Assignment 4

# Assignment 4 - patino(mp25752 Assignment4 luecke(57510 This...

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Unformatted text preview: patino (mp25752) Assignment4 luecke (57510) This print-out should have 15 questions. Multiple-choice 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 3-space 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 right-angled 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. right-angled at R 3. right-angled at P 4. right-angled at Q correct Explanation: Vectors a and b are perpendicular when a b = 0. Thus P QR will be - - - (1) right-angled at P when QP RP = 0, - - - - (2) right-angled at Q when P Q RQ = 0, - - - (3) right-angled 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 right-angled 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 |a|2 = (3)2 + (-2)2 = 13 . Consequently, proja b = 14 3, -2 13 . a = 3, -2 , ab a. |a|2 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 |a|2 = (2)2 + (1)2 + (-3)2 = 14 . Consequently, proja b = 3 (2 i + j - 3 k) . 14 a = 2i+ j - 3k. ab a. |a|2 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 . |u||v| 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 = |a|2 On the other hand, since the unit cube has side-length 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, |a|2 = 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 well-defined because it is the dot product of two vectors. II is not well-defined because each term in the cross product is a dot product, hence a scalar. III is well-defined 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 non-zero 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 xy-plane and b is parallel to k because k is then perpendicular to the xy-plane. 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 b|2 = 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 xy-plane 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| = |a||b| 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 3-space. 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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