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Mathematics
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Mathematics 1229A/B
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Prof
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Fall

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Math 1229A/B Unit 4: Vectors in ℜ m (text reference: Section 2.1) ▯V. Olds 2010 Unit 4 51 m 4 Vectors in ℜ We’ve learnt about ℜ and ℜ , which you’ve seen before. Now, we’re going to extend some of m the same ideas to spaces with more than 3 dimensions. We refer to m-dimensional space as ℜ or m-space. “But”, you’re saying to yourself, “how can you have more than 3 dimensions?”. Well, for starters, time is often referred to as the fourth dimension of physical space. The world we live in has depth, breadth and height, and also time. Where you’re sitting right this minute is a particular location that could be described by x-, y- and z-coordinates. But you’re in that location at this particular instant. At other times you weren’t. However the location still existed. So time is another axis along which space can be measured. And after that? Well, for higher dimensions, it’s just much easier if you don’t try to relate it 5 to physical space. Because it tends to make your brain hurt if you try to actually picture ℜ or ℜ or ℜ , and so forth. But to mathematicians, that’s no reason not to talk about them. The- oretically, there could be more dimensions. And even if there aren’t, there are still situations in which it is useful to use constructs which correspond to things we’ve already seen (points, vectors, etc.) but with more parts to them, as if they were from a higher-dimensional space. For instance, a company which produces 10 different products might have a use for considering a “point” with 10 coordinates, where each coordinate corresponds to a different one of the 10 products and indicates, for instance, how many of that product are in stock at the moment. There are many uses of the kind of mathematical constructs we’ve been working with, other than literally as points in space or directed line segments, in which the kind of arithmetic we’ve been doing still makes sense. And if these constructs aren’t being used to represent physical space, then there’s no reason that they would need to be limited to 3 dimensions. m We start our study of ℜ with a number of definitions, which will all look familiar to you from ℜ and ℜ . These definitions just extend the construct we’ve been using, vectors, to higher dimen- sional space, and then extend (most of) the things we were doing with them, i.e. tell us how to do arithmetic etc. with these higher-dimensional vectors. And along the way we’ll also see some 2 theoretical results. But the arithmetic works just the same in higher dimensions as it did in ℜ and ℜ , and the theorems will look familiar too, so it will be really easy for you to learn this. There’s m not much that’s really new in this unit. It’s just a matter of getting used to the idea ofor m > 3. Definition: For any positive integer m ≥ 2, we use ℜ , also called m-space, to denote the set of all ordered m-tuplesu = (u1,u 2...,um), where each value u iay be any real number (i.e. for alliu ∈ ℜ). For anyu ∈ ℜ , we refer to u as a vector, or an m-vector. The numbers u ,1u ,2..., um are called the components of the m-vector ▯ u. Note: As we have already learnt, the terminology is sometimes a bit different for ℜ and ℜ . For 2 instance, we generally refer to an ordered pair or (rarely) ordered duple in ℜ and an ordered triple in ℜ , rather than saying ordered 2-tuple or ordered 3-tuple. Also note that, as we have been doing 2 3 with vectors in ℜ and ℜ , when a vector is given a certain name, such as ▯u, the components (when not specific numbers) are assumed to be named with the same letter, with subscripts. For instance referring to the vector u we would understand that the first component is called u ▯1, the second component is called u , and so forth. The only exception to this is, again as we’ve already seen, 2 2 3 that often in ℜ we use ▯x = (x,y) and likewise in ℜ we use ▯ x = (x,y,z). Examples: (1,2,3,4) and (−0.2,61.3,0.04,0) are both vectors in ℜ . (5,3,6,3,−1,−10) is a 6-vector. 