Vectors and the Geometry of Space

Transcription

1 Vectors and the Geometry of Space

2 Many quantities in geometry and physics, such as area, volume, temperature, mass, and time, can be characterized by a single real number scaled to appropriate units of measure. These are called scalar quantities, and the real number associated with each is called a scalar. Other quantities, such as force, velocity, and acceleration, involve both magnitude and direction and cannot be characterized completely by a single real number. Vectors are usually denoted by lowercase, boldface letters such as u, v, and w. When written by hand, however, vectors are often denoted by letters with arrows above them, such as u, v and w.

3 Draw a vector as a line with an arrowhead at its tip. The length of the line shows the vector s magnitude. The direction of the line shows the vector s direction. If two vectors have the same direction, they are parallel. We define the negative of a vector as a vector having the same magnitude as the original but the opposite direction.

4 A directed line segment is used to represent vectors. The directed line segment PQ has initial point P and terminal point Q, and its length (or magnitude) is denoted by PQ. Directed line segments that have the same length and direction are equivalent. Two vectors represented in x-y plane are equivalent if they have the same length and the same direction. The length can be calculated using the distance formula and the directions calculating the slope.

5 A directed line segment whose initial point is the origin can be uniquely represented by the coordinates of its terminal point Q(v 1, v 2 ) (This representation of v is said to be in standard position).

6 This definition implies that two vectors u = u 1, u 2 and v = v 1, v 2 are equal if and only if u 1 = v 1 and u 2 = v 2.

7 If P(p 1, p 2 ) and Q(q 1, q 2 ) are the initial and terminal points of a directed line segment, the component form of the vector v represented by PQ is v 1, v 2 = q 1 p 1, q 2 p 2. Moreover, from the Distance Formula you can see that the length (or magnitude) of is v is: v = (q 1 p 1 ) 2 +(q 2 p 2 ) 2 = (v 1 ) 2 +(v 2 ) 2 If v = v 1, v 2, v can be represented by the directed line segment, in standard position, from P(0, 0) to Q(v 1, v 2 ). The length of v is called the norm of v. If v = 1, v is a unit vector. Moreover, v =0 if and only if v is the zero vector 0.

8 Any vector can be represented by an x-component a x and a y-component a y. Use trigonometry to find the components of a vector: a x = acos θ and a y = asin θ, where θ is measured from the + x-axis toward the + y- axis. We can use the components of a vector to find its magnitude and direction: a = a x 2 + a y 2 and θ = tan 1 ( a x a y )

9 Model Problem 1 Finding the Component Form and Length of a Vector 9

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11 Geometrically, the scalar multiple of a vector v and a scalar c is the vector that is c times as long as v. If c is positive, cv has the same direction as v. If c is negative, cv has the opposite direction.

12 The sum of two vectors can be represented geometrically by positioning the vectors (without changing their magnitudes or directions) so that the initial point of one coincides with the terminal point of the other

13 Two vectors may be added graphically using either the parallelogram method or the head-to-tail method.

14 Model Problem 2 Vector Operations Given v = and w =, find each of the vectors. Solution: a. b. 14

15 Model Problem 2 Solution cont d c. Using 2w = you have 15

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17 Any set of vectors (with an accompanying set of scalars) that satisfies the eight properties given in Theorem 11.1 is a vector space. The eight properties are the vector space axioms. So, this theorem states that the set of vectors in the plane (with the set of real numbers) forms a vector space.

18 In Theorem 11.3, u is called a unit vector in the direction of v. The process of multiplying v by to get a unit vector is called normalization of v.

19 Model Problem 3 Finding a unit Vector Find a unit vector in the direction of v = length 1. Solution: and verify that it has 19

20 Generally, the length of the sum of two vectors is not equal to the sum of their lengths. To see this, consider the vectors u and v. By considering u and v as two sides of a triangle, you can see that the length of the third side is u + v, and you have u + v u + v. Equality occurs only if the vectors u and v have the same direction. This result is called the triangle inequality for vectors.

21 The unit vectors and are called the standard unit vectors in the plane and are denoted by as shown in Figure These vectors can be used to represent any vector uniquely, as follows. The vector v = v 1 i + v 2 j is called a linear combination of i and j. The scalars v 1 and v 2 are called the horizontal and vertical components of v.

