MAT 1339-S14 Class 8 July 28, 2014 Contents 7.2 Review Dot Product........................... 2 7.3 Applications of the Dot Product..................... 4 7.4 Vectors in Three-Space......................... 6 7.5 Cross Product.............................. 10 7.6 Applications of the Dot Product and Cross Product........... 13 Review Algebraically we represent a vector by an ordered pair v = [a, b], where its tail is at the origin O(0, 0) and its head is at the point P (a, b) on the Cartesian plane. A geometric vector v can be written in Cartesian form as v = [ v cos θ, v sin θ], where θ is the angle between v and the positive x-axis. Using Cartesian vectors, it is very easy to calculate vector operations. Let u = [u 1, u 2 ] and v = [v 1, v 2 ] be two vectors and let k R be scalar. Then 1
Vector Addition: u + v = [u 1, u 2 ] + [v 1, v 2 ] = [u 1 + v 1, u 2 + v 2 ] Vector Subtraction: u v = [u 1, u 2 ] [v 1, v 2 ] = [u 1 v 1, u 2 v 2 ] Scalar Multiplication: Magnitude of v: k v = k[v 1, v 2 ] = [kv 1, kv 2 ] v = (v 1 ) 2 + (v 2 ) 2 Unit Vectors collinear with v: ± 1 v v Definition 7.1. Vectors are collinear if they lie on a straight line when arranged tail to tail. The vectors are also scalar multiples of each other, which means that they are parallel. Question: Is collinear vectors are parallel?, Is parallel vectors are collinear?, Is opposite vectors are collinear? Example 7.2. Find two unit vectors collinear with v = [5, 7]. Solution: Thus and are two unit vectors collinear with v. 7.2 Review Dot Product v = 5 2 + ( 7) 2 = 74. 1 v v = 1 5 [5, 7] = [, 7 ] 74 74 74 1 v v = 1 [5, 7] = [ 5 7, ] 74 74 74 Definition 7.3. Given two vectors a and b, the dot product is defined as a b = a b cos θ, where θ is the angle between a and b when the vectors are arranged tail to tail, and 0 θ π. The dot product is a scalar, not a vector. Theorem 7.4. Let a = [a 1, a 2 ] and b = [b 1, b 2 ]. Then a b = a 1 b 1 + a 2 b 2. 2
Example 7.5. Calculate dot products of vectors. a) Let u = [1, 2] and v = [0, 0]. Then u v = 1 0 + 2 0 = 0. b) Let u = [1, 2]. Then u u = 1 1 + 2 2 = 5. Notice that u = 1 2 + 2 2 = 5, thus u 2 = 5. c) Let u = [1, 2] and v = [4, 5]. Then u v = 1 4 + 2 5 = 14, v u = 4 1 + 5 2 = 14. d) Let u = [1, 2], v = [4, 5] and w = [1, 3]. Then u ( v + w) = [1, 2] [4 + 1, 5 + 3] = [1, 2] [5, 8] = 1 5 + 2 8 = 21, u v + u w = (1 4 + 2 5) + (1 1 + 2 3) = 14 + 7 = 21. e) Let k = 2, u = [1, 2] and v = [4, 5]. Then (k u) v = [2, 4] [4, 5] = 2 4 + 4 5 = 28, k( u v) = 2 14 = 28, u (k v) = [1, 2] [8, 10] = 1 8 + 2 10 = 28. Theorem 7.6. For any vectors u, v and w and scalar k R, we have a) u 0 = 0 b) u u = u 2 c) u v = v u (commutative property) d) u ( v + w) = u v + u w and ( v + w) u = v u + w u (distributive property) e) (k u) v = k( u v) = u (k v) (associative property). For examples of using properties of the dots product to expand and simplify vector expressions, see textbook example 3 on page 373 and exercise 3 on page 376. 3
7.3 Applications of the Dot Product Angle between Two Vectors Theorem 7.7. The angle 0 θ π between two nonzero vectors u and v is determined by u v cos θ = u v. Example 7.8. Find the angle between u = [ 3, 6] and v = [4, 2]. Solution: cos θ = u v u v = ( 3) 4 + 6 2 ( 3)2 + 6 2 4 2 + 2 2 = 0 45 20 = 0, thus θ = π, and u and v are perpendicular. 2 Definition 7.9. Two nonzero vectors u and v are called perpendicular (orthogonal) to each other if the angle between them is π 2. Theorem 7.10. u 0 and v 0 are perpendicular if and only if u v = 0. Please review the pervious class for more examples about perpendicular vectors! Example 7.11. 1) Find x such that the [1, 3] and [ 1, x] are collinear? 2) Find x such that the [1, 3] and [ 1, x] are perpendicular? 4
