Transcript Module 2 Chapter 14 Vectors in 3D

14 Vectors in Three-dimensional Space
Case Study
14.1
Vectors in Three-dimensional
Rectangular Coordinate System
14.2
Vector Product and Scalar Triple Product
Chapter Summary
Case Study
Captain, please follow
the following flight route.
A plane is approaching Hong Kong
International Airport. The flightcontrol operator of the control tower
is trying to give instructions to the
pilot to provide a safe route for landing.
We are now ready for
To describe the position and the route of
landing. Please indicate
a flight route.
the 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.
P. 2
14.1 Vector in Three-dimensional
Rectangular Coordinate System
A. Vectors in Three-dimensional Space
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 AB . The magnitude of AB is denoted
by AB , which is the same as the vectors on the plane
defined before.
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 example, for the cube ABCDEFGH, we have
1. AB  EF  DC  HG
(equal vectors)
2. AB   BA
(negative vectors)
3. AB // EF // DC // HG
(parallel vectors)
4. AF  AB  BF  AD  DF
(addition of vectors )
5. BE  AE  AB  DE  DB
(subtraction of vectors)
P. 3
14.1 Vector in Three-dimensional
Rectangular Coordinate System
A. Vectors in Three-dimensional Space
Example 14.1T
The figure shows a cube ABCDEFGH. Let AB = a,
AE = b and AD = c. Express the following in terms
of a, b and c.
(a) DE
(b) AG
(c) BD  EB
Solution:
(a) DE  DA  AE
  AD  AE
 c  b
 bc
(b) AG  AB  BC  CG
 AB  AD  AE
 abc
(c)
BD  EB  ( BA  AD)  ( EA  AB)
  AB  AD  AE  AB
 AD  AE
 cb
P. 4
14.1 Vector in Three-dimensional
Rectangular Coordinate System
A. Vectors in Three-dimensional Space
Example 14.2T
The figure shows a cube ABCDEFGH. Prove that
EC  FD  DB  EG  2GC.
Solution:
EC  FD  DB  EG
 EC  EG  FD  DB
 GC  FB
 GC  GC
 2GC
P. 5
14.1 Vector in Three-dimensional
Rectangular Coordinate System
B. Representation of Vectors in Three-dimensional
Rectangular Coordinate System
The three-dimensional Cartesian coordinate system R3
consists of three mutually perpendicular axes: x, y and z.
The directions of these axes are aligned in such a way
that they obey the right-hand rule.
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. 6
14.1 Vector in Three-dimensional
Rectangular Coordinate System
B. Representation of Vectors in Three-dimensional
Rectangular Coordinate System
Every point in space can be represented in the threedimensional 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.
These three values are called x-, y- and z-coordinates of
the point 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 OP as
OP  xi  yj  zk.
By Pythagoras’ theorem, we have
OP  x 2  y 2  z 2
P. 7
14.1 Vector in Three-dimensional
Rectangular Coordinate System
B. Representation of Vectors in Three-dimensional
Rectangular Coordinate System
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 OA from OB which is
the same as we did in the case of R2, then we have
AB  ( x2  x1 )i  ( y2  y1 ) j  ( z2  z1 )k
AB  ( x2  x1 ) 2  ( y2  y1 ) 2  ( z2  z1 ) 2
P. 8
14.1 Vector in Three-dimensional
Rectangular Coordinate System
B. Representation of Vectors in Three-dimensional
Rectangular Coordinate System
Example 14.3T
Given two points P(–3, –2, 8) and Q(0, –5, 4). Find the
unit vector in the direction of PQ .
Solution:
OP  3i  2 j  8k
OQ  5j  4k
PQ  OQ  OP
 (5j  4k )  (3i  2 j  8k )
 3i  3j  4k
PQ  32  (3) 2  (4) 2
 34

3i  3j  4k
\ Unit vector PQ 
34
3
3
4

i
j
k
34
34
34
P. 9
14.1 Vector in Three-dimensional
Rectangular Coordinate System
B. Representation of Vectors in Three-dimensional
Rectangular Coordinate System
Consider two vectors pi + qj + rk and si + tj + uk in R3.
As i, j and k are non parallel vectors, we have
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.
P. 10
14.1 Vector in Three-dimensional
Rectangular Coordinate System
B. Representation of Vectors in Three-dimensional
Rectangular Coordinate System
Example 14.4T
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.
Solution:
Let the coordinates of D be (x, y, z).
(a) If ABCD forms a parallelogram, AB  DC.
AB  [2  (3)]i  (5  1) j  (1  5)k  5i  4 j  6k
DC  (6  x)i  (4  y) j  (3  z)k
\ (6  x)i  (4  y) j  (3  z )k  5i  4 j  6k
 6  x  5 \ x  11