22 Model Problem 4 Writing a Linear Combination of Unit Vectors Let u be the vector with initial point (2, 5) and terminal point ( 1, 3), and let v = 2i j. Write each vector as a linear combination of i and j. a. u b. w = 2u 3v Solution: a. u = b. w = 22

23 Model Problem 5 Finding the Component Form of a Vector in Space Find the component form and magnitude of the vector v having initial point ( 2, 3,1) and terminal point (0, 4, 4). Then find a unit vector in the direction of v. Solution: The component form of v is v = q 1 p 1, q 2 p 2, q 3 p 3 = 0 ( 2), 4 3, 4 1 which implies that its magnitude is = 2, 7, 3 The unit vector in the direction of v is 23

24 Vectors have many applications in physics and engineering. One example is force. A vector can be used to represent force, because force has both magnitude and direction. If two or more forces are acting on an object, then the resultant force on the object is the vector sum of the vector forces.

25 Model Problem 6 Finding the Resultant Force Two tugboats are pushing an ocean liner, as shown in Figure. Each boat is exerting a force of 400 pounds. What is the resultant force on the ocean liner? 25

26 Model Problem 6 Finding the Resultant Force. (Solution) Using Figure 11.12, you can represent the forces exerted by the first and second tugboats as F 1 = = F 2 = The resultant force on the ocean liner is F = F 1 + F 2 = = So, the resultant force on the ocean liner is approximately 752 pounds in the direction of the positive x-axis. = 26

27 Model Problem 8 Finding a Velocity 27

28 Model Problem 8 Finding a Velocity (Solution) 28

29 Many of the formulas established for the two-dimensional coordinate system can be extended to three dimensions. For example, to find the distance between two points in space, you can use the Pythagorean Theorem twice, as shown in Figure. By doing this, you will obtain the formula for the distance between the points (x 1, y 1, z 1 ) and (x 2, y 2, z 2 ).

30 In space, vectors are denoted by ordered triples : v = v 1, v 2, v 3. The zero vector is denoted by 0 = 0,0,0. Using the unit vectors i = 1, 0, 0, j = 0, 1, 0, and k = 0, 0, 1 in the direction of the positive z-axis, the standard unit vector notation for v is: as shown in Figure v = v 1 i + v 2 j + v 3 k

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32 A unit vector has a magnitude of 1 with no units. It only purpose is to point, that is, to describe a direction in space. The unit vector i points in the +x-direction, j points in the +y-direction, and z points in the +z-direction. Any vector can be expressed in terms of its components as a = a x i + a y j + a z z.

33 Model Problem 9 Finding the Component Form of a Vector in Space Find the component form and magnitude of the vector v having initial point ( 2, 3,1) and terminal point (0, 4, 4). Then find a unit vector in the direction of v. Solution: The component form of v is v = q 1 p 1, q 2 p 2, q 3 p 3 = 0 ( 2), 4 3, 4 1 which implies that its magnitude is = 2, 7, 3 The unit vector in the direction of v is 33

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35 Example 10 Parallel Vectors Vector w has initial point (2, 1, 3) and terminal point ( 4, 7, 5). Which of the following vectors is parallel to w? a. u = 3, 4, 1 b. v = 12, 16, 4 Solution: Begin by writing w in component form. w = 4 2, 7 ( 1), 5 3 = 6, 8, 2 a. Because u = 3, 4, 1 = 6, 8, 2 = w, you can conclude that u is parallel to w. 35

36 Example 10 Solution cont d b. In this case, you want to find a scalar c such that 12, 16, 4 = c 6, 8, = 6c c = 2 16 = 8c c = 2 4 = 2c c = 2 Because there is no c for which the equation has a solution, the vectors are not parallel. 36

37 You have studied two operations with vectors vector addition and multiplication by a scalar each of which yields another vector. In this section you will study a third vector operation, called the dot product. This product yields a scalar, rather than a vector.

