Solution to Homework 1

Transcription

1 Solution to Homework 1 Olena Bormashenko September, 0 Section 1.1:, 5b)d), 7b)f), 8a)b), ; Section 1.:, 5, 7, 9, 10, b)c), 1, 15b), 3 Section 1.1. In each of the following cases, find a point that is two-thirds of the distance from the first initial) point to the second terminal) point. a), 7, ), 10, 10, ) Solution: Given two points P and Q, the vector that represents the movement from P to Q is P Q. We are looking for a point A which is two thirds of the distance from P to Q this is equivalent to saying that we start at P, and go two-thirds of the way to Q. Going all the way to Q corresponds to P Q this means that going to A corresponds to going 3 P Q). In this particular case, we see that the vector corresponding to the motion from the first point to the second point is 10, 10, ), 7, ) = [1, 17, 9]. This means that the vector corresponding to going two-thirds of the way is [1, 17, 9] = [8/3, 3/3, 6] 3 Therefore, to get from P to A we need to move according to the above vector. Thus, A P = [8/3, 3/3, 6], and so A = P + [8/3, 3/3, 6] =, 7, ) + [8/3, 3/3, 6] = 16/3, 13/3, 8) b), 1, 0, 7),, 1, 9, ). Solution: Doing the same kind of calculation as in part a), we see that our point is precisely:, 1, 0, 7) +, 1, 9, ), 1, 0, 7)) 3 1

2 which simplifies to 0/3, 1, 6, 1). 5. In each of the following cases, find a unit vector in the same direction as the given vector. Is the resulting normalized)vector longer or shorter than the original? Why? b) [, 1, 0, ] Solution: In order to find a unit vector in the same direction as a given vector, just divide by the length of the vector. This makes intuitive sense: if you had a vector of length, then to get a unit vector a vector of length 1) in the same direction, you would divide by.) Furhermore, it s clear that our unit vector is longer than the original vector if and only if the vector we started with had length less than 1. Let u be our unit vector. Then, [, 1, 0, ] u = [, 1, 0, ] = [, 1, 0, ] ) [ 1 =,, 0, ] Since the length of the original vector 1, the resulting unit vector is shorter than the original vector. d) [ 1 5, 5, 1 5, 1 5, ] 5 Solution: In the same way as before, [ 1 5, 5, 1 5, 1 5, 5] u = ) ) + 1 5) + 1 5) + 5 [ 1 5, 5, 1 5, 1 5, [ 5] 1 5, 5, 1 5, 1 5, 5] = = 5 = [ 1,, 1, 1, Since the length of the original vector is is longer than the original vector. 5 5 ] ), the normalized vector 7. If x = [,, 5], y = [ 1, 0, 3], and z = [, 1, ], find the following b) y Solution: By definition, y = [ 1, 0, 3] = [, 0, 6]

3 f) x + 3 y z Solution: By definition, x + 3 y z = [,, 5] + 3[ 1, 0, 3] [, 1, ] = [, 8, 10] + [ 3, 0, 9] [16,, 8] = [ 3, 1, ] 8. Given x and y as follows, calculate x + y, x y, and y x and sketch x, y, x + y, x y in the same coordinate system. a) x = [ 1, 5], y = [, ] Solution: x + y = [ 1, 5] + [, ] = [1, 1] x y = [ 1, 5] [, ] = [ 3, 9] y x = [, ] [ 1, 5] = [3, 9] b) x = [10, ], y = [ 7, 3] Solution: x + y = [10, ] + [ 7, 3] = [3, 5] x y = [10, ] [ 7, 3] = [17, 1] y x = [ 7, 3] [10, ] = [ 17, 1]. a) Prove that the length of each vector in R n is nonnegative. Assumptions: x is a vector in R n. Need to show: x 0. Let x = [x 1, x,..., x n ]. Then, x = x 1 + x + + x n. Since squares are nonnegative, x 1 + x + + x n 0 Since the square root of a nonnegative number is defined to be nonnegative, we see that x = x 1 + x + + x n 0, so we re done. b) Prove that the only vector in R n of length 0 is the zero vector. 3

4 Section 1. Assumptions: x = 0 Need to show: x = 0 Let x = [x 1, x,..., x n ]. Then, we have that 0 = x = x x n 0 = x x n Since each of the x i is non-negative, and their sum is 0, the only way this is possible is that each is 0. Thus, x i is 0 for each i, and hence x = 0.. Show that the points A 1 9, 19, 16), A, 1, 13), and A 3 1, 3, 10) are the vertices of a right triangle. Hint: Construct vectors between the points and check for an orthogonal pair.) Solution: The vectors that represent the sides of the triangle are A 1 A, A A 3, and A 1 A 3. We had to choose between A 1 A and A A 1, etc., but this is not important.) Accordingly, they are: A 1 A = 9, 19, 16), 1, 13) = [, 7, 3] A A 3 =, 1, 13) 1, 3, 10) = [ 3,, 3] A 3 A 1 = 1, 3, 10) 9, 19, 16) = [5,, 6] Checking, the various pairs, we see that A 1 A ) A A 3 ) = [, 7, 3] [ 3,, 3] = 6 A A 3 ) A 3 A 1 ) = [ 3,, 3] [5,, 6] = 77 A 3 A 1 ) A 1 A ) = [5,, 6] [, 7, 3] = 0 Thus, the last equation shows that A 3 A 1 ) is perpendicular to A 1 A ), and therefore the triangle has a right angle at A Why isn t it true that if x, y, z R n, then x y z) = x y) z? Solution: Since y z is a scalar, the dot product x y z) is simply not defined it s impossible to calculate the dot product of a vector and a scalar. Therefore, neither left-hand side nor right-hand side are defined, so the statement is not true. 7. Does the Cancellation Law of algebra always hold for the dot product:that is, assuming that z 0, does x z = y z always imply that x = y?

