Unformatted text preview: patino (mp25752) Assignment4 luecke (57510) 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 002 10.0 points 1 A triangle P QR in 3space has vertices P (2, 0, 1), Q(2, 2, 2), R(3, 1, 4) . Use vectors to decide which one of the following properties the triangle has. 1. not rightangled at P, Q, or R Find the angle between the vectors a = 2 3, 1 , b = 7,  3 . 5 1. angle = correct 6 3 2. angle = 4 3. angle = 3 2 4. angle = 3 5. angle = 6 6. angle = 4 Explanation: Since the dot product of vectors a and b can be written as a.b = a b cos , 0 , 2. rightangled at R 3. rightangled at P 4. rightangled at Q correct 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 (2, 0, 1), Q(2, 2, 2), R(3, 1, 4) we see that   P Q = 4, 2, 1 , while Thus    QP RP = 21, and     P Q RQ = 0,  RP =   QR = 1, 3, 2 , But for the vertices where is the angle between the vectors, we see that a.b cos = , a b 0 . But for the given vectors, a b = (2 3)(7) + (1)( 3) = 13 3 , while a = Consequently, 13 3 3 cos =  =  2 13 2 13 where 0 . Thus angle = 5 . 6 13 , b = 52 . 5, 1, 3 .    P R QR = 14 . Consequently, P QR is rightangled at Q . keywords: vectors, dot product, right triangle, perpendicular, patino (mp25752) Assignment4 luecke (57510) 003 10.0 points b = 3i+ 3j+ 2k, a = 2i+ j  3k. 2 Find the vector projection of b onto a when Find the vector projection of b onto a when b = 4, 1 , a = 3, 2 . 1. vector proj. = 2. vector proj. = 3. vector proj. = 4. vector proj. = 5. vector proj. = 6. vector proj. = 7 (3 i + 3 j + 2 k) 11 3 (2 i + j  3 k) 22 3 (3 i + 3 j + 2 k) 22 3 (3 i + 3 j + 2 k) 14 3 (2 i + j  3 k) correct 14 7 (2 i + j  3 k) 11 14 4, 1 1. vector proj. = 13 2. vector proj. = 3. vector proj. = 14 4, 1 13 16 3, 2 13 16 4, 1 4. vector proj. = 13 16 5. vector proj. = 3, 2 13 6. vector proj. = 14 3, 2 correct 13 Explanation: The vector projection of b onto a is given in terms of the dot product by proja b = But when b = 4, 1 , we see that a b = (3)(4) + (2)(1) = 14 , while a2 = (3)2 + (2)2 = 13 . Consequently, proja b = 14 3, 2 13 . a = 3, 2 , 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 = 3i+ 3j+ 2k, we see that a b = (2)(3) + (1)(3) + (3)(2) = 3 , while a2 = (2)2 + (1)2 + (3)2 = 14 . Consequently, proja b = 3 (2 i + j  3 k) . 14 a = 2i+ j  3k. ab a. a2 keywords: vector projection, vectors in space, 005 10.0 points keywords: 004 10.0 points The box shown in patino (mp25752) Assignment4 luecke (57510) z C D B 3   while C = (0, 0, 1). In this case AB is a directed line segment determining the vector u = 1, 0, 1 ,  while AC determines v = y 1, 1, 1 . A 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 correct 3 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 2. cos = 2 1 3. cos = 3 3 4. cos = 2 5. cos = 1 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 1. vector projection =  (j  k) correct 2 On the other hand, since the unit cube has sidelength 1, A = (1, 1, 0), B = (0, 1, 1) , patino (mp25752) Assignment4 luecke (57510) 2 2. vector projection = (i + j  k) 3 1 3. vector projection = (i  k) 2 4. vector projection = 1 (j  k) 2 Find the value of the determinant 1 D = 3 2 1. D = 8 2. D = 12 correct 3. D = 14 4. D = 10 5. D = 6 Explanation: For any 3 3 determinant A a1 a2 B b1 b2 C c1 c2 B Thus 1 D = 3 2 = 3 3 2 3 2 1 3 2 2 3 3 3 2 3 3 2 3 3 a1 a2 = A b1 b2 c1 c2 c1 c2 a1 a2 b1 b2 2 1 3 2 3 3 . 4 1 5. vector projection =  (i  k) 2 2 6. vector projection =  (i + j  k) 3 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) , +C .   while D = (1, 0, 1). In this case AB is a directed line segment determining the vector a = 1, 0, 1 ,   while AD determines the vector b = 0, 1, 1 . For these choices of a and b, ab = 1, a2 = 2 . 2  = (3)(2)  (3)(3)  2 ((3)(2)  (2)(3))  ((3)(3)  (2)(3)) . Consequently, D = 12 .   Consequently, the vector projection of AD   onto AB is given by 1 proja b =  (j  k) . 2 keywords: vector projection, dot product, unit cube, component, 007 10.0 points keywords: determinant 008 10.0 points patino (mp25752) Assignment4 luecke (57510) Find the cross product of the vectors a = 2i  3j  k , b = 2i  2j  3k . 