38 We can calculate the scalar product A B directly if we know the x-, y-, and z- components of A and B. A B = A x B x + A y B y + A z B z

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40 Model Problem 11 Finding Dot Products Given and, find each of the following. Solution: 40

41 Model Problem 11 Solution cont d Notice that the result of part (b) is a vector quantity, whereas the results of the other three parts are scalar quantities. 41

42 The angle between two nonzero vectors is the angle θ, 0 θ π, between their respective standard position vectors, as shown in Figure. If the angle between two vectors is known, rewriting Theorem 11.5 in the form: produces an alternative way to calculate the dot product.

43 From Theorem 11.5, you can see that two nonzero vectors meet at a right angle if and only if their dot product is zero. Two such vectors are said to be orthogonal.

44 The scalar product (also called the dot product ) of two vectors is A B = ABcosϕ The scalar product can be positive, negative or zero. For any two vectors A and B, ABcosϕ = BAcosϕ (A B = B A)

45 Model Problem 12 Finding the Angle Between Two vectors For u = 3, 1, 2, v = 4, 0,2, w = 1, 1, 2, and z = 2, 0, 1, find the angle between each pair of vectors. a. u and v b. u and w c. v and z Solution: Because u. v < 0, 45

46 Model Problem 12 Solution cont d Because u. w = 0, u and w are orthogonal. So, θ = π/2. Consequently, θ = π. Note that v and z are parallel, with v = 2z. 46

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48 The projection of u onto v can be written as a scalar multiple of a unit vector in the direction of v.that is, The scalar k is called the component of u in the direction of v.

49 Model Problem 13 Finding a Vector Component of u Orthogonal to v Find the vector component of, given that and that is orthogonal to Solution: Because u = w 1 + w 2, where w 1 is parallel to v, it follows that w 2 is the vector component of u orthogonal to v. So, you have 49

50 Model Problem 13 Solution cont d Check to see that w 2 is orthogonal to v, as shown in Figure Figure

51 Many applications in physics, engineering, and geometry involve finding a vector in space that is orthogonal to two given vectors. The product that will yield such a vector is called the cross product, and it is most conveniently defined and calculated using the standard unit vector form. Because the cross product yields a vector, it is also called the vector product.

52 The vector product ( cross product ) of two vectors is a vector with direction perpendicular to the plane formed by the two vectors and with a magnitude equal to ABsin ϕ. ( A B ABsin ).

53 Vectors A and B and their vector product C = A B The vector product A B = B A is anticonmutative.

54 We can calculate the vector product A B directly if we know the x-, y-, and z- components of A and B.

55 The vector product can also be expressed in determinant form = A y B z i + A z B x j + A x B y k A y B x k A x B z j A z B y i = (A y B z A z B y )i + (A z B x A x B z )j + (A x B y A y B x )k

56 Another way to calculate u v is to use the following determinant form with cofactor expansion.

57 Note the minus sign in front of the j-component. Each of the three 2 2 determinants can be evaluated by using the following diagonal pattern. Here are a couple of examples.

58 Model Problem 14 Finding the Cross Product Given u = i 2j + k and v = 3i + j 2k, find each of the following. a. u v b. v u c. v v Solution: 58

59 Model Problem 14 Solution cont d Note that this result is the negative of that in part (a). 59

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61 Both u v and v u are perpendicular to the plane determined by u and v. One way to remember the orientations of the vectors u, v and u v is to compare them with the unit vectors i, j, and k = i j, as shown in Figure

62 Model Problem 15 Using the Cross Product Find a unit vector that is orthogonal to both u = i 4j + k and v = 2i + 3j. Solution: The cross product u v, as shown in Figure 11.37, is orthogonal to both u and v. Figure

63 Model Problem 15 Solution cont d Because a unit vector orthogonal to both u and v is 63

64 If the vectors u, v, and w do not lie in the same plane, the triple scalar product u (v w) can be used to determine the volume of the parallelepiped (a polyhedron, all of whose faces are parallelograms) with u, v, and w as adjacent edges, as shown in Figure

65 Model Problem 16 Volume by the Triple Scalar Product Find the volume of the parallelepiped shown in Figure having u = 3i 5j + k, v = 2j 2k, and w = 3i + j + k as adjacent edges. Solution: By Theorem 11.10, you have Figure