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Unformatted text preview: miller (zdm77) assignment 4 luecke (58600) This printout should have 15 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A triangle P QR in 3space has vertices P (1, 6, 3), Q(2, 3, 1), R(6, 5, 2) . 1 keywords: vectors, dot product, angle between vectors 002 10.0 points Find the angle between the vectors a = 1, 2 3 , b =  3, 7 . 1. angle = correct 6 3 2. angle = 4 3. angle = 4 5 4. angle = 6 2 5. angle = 3 6. angle = 3 Explanation: Since the dot product of vectors a and b can be written as a.b = a b cos , 0 , Use vectors to decide which one of the following properties the triangle has. 1. rightangled at R 2. not rightangled at P, Q, or R correct 3. rightangled at P 4. rightangled at Q Explanation: Vectors a and b are perpendicular when a b = 0. Thus P QR will be    (1) rightangled at P when QP RP = 0,     (2) rightangled at Q when P Q RQ = 0,    (3) rightangled at R when P R QR = 0. P (1, 6, 3), Q(2, 3, 1), R(6, 5, 2) we see that     P Q = 1, 3, 2 , QR = 4, 2, 1 , while Thus  RP = 5, 1, 1 .     P Q RQ = 4, where is the angle between the vectors, we see that cos = a.b , a b 0 . But for the vertices But for the given vectors, a b = (1)( 3) + (2 3)(7) = 13 3 , while a = Consequently, 3 13 3 cos = = 2 13 2 13 where 0 . Thus angle = . 6 13 , b = 52 .    QP RP = 10, and    P R QR = 17 . Consequently, P QR is not rightangled at P, Q, or R . miller (zdm77) assignment 4 luecke (58600) keywords: vectors, dot product, right triangle, perpendicular, 003 10.0 points 2 Find the vector projection of b onto a when b = i + 3j k, a = 2i+ j + 3k. 3 (2i+ j + 3k) 14 2 ( i + 3j  k) 7 3 ( i + 3j k) 14 1 ( i + 3j  k) 7 Find the vector projection of b onto a when b = 2, 4 , a = 1, 4 . 1. vector projection = 2. vector projection = 3. vector projection = 4. vector projection = 1. vector projection = 1 1, 4 18 2, 4 2. vector projection = 17 17 1, 4 3. vector projection = 17 17 2, 4 4. vector projection = 17 5. vector projection =  18 2, 4 17 1 5. vector projection = (2i + j + 3k) 7 correct 6. vector projection = 2 (2i+ j + 3k) 7 18 1, 4 cor6. vector projection =  17 rect Explanation: The vector projection of b onto a is given in terms of the dot product by proja b = Now when b = 2, 4 , we see that a b = 18 , Consequently, proja b =  keywords: 004 10.0 points 18 1, 4 17 . a = (1)2 + (4)2 . a = 1, 4 , ab a. a2 Explanation: The vector projection of b onto a is given in terms of the dot product by proja b = Now when b = i + 3j k, we see that ab = 2, Consequently, proja b = 1 (2i+ j + 3k) . 7 a2 = (2)2 + (1)2 + (3)2 . a = 2i+ j + 3k. ab a. a2 keywords: 005 The box shown in 10.0 points miller (zdm77) assignment 4 luecke (58600) z A 3   while C = (1, 1, 0). In this case AB is a directed line segment determining the vector u = 1, 0, 1 ,  while AC determines v = D B x is the unit cube having one corner at the origin and the coordinate planes for three of its faces. Find the cosine of the angle between AB and AC. 1. cos = 2. cos = 1 2 2 correct 3 C y 1, 1, 1 . For these choices of u and v, u v = 2 = 2 3 cos . Consequently, the cosine of the angle between AB and AC is given by cos = uv = u v 2 . 