5 Solution: This does not hold. The simplest example one can come up with is an example where z is perpendicular to both y and x in that case, both the dot products are 0. For example, let x = [0, 1, 1], y = [0, 1, 0] and let z = [1, 0, 0]. In that case, but x y. x z = 0 = y z 9. Prove that if x + y) x y) = 0 then x = y. Assumptions: x + y) x y) = 0 Need to show: x = y As usual, we use the assumption to show the desired result. Here, expanding things out, 0 = x + y) x y) = x x y x + x y y y = x x y y = x y using the fact that x y = y x and the identity for length in terms of the dot product. Rearranging, this yields x = y. Since the lengths are both nonnegative, we can take the square roots of both sides to conclude that x = y, as required. 10. Prove that 1 x + y + x y ) = x + y. Assumptions: No real assumptions. Need to show: 1 x + y + x y ) = x + y Here, there are no assumptions, so we just need to check that the lefthand side and the right-hand side are equal in general. Since we have squares of lengths, we rewrite everything in terms of dot products. We start from the left-hand side and manipulate it: 1 x + y + x y ) = 1 x + y) x + y) + x y) x y)) = 1 x x + x y + y y + x x x y + y y) as required. = 1 x x + y y) = x + y 5

6 . b) Prove that if x, y, z are mutually orthogonal vectors in R n, then x + y + z = x + y + z Assumptions: x, y and z are mutually orhogonal: that is, x y = 0, x z = 0, and y z = 0 Need to show: x + y + z = x + y + z Again, rewriting everything in terms of dot products: x + y + z = x + y + z) x + y + z) = x x + y y + z z + x y + y z + x z = x x + y y + z z = x + y + z using the assumption, and the identity for the length of a vector. c) Prove that x y = 1 x + y x y ). Assumptions: None. Need to show: x y = 1 x + y x y ). Using dot products once again, and starting from the right-hand side: 1 x + y x y ) = 1 x + y) x + y) x y) x y)) as required. = 1 x x + x y + y y x x x y + y y)) = 1 x y) = x y 1. Given x, y, z in R n, with x orthogonal to both y and z, prove that x is orthogonal to c 1 y + c z where c 1, c R. Assumptions: x orthogonal to both y and z: that is, x y = 0 and x z = 0. Need to show: x c 1 y + c z) = 0 The trickiest part here is writing everything down in terms of dot products! As soon as that s done, it s very easy. Using vector identities. and so we re done. x c 1 y + c z) = x c 1 y) + x c z) = c 1 x y + c x z = 0 6

7 15. Calculate proj a b in each case, and verify that b proj a b is orthogonal to a. b) a = [ 5, 3, 0], b = [3, 7, 1]. Solution: By definition, ) proj a b a b = a a Therefore, here we have proj [ 5, 3, 0] [3, 7, 1] a b = [ 5, 3, 0] 36 = 5) = 36 [ 5, 3, 0] = 3 ) [ 5, 3, 0] ) [ 5, 3, 0] [ ] 90 17, 5 17, 0 Now, verifying that b proj a b is orthogonal to a: ) [ ]) a b proj a 90 b = [ 5, 3, 0] [3, 7, 1] 17, 5 17, 0 [ = [ 5, 3, 0] 39 ] 17, 65 17, 1 3. True or False: = = 0 Since the dot product is 0, they are indeed orthogonal. a) For any vectors x and y, and any scalar d, x d y) = d x) y. TRUE: This follows from Identity ) in Theorem 1.5. b) For all x, y in R n with x 0, x y/ x y. TRUE: This follows from Cauchy-Schwarz. c) For all x, y in R n, x y x y. FALSE: In fact, the opposite of this is true, which can be shown using Triangle Inequality: x y x y. Let s provide a counterexample: let x = [1, 0] and y = [0, 1]. In that case, x y = [1, 1] =, x = 1, y = 1 7

8 and it s not true that 1 1 = 0. d) If θ is the angle between x and y in R n, and θ > π, then x y > 0. FALSE: From Theorem 1.8, x y > 0 if and only if the angle θ is acute. e) The standard unit vectors in R n are mutually orthogonal. TRUE: The standard unit vectors are e 1 = [1, 0,..., 0], e = [0, 1, 0,... ], etc., and it s easy to check that any pair of these has dot product 0. f) If proj a b = b, then a is perpendicular to b. FALSE: It can be checked that a is perpendicular to b if and only if proj a b = 0. Furthermore, proj a b = b if and only if a and b are parallel. We will not prove these statements here, but we will provide a counterexample to the assertion. Take a = b = [1, 0]. In that case, it can be checked that proj a b = b, but a and b are certainly not orthogonal. 8