\ We have  4  y  4 \ y  0
 3  z  6 \ z  9

\ The coordinates of D are (11, 0, 9).
P. 11
14.1 Vector in Three-dimensional
Rectangular Coordinate System
B. Representation of Vectors in Three-dimensional
Rectangular Coordinate System
Example 14.4T
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.
Solution:
(b) If ABDC forms a parallelogram, AB  CD.
AB  5i  4 j  6k
CD  ( x  6)i  ( y  4) j  ( z  3)k
\ ( x  6)i  ( y  4) j  ( z  3)k  5i  4 j  6k
 x  6  5 \ x  1

\ We have  y  4  4 \ y  8
 z  3  6 \ z  3

\ The coordinates of D are (1, 8, 3).
P. 12
14.1 Vector in Three-dimensional
Rectangular Coordinate System
B. Representation of Vectors in Three-dimensional
Rectangular Coordinate System
Example 14.5T
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.
Solution:
Let m = a a + b b + g c.
a (3i  4 j  2k )  b (i  3j  k )  g (5i  2 j  k )  16i  8 j  7k
(3a  b  5g )i  (4a  3b  2g ) j  (2a  b  g )k  16i  8 j  7k
 3a  b  5g  16

\  4a  3b  2g  8

3
1 5
 2a  b  g  7
Consider the determinant of the coefficient matrix:    4  3 2  55
By Cramer’s rule,
2 1 1
a  55,  b  110 and g  165
 55
110
 165
\a 
 1, b 
 2, g 
 3
55
55
55
\m  a  2b  3c
P. 13
14.1 Vector in Three-dimensional
Rectangular Coordinate System
B. Representation of Vectors in Three-dimensional
Rectangular Coordinate System
Example 14.6T
Given two points A and B with OA = 3i – 8j + 5k and
OB = 6i + 2j – 7k. C is a point on the line segment AB.
Find OC if
(a) C is the mid-point of AB,
(b) C divides AB in the ratio 2 : 1.
Solution:
(a)
(b) OC
OC
OA  OB
1OA  2OB


2
1 2
1
1
 [(3i  8 j  5k )  (6i  2 j  7k )]  [(3i  8 j  5k )  2(6i  2 j  7k )]
2
3
9
4
 i  3j  k

5
i

j  3k
2
3
P. 14
14.1 Vector in Three-dimensional
Rectangular Coordinate System
C. Scalar 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=0
jk=kj=0
ik=ki=0
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)
If a = x1i + y1j + z1k and b = x2i + y2j + z2k are two non-zero vectors,
then
a  b = x1x2 + y1y2 + z1z2,
x1 x2  y1 y2  z1 z2
cos 
,
2
2
2
2
2
2
x1  y1  z1 x2  y2  z2
where  is the angle between a and b.
P. 15
14.1 Vector in Three-dimensional
Rectangular Coordinate System
C. Scalar Product
Example 14.7T
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.
Solution:
(a)
r  s  (2i  3j  k )  (i  2k )
 (2)(1)  (3)(0)  (1)(2)
0
(b)  r s  0
\ r s
\ The angle between r and s is 90.
P. 16
14.1 Vector in Three-dimensional
Rectangular Coordinate System
C. Scalar Product
Example 14.8T
If the vectors ci + 5j – 3k and 2ci + cj + k are
perpendicular to each other, find the value(s) of c.
Solution:
 ci + 5j – 3k and 2ci + cj + k are perpendicular to each other.
\ (ci  5 j  3k )  (2ci  cj  k )  0
(c)(2c)  (5)(c)  (3)(1)  0
2c 2  5c  3  0
(c  3)(2c  1)  0
1
c   3 or
2
P. 17
14.1 Vector in Three-dimensional
Rectangular Coordinate System
C. Scalar Product
Example 14.9T
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.)
Solution:
AB  (5  2)i  (3  1) j  (5  6)k  7i  2 j  k
BC  [0  (5)i  (6  3) j  (5  5)k  5i  9 j
AC  (0  2)i  (6  1) j  (5  6)k  2i  7 j  k
AB  (7) 2  2 2  (1) 2  3 6
BC  5 2  (9) 2  0 2  106
AC  (2) 2  (7) 2  (1) 2  3 6
(7)(2)  (2)(7)  (1)(1) 1

cos BAC 
54
(3 6 )(3 6 )
BAC  88.9389
 89 (cor. to the nearest degree)
P. 18
14.1 Vector in Three-dimensional
Rectangular Coordinate System
C. Scalar Product
Example 14.9T
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.)
Solution:
53
(7)(5)  (2)(9)  (1)(0)

