapter Chapter SummarySummary
2. Volume of the parallelepiped with sides a, b and c = |a (b c)|.
1. If a = x1i + y1j + z1k and b = x2i + y2j + z2k and c = x3i + y3j + z3k are non-zero vectors, then
.)(
333
222
111
zyxzyxzyx
cba