3 keywords: vectors, dot product, unit cube, cosine, angle between vectors 006 The box shown in z C D B 10.0 points 1 3. cos = 2 1 4. cos = 3 3 5. cos = 2 6. cos = 0 Explanation: To use vectors we shall replace a line segment with the corresponding directed line segment. Now the angle between any pair of vectors u, v is given in terms of their dot product by cos = uv . uv y A x is the unit cube having one corner at the origin and the coordinate planes for three of its adjacent faces.   Determine the vector projection of AD on   AB. 1. vector projection = 1 (i  k) 2 On the other hand, since the unit cube has sidelength 1, A = (0, 0, 1), B = (1, 0, 0) , miller (zdm77) assignment 4 luecke (58600) 2 2. vector projection =  (i + j  k) 3 1 3. vector projection =  (j  k) 2 4. vector projection = 5. vector projection = 2 (i + j  k) 3 1 (j  k) 2 Find the value of the determinant 1 D = 3 3 1. D = 2 2. D = 4 3. D = 2 4. D = 0 5. D = 6 correct 2 3 1 1 . 3 1 4 1 6. vector projection =  (i  k) correct 2 Explanation: The vector projection of a vector b onto a vector a is given in terms of the dot product by ab a. proja b = a2 On the other hand, since the unit cube has sidelength 1, A = (1, 1, 0), B = (0, 1, 1) , Explanation: For any 3 3 determinant A a1 a2 B b1 b2 C c1 c2 B Thus 1 D = 3 2 1 1 3 1 a1 a2 = A b1 b2 c1 c2 c1 c2 a1 a2 b1 b2 +C .   while D = (1, 0, 1). In this case AB is a directed line segment determining the vector a = 1, 0, 1 = i + k ,   while AD determines the vector b = 0, 1, 1 = j + k . 3 3 = 3 1 1 3 For these choices of a and b, ab = 1, a2 = 2 . 2 3 3 1 3  3 3 3 1 = (1)(3)  (3)(1)  2 ((3)(3)  (3)(1))  ((3)(3)  (3)(1)) . Consequently, D = 6 . keywords: determinant 008 10.0 points   Consequently, the vector projection of AD   onto AB is given by 1 proja b =  (i  k) . 2 keywords: vector projection, dot product, unit cube, component, 007 10.0 points miller (zdm77) assignment 4 luecke (58600) Find the cross product of the vectors a = i + j + 3k , b = 3i + 3j  2k . 1. a b = 2. a b = 3. a b = 3, 3, 6 correct 2, 3, 6  3, 4, 6 5 1. a b = 10i + 8j  7k 2. a b = 11i + 7j  7k 3. a b = 10i + 7j  7k 4. a b = 10i + 7j  6k 5. a b = 11i + 8j  6k Explanation: One way of computing the cross product (i + j + 3k) (3i + 3j  2k) is to use the fact that i j = k, while ii = 0, For then a b = 11i + 7j  6k . Alternatively, we can use the definition ab = 1 3 i 1 3 j k 1 3 3 2 3 j 2 j j = 0, k k = 0. j k = i, k i = j, 6. a b = 11i + 7j  6k correct 4. a b =  3, 4, 7 5. a b = 6. a b = Explanation: By definition i 1 1 j 3 3 k 1 2 1 1 j+ 1 2 3 k. 3  2, 4, 7  2, 3, 7 ab = 3 3 = 1 1 i 1 2 Consequently, a b =  3, 3, 6 . keywords: vectors, cross product 010 10.0 points Find the value of f (2) when f (x) = 3 1 3 3 x2 + 3 3 3 1 x. = 1 3 i 3 2 + 1 3 1. f (2) = 36 correct 2. f (2) = 34 3. f (2) = 32 1 k 3 to determine a b. 009 10.0 points Find the cross product of the vectors a = 1, 3, 1 , b = 1, 3, 2 . 4. f (2) = 30 5. f (2) = 28 Explanation: miller (zdm77) assignment 4 luecke (58600) For any 2 2 determinant a c Thus f (x) = 3 1 3 3 x2 + 3 3 3 1 x b d = ad  bc . 