Vector Projections Definition 7.12. Given two vectors u and v. The projection of v on u, denoted by proj u v, is the rectangular component of v in the direction of u. Theorem 7.13. The projection of v on u is the vector given by ( ) v u proj u v = u. u u Proof. There are two cases. ( ) 1 If 0 θ π, then proj 2 u v = v cos θ u u v u = v u v ( If π θ π, then proj 2 u v = v cos θ 1 ) u u = v ( ) 1 u u = v u u v ( v u ( ) 1 u u = ) u. u u ( v u u u ) u. 5
Example 7.14. Determine the projection of v = [2, 3] on u = [1, 4]. Solution: ( ) v u proj u v = u = u u ( [2, 3] [1, 4] [1, 4] [1, 4] ) u = 2 1 + ( 3) 4 u = 10 [1, 4] = [ 10 1 2 + 4 2 17 17, 40 17 ]. Example 7.15 (Problem 9 in DGD 4 part c)). c) Discuss finding proj u v and proj p q in a different way using part if we know cos(θ) 7.4 Vectors in Three-Space A vector in three-space (a 3-D vector) can be represented algebraically by an ordered triple v = [a, b, c], where its tail is at the origin O(0, 0, 0) and its head is at the point P (a, b, c) on the 3-D Cartesian system. Hence v = OP. Unit Vectors in 3-D Cartesian Vectors: As in two dimension, we define the unit vectors along the axes. The unit vector along the x axes is i = [1, 0, 0], the unit vector along the y axes is j = [0, 1, 0] and the unit vector along the z axes is k = [0, 0, 1]. Hence i, j, and k all have magnitude, or length, one. Any 3-D vector a can be written as [a x, a y, a z ]. 6
From the diagram we can see that [a x, a y, a z ] = [a x, 0, 0] + [0, a y, 0] + [0, 0, a z ] = a x i + a y j + a z k Many of the tools we developed for 2-D Cartesian vectors can be easily generalized to 3-D Cartesian vectors. Let u = [u 1, u 2, u 3 ] and v = [v 1, v 2, v 3 ] be two vectors and let k R be scalar. Then Magnitude of a Cartesian Vector u: Vector Addition: u = u 2 1 + u2 2 + u2 3 u + v = [u 1, u 2, u 3 ] + [v 1, v 2, v 3 ] = [u 1 + v 1, u 2 + v 2, u 3 + v 3 ] Vector Subtraction: u v = [u 1, u 2, u 3 ] [v 1, v 2, v 3 ] = [u 1 v 1, u 2 v 2, u 3 v 3 ] Scalar Multiplication: k v = k[v 1, v 2, v 3 ] = [kv 1, kv 2, kv 3 ] Example 7.16. Find b and c such that [ 2, b, 7] and [c, 6, 21] are collinear. 7
Vector Between Two Points: The vector P 1 P 2 from the point P 1 (x 1, y 1, z 1 ) and P 2 (x 2, y 2, z 2 ) is P 1 P 2 = [x 2 x 1, y 2 y 1, z 2 z 1 ] Magnitude of a Vector Between Two Points (Distance Between Two Points): The magnitude of the vector P 1 P 2 between the points P 1 (x 1, y 1, z 1 ) and P 2 (x 2, y 2, z 2 ) is P 1 P 2 = (x2 x 1 ) 2 + (y 2 y 1 ) 2 + (z 2 z 1 ) 2 Example 7.17 (Example 5, page 394). a) Given the points A(3, 6, 1) and B( 1, 0, 5), express AB as an ordered triple. Then write AB in terms of i, j and k. b) Determine the magnitude of AB. c) Determine the unite vector, u, in the direction of AB. The Dot Product for 3 D Cartesian Vectors: For two 3 D Cartesian Vectors u = [u 1, u 2, u 3 ] and v = [v 1, v 2, v 3 ], u v = u 1 v 1 + u 2 v 2 + u 3 v 3 Example 7.18. Let u = [2, 3, 5], v = [8, 4, 3] and w = [ 6, 2, 0], find: a) 3 v b) u + v + w c) u v d) 3 v u v 8
