94 CHAPTER 3. VECTORS AND THE GEOMETRY OF SPACE
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1 94 CHAPTER 3. VECTORS AND THE GEOMETRY OF SPACE 3.3 Dot Product We haven t yet de ned a multiplication between vectors. It turns out there are di erent ways this can be done. In this section, we present one way, called the dot product De nitions and Properties De nition 205 The dot product, also called inner product, is denoted with the symbol : We can only take the dot product of two vectors having the same dimension. The result is a number, obtained by multiplying the corresponding coordinates of the two vectors, then adding all the products. If ~u = (a; b) and ~v = (c; d) are two vectors, then ~u ~v = ac + bd. You will note that the answer is a number, not a vector. We have a similar de nition for vectors in higher dimensions. In general, if ~u = (x 1 ; x 2 ; ; x n ) and ~v = (y 1 ; y 2 ; ; y n ), then P ~u ~v = n x i y i. i=1 Remark 206 If ~u = (x 1 ; x 2 ; ; x n ) (a 1 n matrix) and ~v = (y 1 ; y 2 ; ; y n ) (also a 1 n matrix), then ~v T is an n 1 matrix and we can use matrix multiplication to de ne the dot product as follows: ~u ~v = ~u~v T (3.8) Example 207 If ~u = (1; 1; 1) and ~v = (1; 2; 3) then ~u ~v = (1) (1) + (1) (2) + (1) (3) = 6. Proposition 208 The dot product satis es the following properties: 1. ~u ~u = k~uk 2 2. ~u ~0 = ~0 ~u = 0 3. ~u ~v = ~v ~u 4. ~u ~v = (~u ~v) for any scalars and. 5. ~u (~v + ~w) = ~u ~v + ~u ~w 6. If is the smallest angle (between 0 and ) between two non-zero vectors ~u and ~v, then ~u ~v = k~uk k~vk cos. 7. ~u? ~v if and only if ~u ~v = 0. Again, we assume that ~u and ~v are two non-zero vectors. 8. j~u ~vj k~uk k~vk Proof. We give an idea of a proof for some of these properties.
2 3.3. DOT PRODUCT We see why this is true for a 3-D vector. A general proof is similar. Suppose that ~u = ha; b; ci. Then, by de nition ~u ~u = a 2 + b 2 + c 2 Also by de nition, Thus k~uk = p a 2 + b 2 + c 2 k~uk 2 = a 2 + b 2 + c 2 = ~u ~u 2. Left as an exercise. 3. Left as an exercise. 4. Left as an exercise. 5. Left as an exercise. 6. Consider gure 6. Using the law of cosines, we see that k~u ~vk 2 = k~uk 2 + k~vk 2 2 k~uk k~vk cos Thus 2 k~uk k~vk cos = k~u ~vk 2 + k~uk 2 + k~vk 2 = (~u ~v) (~u ~v) + ~u ~u +! v! v = (~u ~u 2~u ~v +! v! v ) + ~u ~u +! v! v = ~u ~u + 2~u ~v! v! v + ~u ~u +! v! v 2 k~uk k~vk cos = 2~u ~v Hence ~u ~v = k~uk k~vk cos
3 96 CHAPTER 3. VECTORS AND THE GEOMETRY OF SPACE 7. If the vectors are perpendicular, then the angle between them is 90. Thus, cos = 0. The result follows from property 6. Conversely, if ~u and ~v are two non-zero vectors such that ~u~v = 0 then k~uk k~vk cos = 0. Since k~uk 6= 0 and k~vk 6= 0 (~u and ~v are two non-zero vectors), it must be that cos = 0 thus = 90 hence ~u? ~v. 8. From 6, we see that since jcos j 1: j~u ~vj = jk~uk k~vk cos j = k~uk k~vk jcos j k~uk k~vk Remark 209 We can also use the dot product to prove the triangle inequality k~u + ~vk k~uk + k~vk. Using the fact that ~u ~u = k~uk 2, we have k~u + ~vk 2 = (~u + ~v) (~u + ~v) = ~u ~u + 2~u ~v + ~v ~v using properties of the dot product = k~uk 2 + 2~u ~v + k~vk 2 k~uk j~u ~vj + k~vk 2 property of absolute value k~uk k~uk k~vk + k~vk 2 by property 8 above k~u + ~vk 2 (k~uk + k~vk) 2 Since both k~u + ~vk and k~uk + k~vk are positive, we can take the square root on both side of the last equality to obtain the result. Remark 210 Property 6 in proposition 208 can also be used to determine how close two non-zero vectors are to pointing in the same direction. Given two non-zero vectors ~u and ~v, consider P, the plane passing through the origin and perpendicular to ~u. ~v will be on the same side of P as ~u if and only if ~u ~v > 0 because in that case, the angle between the two vectors will be less than 90 and its cosine will be positive. Remark 211 Property 7 in proposition 208 is also very important, it gives us a quick way to check if two vectors are perpendicular. Another term for perpendicular is orthogonal. We de ne the zero vector ~0 = (0; 0; :::) to be orthogonal to every vector. Remark 212 It is important to understand that when we talk about the angle between two vectors, we are talking about the smallest positive angle between them in other words an angle such that 0. We illustrate some of these properties with examples.