3 636
(3 6 )( 106)
ABC  45.5305
 46 (cor. to the nearest degree)
cos ABC 
ACB  180  BAC  ABC
 180  88.9389  45.5305
 46 (cor. to the nearest degree)
P. 19
14.1 Vector in Three-dimensional
Rectangular Coordinate System
C. Scalar Product
Example 14.10T
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.
Solution:
(a)
x  y  (6 j  5k )  (3i  4 j)  (0)(3)  (6)(4)  (0)(5)  24
Let  be the angle between x and y.
xy
 24
 24

cos 

xy
0 2  6 2  (5) 2 32  (4) 2  0 2 5 61
  128(cor. to the nearest degree)
\ The angle between x and y is 128.
(b)
The length of the projection of y on x
xy
24
24
 y cos 


2
2
2
x
61
0  6  (5)
P. 20
14.2 Vector Product and Scalar
Triple Product
A. Definition of Vector Product
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°).
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,
a  b = |a||b|sin  nˆ ,
where  is the angle between a and b, and nˆ is a unit
vector whose direction is defined by the right-hand rule.
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.
P. 21
14.2 Vector Product and Scalar
Triple Product
A. Definition of Vector Product
For and two non-zero vectors a and b, a  b = 0 if and only if
a and b are parallel to each other.
In particular, if b = a, we have
a  a = 0.
For the unit vectors i, j and k:
ii=jj=kk=0
ij=k
jk=i
ki=j
j  i = k
k  j = i
i  k = j
P. 22
14.2 Vector Product and Scalar
Triple Product
B. Properties of Vector Product
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
Proof of (a):
a  b  a b sin  nˆ
b  a  b a sin  (nˆ )
  a b sin  nˆ
\ b  a = (a  b)
P. 23
14.2 Vector Product and Scalar
Triple Product
B. Properties 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.
Let  be the angle between a and b, and nˆ be the unit vector in the
direction of a × b.
When k > 0, ka b  ka b sin  nˆ
 k a b sin  nˆ
 k a  b 
When k < 0, ka b  ka b sin   nˆ 
 k a b sin   nˆ 
 k a b sin  nˆ
 k a  b 
Similarly, it can be proved that a × (kb) = k(a × b).
\ (ka) × b = a × (kb) = k(a × b)
P. 24
14.2 Vector Product and Scalar
Triple Product
B. Properties of Vector Product
Proof of (e):
Since |a × b| = |a||b|sin ,
|a × b|2 = (|a||b|sin )2
= |a|2|b|2sin2
= |a|2|b|2 − |a|2|b|2cos2
= |a|2|b|2 – (a  b)2
sin2 = 1 – cos2
a – b = |a||b|cos 
P. 25
14.2 Vector Product and Scalar
Triple Product
C. Calculation of Vector Product
We can use the determinant to represent the vector
product:
If a = x1i + y1j + z1k and b = x2i + y2j + z2k,
i
then a  b  x1
x2
j
y1
y2
k
z1 .
z2
P. 26
14.2 Vector Product and Scalar
Triple Product
C. Calculation of Vector Product
Example 14.11T
For the following pairs of vectors m and n, find the
vector products m × n.
(a) m = 3i + 8j, n = 6k
1 1 3
n

i  j k
(b) m = –4i + 2j + 6k,
2 4 4
Solution:
i j k
8 0 3 0 3 8
(a) m  n  3 8 0 
i
j
k  48i  18j
0 6 0 6 0 0
0 0 6
i
j
k
2
6
4 6
4 2
(b) m  n   4 2
1k
3i 1
3 j 1
6  1




1
1
3
2
4
4
4
2
4


2
4
4
 0i  0 j  0k  0
P. 27
14.2 Vector Product and Scalar
Triple Product
C. Calculation of Vector Product
Example 14.12T
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 PQ and PR .
Solution:
PQ  2 j  (i  j  k )  i  3j  k
PR  (i  3j  k )  (i  j  k )  2i  2 j  2k
i
j
k
 3 1
1 1
1  3

i

j

k  8i  8k
PQ PR   1  3  1
2 2 2 2 2 2
2 2 2
PQ  PR  82  02  (8)2  8 2
Unit vectors which are perpendicular to PQ and PR
PQ  PR  8i  8k 
1 
 1