Example 7.19. Determine if the vectors a = [6, 2, 4] and b = [9, 3, 6] are collinear. Example 7.20. find a vector that is Orthogonal to [3, 4, 5] Example 7.21 (Example 9, page 397). a crane lifts a steel beam upward with a force of 5000 N. At the same time, two workers push teh beam with forces of 600 N toward the west and 650 N toward the north. Determine the magnitude of the resultant force acting on the beam 9
7.5 Cross Product Today we will introduce a new operation of 3-D vectors, called the cross product. Definition of the Cross Product Definition 7.22. Given two 3-D vectors a and b, the cross product of a and b, denoted by a b, is a new 3-D vector defined by a b = ( a b sin θ)ˆn, where 0 θ π is the angle between a and b, and n is the unit vector perpendicular to both a and b such that a, b, ˆn follow the right-hand rule as illustrated. Theorem 7.23. For any vectors a and b, we have a b = b a. product is NOT commutative, but anti-commutative. Recall that a b = b a. Theorem 7.24. a b = a b sin θ. Proof. Since ˆn is a unit vector, we have ˆn = 1. Thus So the cross a b = a b sin θ ˆn = a b sin θ. 10
From the definition of cross product, we can see that a b is a new vector with Magnitude: a b = a b sin θ. Direction: a b is perpendicular to both a and b in the direction of ˆn where a, b and ˆn form a right-handed system. The Cross Product in Cartesian Form Theorem 7.25. Let a = [a 1, a 2, a 3 ] and b = [b 1, b 2, b 3 ], then a b = [a 2 b 3 a 3 b 2, a 3 b 1 a 1 b 3, a 1 b 2 a 2 b 1 ]. Here is a visual way to remember how to perform these oporation: Proof. Once can check that a b = [a 2 b 3 a 3 b 2, a 3 b 1 a 1 b 3, a 1 b 2 a 2 b 1 ] and [a 2 b 3 a 3 b 2, a 3 b 1 a 1 b 3, a 1 b 2 a 2 b 1 ] is perpendicular to both a and b by using the dot product. For details you can see the textbook page 405-406. Example 7.26. Let a = [7, 1, 2] and b = [4, 3, 6]. Then a b = [a 2 b 3 a 3 b 2, a 3 b 1 a 1 b 3, a 1 b 2 a 2 b 1 ] = [1 6 ( 2)3, ( 2)4 7 6, 7 3 1 4] = [12, 50, 17]. or by the visual way: 11
Geometric Meaning The geometric meaning of cross product is that the magnitude of a b represents the area of the parallelogram formed by a and b. Example 7.27. a) Determine the area of the parallelogram defined by the vectors. u = [4, 5, 2] and v = [3, 2, 7] b) Determine the angle between vectors u and v. Properties of the Cross Product Theorem 7.28. Let a and b be non-zero vectors. Then a b = 0 if and only if a and b are collinear. Proof. a b = 0 if and only if sin θ = 0 if and only if θ = 0 or θ = π if and only if a and b are collinear. Recall that a b = 0 if and only if a and b are perpendicular. The following properties of the cross product are similar to those of the dot product. Theorem 7.29. For any vectors u, v and w and scalar k R, we have a) u ( v + w) = u v + u w and ( v + w) u = v u + w u (distributive property), b) (k u) v = k( u v) = u (k v) (associative property). 12
7.6 Applications of the Dot Product and Cross Product Triple Scalar Product The triple scalar product is a combination of the dot product and the cross product. Definition 7.30. Given three 3-D vectors a, b and c, the triple scalar product is a scalar defined by a ( b c). The geometric meaning of the triple scalar product is that a ( b c) equals the volume of the parallelepiped defined by the three vectors. Note that: a ( b c) = b ( a c) = c ( a b) Example 7.31. Find the volume of the parallelepiped defined by u = [4, 5, 2], v = [3, 2, 7], and w = [1, 2, 7]. 13