6 The cross product is defined only for two vectors, and its value is a vector; on the other hand, the dot product is defined only for two vectors, and its value is a scalar. For the three given expressions, therefore, we see that I is not welldefined because each term in the cross product is a dot product, hence a scalar. II is welldefined because it is the cross product of two vectors. III is welldefined because it is the dot product of two vectors. keywords: vectors, dot product, cross product, T/F, length, 012 10.0 points = ((3) (3)  (1) (3)) x2 + ((3) (1)  (3) (3)) x . Consequently, f (x) = 6x2  6x , and so f (2) = 36 . keywords: determinant 011 10.0 points Determine all unit vectors v orthogonal to a = 4 i + 3 j + k, 1. v = 2. v = 3. v = 4. v = b = 10 i6 j + 3 k . Which of the following expressions are welldefined for all vectors a, b, c, and d? I II (a b) (c d) , a (b c) , 6 2 3 i+ j k 7 7 7 2 6 3 i  j  k correct 7 7 7 III a (b c) . 6 2 3 i+ j k 7 7 7 3 2 6 i j k 7 7 7 1. II and III only correct 2. I only 3. I and III only 4. all of them 5. I and II only 6. III only 7. II only 8. none of them Explanation: 5. v = 3 i  2 j  6 k 6. v = 6 i + 2 j  3 k Explanation: The nonzero vectors orthogonal to a and b are all of the form v = (a b) , = 0, with a scalar. The only unit vectors orthogonal to a, b are thus v = ab . a b miller (zdm77) assignment 4 luecke (58600) But for the given vectors a and b, ab = i j 4 3 10 6 k 1 3 7 But = /2 in the case when a is parallel to the xyplane and b is parallel to k becaus k is then perpendicular to the xyplane. Consequently, for the given vectors, a b = 20 . 4 3 4 1 3 1 k j+ i = 10 6 10 3 6 3 = 3i 2j  6k. In this case, a b2 = 49 . Consequently, v = 3 2 6 i j k 7 7 7 . keywords: 014 10.0 points Find a vector v orthogonal to the plane through the points P (4, 0, 0), Q(0, 2, 0), R(0, 0, 5) . 1. v = 10, 20, 8 correct keywords: vector product, cross product, unit vector, orthogonal, 013 10.0 points 2. v = 5, 20, 8 3. v = 2, 20, 8 4. v = 10, 5, 8 5. v = 10, 4, 8 Explanation: Because the plane through P , Q, R con   tains the vectors P Q and P R, any vector v orthogonal to both of these vectors (such as their cross product) must therefore be orthogonal to the plane. Here   P Q = 4, 2, 0 , Consequently,    v = P Q P R = 10, 20, 8 is othogonal to the plane through P, Q and R. 015 10.0 points  P R = 4, 0, 5 . If a a vector parallel to the xyplane and b is a vector parallel to k, determine a b when a = 4 and b = 5. 1. a b = 10 2 2. a b = 10 3. a b = 10 4. a b = 20 correct 5. a b = 10 2 6. a b = 20 7. a b = 0 Explanation: For vectors a and b, a b = ab sin when the angle between them is . miller (zdm77) assignment 4 luecke (58600) Compute the volume of the parallelopiped with adjacent edges OP , OQ, and OR determined by vertices P (3, 1, 1) , Q(2, 3, 2) , R(2, 3, 4) , 8 where O is the origin in 3space. 1. volume = 18 2. volume = 22 correct 3. volume = 20 4. volume = 19 5. volume = 21 Explanation: The parallelopiped is determined by the vectors   a = OP = 3, 1, 1 ,   b = OQ =   c = OR = 2, 3, 2 , 2, 3, 4 . Thus its volume is given in terms of a scalar triple product by vol = a (b c) . But 3 1 2 2 + 2 2 3 3 2 4 1 2 4 + 2 3 2 3 . a (b c) = 3 2 3 4 = 3 Consequently, the parallelopiped has volume = 22 . keywords: determinant, cross product scalar triple product, parallelopiped, volume, ...
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This note was uploaded on 09/30/2008 for the course M 408M taught by Professor Gilbert during the Fall '07 term at University of Texas at Austin.
 Fall '07
 Gilbert

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