1. a b = 5, 6, 7 5 2. a b = 5, 1, 8 3. a b = 4, 1, 8 4. a b = 4, 6, 7 correct 1. a b = 7i  4j  3k 2. a b = 7i  3j  2k 3. a b = 7i  4j  2k correct 4. a b = 8i  4j  3k 5. a b = 8i  4j  2k Explanation: One way of computing the cross product (2i  3j  k) (2i  2j  3k) is to use the fact that i j = k, while ii = 0, For then a b = 7i  4j  2k . Alternatively, we can use the definition ab = i 2 2 3 2 j 3 2 k 1 3 1 j 3 j j = 0, k k = 0. j k = i, k i = j, 6. a b = 8i  3j  3k 5. a b = 5, 6, 8 6. a b = 4, 1, 7 Explanation: By definition i j 2 1 1 3 k 2 2 2 2 j+ 1 2 1 k. 3 ab = 1 3 = 2 2 i 1 2 Consequently, a b = 4, 6, 7 . keywords: vectors, cross product 010 10.0 points Find the value of f (2) when f (x) = 3 3 2 3 x2 + 2 3 3 3 x. = 2 1 i 2 3 + 2 2 1. f (2) = 20 2. f (2) = 16 3. f (2) = 14 4. f (2) = 12 5. f (2) = 18 correct Explanation: 3 k 2 to determine a b. 009 10.0 points Find the cross product of the vectors a = 2, 1, 2 , b = 1, 3, 2 . patino (mp25752) Assignment4 luecke (57510) For any 2 2 determinant a c Thus f (x) = 3 3 2 3 x2 + 2 3 3 3 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 welldefined because it is the dot product of two vectors. II is not welldefined because each term in the cross product is a dot product, hence a scalar. III is welldefined because it is the cross product of two vectors. keywords: vectors, dot product, cross product, T/F, length, 012 10.0 points = ((3) (3)  (3) (2)) x2 + ((2) (3)  (3) (3)) x . Consequently, f (x) = 3x2  3x , and so f (2) = 18 . keywords: determinant 011 10.0 points Determine all unit vectors v orthogonal to a = 3 i + j + 4 k, 1. v = b = 12 i + 3 j + 14 k . Which of the following expressions are welldefined for all vectors a, b, c, and d? I II a (b c) , (a b) (c d) , 6 3 2 i + j  k correct 7 7 7 2 3 6 2. v =  i + j  k 7 7 7 3. v = 3 6 2 i j+ k 7 7 7 III a (b c) . 1. none of them 2. II only 3. I and II only 4. III only 4. v = 2 i + 3 j  6 k 5. v = 2 i + 6 j  3 k 6. v = 2 6 3 i+ j k 7 7 7 5. II and III only 6. all of them 7. I and III only correct 8. I only Explanation: 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 patino (mp25752) Assignment4 luecke (57510) But for the given vectors a and b, ab = 1 = 3 i 3 12 j k 1 4 3 14 7 But = /2 in the case when a is parallel to the xyplane and b is parallel to k because k is then perpendicular to the xyplane. Consequently, for the given vectors, a b = 2 . 3 1 3 4 4 k j+ i 12 3 12 14 14 = 2i + 6j  3k. keywords: cross product, length, angle, 014 10.0 points In this case, a b2 = 49 . Consequently, v = 2 6 3 i+ j k 7 7 7 . Find a vector v orthogonal to the plane through the points P (3, 0, 0), Q(0, 2, 0), R(0, 0, 4) . 1. v = 2, 12, 6 keywords: vector product, cross product, unit vector, orthogonal, 013 10.0 points 2. v = 8, 12, 6 correct 3. v = 8, 4, 6 4. v = 4, 12, 6 5. v = 8, 3, 6 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 = 3, 2, 0 , Consequently,    v = P Q P R = 8, 12, 6 is othogonal to the plane through P, Q and R. 015 10.0 points  P R = 3, 0, 4 . If a is a vector parallel to the xyplane and b is a vector parallel to k, determine a b when a = 1 and b = 2. 1. a b = 1 2. a b = 1 3. a b = 1 2 4. a b = 2 5. a b = 1 2 6. a b = 2 correct 7. a b = 0 Explanation: For vectors a and b, a b = ab sin when the angle between them is , 0 < . patino (mp25752) Assignment4 luecke (57510) Compute the volume of the parallelopiped with adjacent edges OP , OQ, and OR determined by vertices P (2, 1, 2) , Q(1, 2, 1) , R(1, 3, 1) , 8 where O is the origin in 3space. 1. volume = 8 2. volume = 7 3. volume = 6 correct 4. volume = 9 5. volume = 5 Explanation: The parallelopiped is determined by the vectors   a = OP = 2, 1, 2 ,   b = OQ =   c = OR = 1, 2, 1 , 1, 3, 1 . Thus its volume is given in terms of a scalar triple product by vol = a (b c) . But 2 1 a (b c) = 2 1 3 1 1 2 1 3  1 1 2 1 1 1 1 = 2 2 1 2 1 3 . Consequently, the parallelopiped has volume = 6 . keywords: determinant, cross product scalar triple product, parallelopiped, volume, ...
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This note was uploaded on 10/21/2009 for the course M 53215 taught by Professor Lueke during the Spring '09 term at University of Texas.
 Spring '09
 Lueke

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