5 Ifv ∈ ℜ , then ▯v = (v1,v2,v3,v4,v5). 52 Unit 4 m Definition: The zero vector in ℜ is the m-vector whose components are all 0. Also, two m-vectors are equal if and only if their corresponding components are identical. m That is, for anu,▯v ∈ ℜ , u = v if and only i1 u =1v 2 u =2v , ..., amd= vm. Note: Only vectors which have the same number of components can be equal. That isu can only be equal tov if they are both m-vectors, for the same value of m. (And of course will only actually be equal if their corresponding components are identical.) 4 12 Examples: (0,0,0,0) is the zero vector in ℜ , and for 0we have 0 = (0,0,0,0,0,0,0,0,0,0,0,0). If we haveu = (1,2,c,3,−1) and ▯v = (a,2,8,d,−1) and we know that ▯u = v, then we must have a = 1, c = 8 and d = 3. Definition: The distance between two m-vectors ▯ u and v is given by the distance formula: p d(u,v) = (v1− u1) + (v2− u 2 + ... + (m − um)2 Also, the magnitude or length of an m-vector, sometimes referred to as the norm of the vector, is p ||u|| = (u1) + (u2) + ...(m )2 An m-vector whose magnitude is 1 is called a unit vector. Note: These are precisely analogous to the notation, terminology and formulas we had for the dis- 2 3 tance between 2 vectors, and the magnitude of a vector, in ℜ and in ℜ . Example 4.1. (a) Show that (0,0,0,0,−1,0,0,0) is a unit vector. (b) Show that there are no real numbers a and b for whiu = (1,1,a,b) is a unit vector. Solution: (a) We have p ||(0,0,0,0,−1,0,0,0)||= 02 + 0 + 0 + 0 + (−1) + 0 + 0 + 0 2 √ = 0 + 0 + 0 + 0 + 1 + 0 + 0 + 0 √ = 1 = 1 so this vector is a unit vector. (b) We find the magnitude of u: p p p |u|| = ||(1,1,a,b)|| =1 + 1 + a + b =2 1 + 1 + a + b = 2 + a + b2 2 2 2 2 √ 2 2 √ Since a ≥ 0 and b ≥ 0 for √ll real numbers a and b, then 2+a +b ≥ 2 and so 2 + a + b ≥ 2. But then we see that u|| ≥ 2 > 1, so u||▯=1 and u cannot be a unit vector. Unit 4 53 Example 4.2. Ifu = (1,2,3,4) andv = (0,−1,12,7), find the distance betweeu and v and also find the magnitude of each. Solution: The distance betweenu andv is p p du,v) = (0 − 1)+ (−1 − 2) + (12 − 3) + (7 − 4) = (−1)2+ (−3) + (9) + (3)2 √ √ = 1 + 9 + 81 + 9 = 100 = 10 For the magnitudes ou and v we get p √ √ |u|| = ||(1,2,3,4)|| =1 + 2 + 3 + 4 = 1 + 4 + 9 + 16 = 30 p √ √ and |v|| = ||(0,−1,12,7)|| =02 + (−1) + (12) + 7 = 0 + 1 + 144 + 49 = 194 Example 4.3. How long is the 5-vector (1,1,−1,0,−1)? Solution: We are being asked to find the length, i.e. the magnitude, of this vector. We get: p √ √ ||(1,1,−1,0,−1)|| = 12+ 1 + (−1) + 0 + (−1) = 1 + 1 + 1 + 0 + 1 =4 = 2 We see that the vector (1,1,−1,0,−1) is 2 units long. Definition: For any scalar c and any m-vectu, the scalar multiple ou by c is cu = (cu1,cu2,...,mu ) For the scalar −1, the scalar multiple u ∈ ℜy ▯by −1 is called the negativeu,f ▯ denoted −u, so that −▯u = (−u1,−u2,...,−m ) If two m-vectors are scalar multiples of one another, we say that they are parallel. Note: Again, this is the same terminology and notation, and similar calculations, as we had in ℜ 3 and ℜ . To find a scalar multiple of an m-vector, we multiply each component of that vector by the scalar. And to find the negative of an m-vector, we multiply each component by −1, i.e. switch the sign of each component. Also, we use the term parallel to describe two m-vectors which have the same property that would lead us to conclude that they were parallel if they were vectors with fewer components. Example 4.4. Foru = (1,2,−1,2),v = (2,0,3,0,1) an▯ = (1,0,1,0,0,3,2), findu, 5v and −2▯. Solution: We simply need to multiply the