4 3.3. DOT PRODUCT 97 Figure 3.11: Compute! u! v if k! u k = 2 and k! v k = 3 Example 213 Suppose that the length of a vector! u is 2, the length of a vector! v is 3 and they make an angle of 135 as shown in gure Find ~u! v. ~u! v = k~uk k~vk cos 135 p! 2 = (2) (3) 2 = 3 p 2 Example 214 What is the angle in degrees between! u = (1; 1; 1) and! v = (2; 1; 0)? From the formula ~u ~v = k~uk k~vk cos, we can compute cos and therefore. First, let us compute the various quantities needed. ~u ~v = (2) (1) + (1) (1) + (1) (0) = 3 k~uk = p = p 3 k! v k = p = p 5
5 98 CHAPTER 3. VECTORS AND THE GEOMETRY OF SPACE It follows that ~u ~v = cos 1 k~uk k~vk = cos 1 3 p 15 = 39: Projections It often happens that we need to decompose a vector into components that are parallel and perpendicular to a given vector. Or, as shown in gure 3.12, given two non-zero vectors ~a and ~ b, one needs to nd the projection of ~ b onto ~a and the projection of ~ b onto a vector perpendicular to ~a. Let us denote proj ~a ~ b the projection of ~ b onto ~a, comp ~a ~ b the component of ~ b along ~a (also known as the scalar projection of ~ b onto ~a) and orth ~a ~ b the projection of ~ b onto a vector perpendicular to ~a.let us call the angle between ~a and ~ b. Then, we have: Figure 3.12: proj ~a ~ b and orth~a ~ b cos = comp ~ ~a b ~ b
6 3.3. DOT PRODUCT 99 Therefore comp ~a ~ b = ~ b cos Using Property 6 in proposition 208, we can write ~ b~a ~ b comp ~ ~a b = k~ak ~ b (3.9) It follows that comp ~a ~ b = ~a ~ b k~ak To nd proj ~ ~a b, we simply multiply comp~a ~ b by a unit vector in the direction of ~a ~a. Such a vector is and therefore k~ak proj ~ ~a ~ b ~a b = 2~a (3.10) k~ak Finally, orth ~a ~ b = ~ b proj ~a ~ b and therefore orth ~a ~ b = ~ b ~a ~ b 2~a (3.11) k~ak Example 215 Find the scalar projection and vector projection of! b = (1; 1; 2) onto! a = ( 2; 3; 1). Let us rst compute the various quantities involved.! a! b = j! a j = = 3 q ( 2) = p 14 Therefore and comp ~a ~ b = 3 p 14 proj ~ 3 ~a b = h 2; 3; 1i 14
7 100 CHAPTER 3. VECTORS AND THE GEOMETRY OF SPACE Applications The dot product has many applications including: 1. Use it to see if two non-zero vectors are perpendicular. Recall that two non-zero vectors are perpendicular if and only if their dot product is Use it to compute the angle between two vectors. From the formula ~u~v = k~uk k~vk cos, if we are given ~u and ~v, we have = cos 1 ~u ~v k~uk k~vk 3. Use it to quickly determine if the angle between two vectors is acute (between 0 and 90 degrees) or obtuse (between 90 and 180 degrees). Given ~u and ~v, if the angle between them is acute, then ~u~v > 0 and if it is obtuse, ~u ~v < Given a vector and a plane perpendicular to it, use it to determine if a second vector is on the same side of the plane or the opposite side. If the two vectors are on the same side of the plane, the angle between them will be acute. If they are on opposite sides, it will be obtuse. So, we can use the technique describe above. 5. Use it to nd the projection of a vector onto another one. 6. Use it to compute the work done by a force acting on an object and displacing it. In physics, the work W done by a force! F moving an object along the displacement vector! r is de ned to be the product of the component of! F along! r by the magnitude of the displacement. This gives us: W = comp! r! F j! r j =! F! r j! r j =! F! r j! r j We illustrate that last application with an example. Example 216 A crate is displaced by 8m up a ramp by a constant force! F of 200N applied at an angle of 25 degrees to the ramp. Find W, the work done.
8 3.3. DOT PRODUCT 101 Let! D be the displacement vector. Then! D = 8 and F = 200: W =! F! D =! D F cos 25 = (8) (200) cos N:m 1450J Note that in the metric system, distances are in meters (m), forces are in Newton (N), work is expressed in Joules (J) Problems 1. Show that the distance d between a point P 0 of coordinates (x 0 ; y 0 ) and the line ax + by + c = 0 is given by d = jax 0 + by 0 + cj p a2 + b 2 (hint: what is vector vector perpendicular to the line?) 2. Do # 1, 2, 3, 4, 5, 6, 8, 9, 12, 13, 16, 17, 21, 25, 27, 31 on pages in your book.
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