i
k
 
  
2 
PQ  PR  8 2 
 2
P. 28
14.2 Vector Product and Scalar
Triple Product
D. Applications of Vector Product
Consider a parallelogram ABCD.
Area of the parallelogram ABCD = AB  AD
Since the area of ABD is half that of parallelogram ABCD, we can
obtain a formula for the area of triangle:
1
Area of ABD =  AB  AD
2
The above formula can be further rewritten as
1
Area of ABD = 
2
2
2
AB AD  ( AB  AD) 2
P. 29
14.2 Vector Product and Scalar
Triple Product
D. Applications of Vector Product
Example 14.13T
Find the area of the triangle formed by vertices X(2, 1, 1),
Y(0, –1, 0) and Z(–2, 1, –1).
Solution:
XY   j  (2i  j  k )  2i  2 j  k
XZ  (2i  j  k )  (2i  j  k )  4i  2k
i
j
XY  XZ   2  2
k
 2 1  2 1
2 2
i
j
k
1 
0 2 4 2 4 0
2
4 0
 4i  8k
1
Area of XYZ   XY  XZ
2
1
  4 2  0 2  (8) 2
2
2 5
P. 30
14.2 Vector Product and Scalar
Triple Product
E. Scalar Triple Product
The volume of a parallelepiped (a prism with all
faces are parallelograms) with sides a, b and c is given
by |a  (b × c)|.
The expression a  (b  c) is called the scalar triple product of a, b and c.
Since the scalar product of two vectors is a scalar, thus a  (b  c)
is a scalar as the name suggests.
In the three-dimensional rectangular coordinate system, suppose
a = x1i + y1j + z1k, b = x2i + y2j + z2k and c = x3i + y3j + z3k, then
x1
a  (b  c)  x2
x3
y1
y2
y3
z1
z2 .
z3
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. 31
14.2 Vector Product and Scalar
Triple Product
E. Scalar Triple Product
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:
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)
P. 32
14.2 Vector Product and Scalar
Triple Product
E. Scalar Triple Product
Example 14.14T
If p = 2i + j + 3k, q = 3i – j – 2k and r = –i + 2j – k, find
(a) r × p, and
(b) q  (r × p).
Solution:
i j k
(a) r  p   1 2  1
2

1
2 1
3
1 1
i
1 3
2
 7i  j  5k
3
j
1 2
2
1
k
(b) q  (r  p)  q  (7i  j  5k )
 (3i  j  2k )  (7i  j  5k )
 (3)(7)  (1)(1)  (2)(5)
 30
P. 33
14.2 Vector Product and Scalar
Triple Product
E. Scalar Triple Product
Example 14.15T
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 AB , AC
and AD .
Solution:
AB  (3  2)i  (4  1) j  (5  0)k
 5i  3 j  5k
AC  (0  2)i  (2  1) j  (4  0)k
 2i  3 j  4k
AD  (1  2)i  (2  1) j  (5  0)k
 i  j  5k
\ Volume of the parallelepiped
5 3 5
 2 3 4
1
 88
1
5
P. 34
Chapter Summary
14.1 Vectors in Three-dimensional
Rectangular Coordinate System
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
AB  ( x1  x2 ) 2  ( y1  y2 ) 2  ( z1  z2 ) 2 .
P. 35
Chapter Summary
14.1 Vectors in Three-dimensional
Rectangular Coordinate System
1.
2.
3.
The rules of operations and properties of vectors in
the space are the same as vectors on a plane.
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.
For a point P(x, y, z) in R3, the position vector can be expressed
2
2
2
as OP  xi  yj  zk , where OP  x  y  z .
P. 36
Chapter Summary
14.1 Vectors in Three-dimensional
Rectangular Coordinate System
Scalar Product
If a = x1i + y1j + z1k and b = x2i + y2j + z2k, are two non-zero
vectors, then
a b  x1x2  y1 y2  z1z2 ,
x1 x2  y1 y2  z1 z2
cos 
,
2
2
2
2
2
2
x1  y1  z1 x2  y2  z2
where  is the angle between a and b.
P. 37
Chapter Summary
14.2 Vector Product and Scalar Triple Product
Vector Product
1. If a = x1i + y1j + z1k and b = x2i + y2j + z2k, are non-zero
vectors and  is the angle between them, then
a  b | a || b | sin  nˆ
i
j k
2.
 x1
y1
z1
x2
y2
z2
Area of ABC
1
  AB  AC
2
2
2
1
  AB AC  ( AB  AC ) 2
2
P. 38
Chapter Summary
14.2 Vector Product and Scalar Triple Product
Scalar Triple Product
1. If a = x1i + y1j + z1k and b = x2i + y2j + z2k and
c = x3i + y3j + z3k are non-zero vectors, then
x1 y1 z1
a  (b  c)  x2 y2 z2 .
x3 y3 z3
2.
Volume of the parallelepiped with sides a, b and c = |a  (b  c)|.
P. 39