components of each vector by the corresponding scalar: −▯u = −(1,2,−1,2) = (−1,−2,−(−1),−2) = (−1,−2,1,−2) 5v = 5(2,0,3,0,1) = (5 × 2,5 × 0,5 × 3,5 × 0,5 × 1) = (10,0,15,0,5) −2▯ = −2(1,0,1,0,0,3,2) = (−2 × 1,−2 × 0,−2 × 1,−2 × 0,−2 × 0,−2 × 3,−2 × 2) = (−2,0,−2,0,0,−6,−4) Of course, we could have founu more quickly by realizing that we just needed to switch the sign of each component: −▯u = −(1,2,−1,−2) = (−1,−2,1,2) 54 Unit 4 Definition: Two m-vectors are said to be collinear if and only if each is a scalar multiple of the other. That iu and v ∈ ℜm are collinear if and only if there is some scalar c such thu = cv. If two non-zero m-vectou and v are collinear, so u = cv, then they are said to have the same direction if c > 0 and are said to have opposite directions if c < 0. Examples: The vectors (1,2,3,4) and (2,4,6,8) are collinear because (2,4,6,8) = 2(1,2,3,4). These vectors have the same direction. Also,u = (1,0,−1,0,1) and v = (−10,0,10,0,−10), v = −10u and so u andv are collinear, with directions opposite to one another. Theorem 4.1. For any m-vector ▯u and any scalar c, ||cu|| = |cu||▯ That is, the magnitude of the scalar multipu is simply the magnitude ofu times the absolute value of the scalar multiplier. Example 4.5. Find the magnitude of (4,8,−20,12). Solution: The arithmetic is easier if we realize that (4,8,−20,12) = 4(1,2,−5,3). We get p 2 2 2 2 √ √ ||(4,8,−20,12)|| = ||4(1,2,−5,3)|| = |4|||(1,2,−5,3)||1 + 2 + (−5) + 3 = 4 1 + 4 + 25 + 9 = 439 Example 4.6. Ifu = (1,−1,2,−2,3,−3) and ▯v = −5u, find |v||. Solution: It will be easier to find the magnitudeu and use that to find the magnitude v, rather than to find the magnitude ofv directly. We get: |v|| = |p −u|| = | − 5u|| = 5||(1,−1,2,−2,3,−3)|| = 5 1 + (−1) + 2 + (−2) + 3 + (−3) 2 √ = 5 1 + 1 + 4 + 4 + 9 + 9 √ √ √ √ √ √ = 5 28 = 5 4 × 7 = 5 4 7 = 5(2) 7 = 10 7 Example 4.7. Ifu = (1,−1,1,0), find a unit vector in the opposite directiu. to ▯ Solution: Let v be a unit vector in the opposite directiu. Then ▯v is a scalar multipleu, i.ev = cu, for some scalar value c < 0, such thv|| = 1. Sincv = u, then we get ||v|| = |u|| = |cu||▯ and so |v|| = 1 gives |u|| = 1, which means that we must have |c| |▯|| (Notice that we have 2 3 previously observed this in ℜ and ℜ .) Fu = (1,−1,1,0), we have p √ √ ||u|| = 12+ (−1) + 1 + 0 =2 1 + 1 + 1 + 0 = 3 so we see that we need |c| = in order fov = cu to be a unit vector, and we need c < 0 in order 3 1 forv = cu to have the opposite directionu. Therefore we need c = −√ 3. So we see that a unit vector in the opposite directionu is▯ 1 1 1 1 v = cu = − √ (1,−1,1,0) = − √ ,√ ,− √ ,0 3 3 3 3 Unit 4 55 m Definition: Consider any ▯u andv in ℜ . The vector sum of the vectors is u + v = (u1,u2,...mu ) + 1v 2v ,.m.,v ) 1 (u 1 v2,u 2 v ,.m.+ vm) and the vector difference of u and v is u−▯ v =u+(−▯ v) = (u1,u2,...,m )+(−v 1−v 2...,−vm) = (u1−v 1u 2v ,.2.,u m −v m Note: This says that, as we do in ℜ and ℜ , we find the sum of 2 vectors as the vector whose com- ponents are the sums of the corresponding components of the 2 vectors, and we find the difference of 2 vectors, which can be considered to be the sum of the first vector and the negative of the second vector, as the vector whose components are the differences of the corresponding components. Example 4.8. Ifu = (1,2,3,4) andv = (0,−1,3,−2), find u + v and u −v. Solution: We get u + v = (1,2,3,4) + (0,−1,3,−2) = (1 + 0,2 + (−1),3 + 3,4 + (−2)) = (1,1,6,2) and u − v = (1,2,3,4) − (0,−1,3,−2) = (1 − 0,2 − (−1),3 − 3,4 − (−2)) = (1,3,0,6) Example 4.9. If u = (1,0,1,0,−1,0) and ▯v = (0,1,1,0,−1,−1), find w▯ where it is known that ▯ = 3u − 2v. Solution: We simply carry out the specified arithmetic operations. We get: w▯ = 3u − v▯ = 3(1,0,1,0,−1,0)− 2(0,1,1,0,−1,−1) = (3,0,3,0,−3,0)− (0,2,2,0,−2,−2) = (3 − 0,0 − 2,3 − 2,0 − 0,−3 − (−2),0 − (−2))
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