assignment 4-solutions

# assignment 4-solutions - Mirzoyeva(tim89 assignment 4...

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Unformatted text preview: Mirzoyeva (tim89) assignment 4 luecke (58600) 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 A triangle P QR in 3-space has vertices P (-1, -3, 3), Q(3, -1, 2), R(-2, 0, 5) . 1 keywords: vectors, dot product, angle between vectors 002 10.0 points Find the angle between the vectors a = 2 3, -1 , b = 3 3, 5 . 1. angle = 6 correct 2. angle = 3 3 3. angle = 4 4. angle = 4 5 5. angle = 6 2 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. right-angled at P correct 2. right-angled at R 3. not right-angled at P, Q, or R 4. right-angled at Q 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 (-1, -3, 3), Q(3, -1, 2), R(-2, 0, 5) we see that - - - - P Q = 4, 2, -1 , QR = -5, 1, 3 , while Thus - RP = 1, -3, -2 . - - - - P Q RQ = 21, But for the vertices where is the angle between the vectors, we see that cos = a.b , |a| |b| 0 . But for the given vectors, a b = (2 3)(3 3) + (-1)(5) = 13 , while |a| = Consequently, cos = 13 1 = 2 13 2 13 . 3 13 , |b| = 52 . - - - QP RP = 0, and - - - P R QR = -14 . Consequently, P QR is right-angled at P . where 0 . Thus angle = Mirzoyeva (tim89) assignment 4 luecke (58600) keywords: vectors, dot product, right triangle, perpendicular, 003 10.0 points 1. vector projection = 2. vector projection = 3. vector projection = cor4. vector projection = 5. vector projection = correct 6. vector projection = 4 ( i + 2j + 2k) 3 11 ( i + 2j+ 2k) 9 10 (4i+ j + 2k) 9 11 (4i+ j + 2k) 9 2 Find the vector projection of b onto a when b = 4i+ j + 2k, a = i + 2j + 2k. Find the vector projection of b onto a when b = -2, -2 , a = 2, 3 . 1. vector projection = - 10 -2, -2 13 10 2. vector projection = - 2, 3 13 rect 3. vector projection = - 9 2, 3 13 10 ( i + 2j + 2k) 9 4 (4i+ j + 2k) 3 -10 -2, -2 4. vector projection = 13 -9 5. vector projection = 2, 3 13 -9 6. vector projection = -2, -2 13 Explanation: The vector projection of b onto a is given in terms of the dot product by proja b = Now when b = -2, -2 , we see that a b = -10 , Consequently, proja b = - keywords: 004 10.0 points 10 2, 3 13 . |a| = (2)2 + (3)2 . a = 2, 3 , 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 = 4i+ j + 2k, we see that a b = 10 , Consequently, proja b = 10 ( i + 2j+ 2k) . 9 |a|2 = (1)2 + (2)2 + (2)2 . a = i + 2j + 2k. ab a. |a|2 keywords: 005 The box shown in 10.0 points Mirzoyeva (tim89) assignment 4 luecke (58600) z C D B 3 - - while D = (1, 0, 1). In this case AB is a directed line segment determining the vector u = -1, 0, 1 , - - while AD determines v = y 0, -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 AD. 1. cos = 2 3 For these choices of u and v, u v = 1 = 2 2 cos . Consequently, the cosine of the angle between AB and AD is given by cos = 1 uv = . |u| |v| 2 keywords: vectors, dot product, unit cube, cosine, angle between vectors 006 The box shown in z 10.0 points 1 2. cos = 3 3. cos = 0 4. cos = 3 2 A 1 5. cos = 2 6. cos = 1 correct 2 D B 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 AB onto - AC. 1. vector projection = 1 (j - k) 2 C y 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| On the other hand, since the unit cube has sidelength 1, A = (1, 1, 0), B = (0, 1, 1) , Mirzoyeva (tim89) assignment 4 luecke (58600) 1 2. vector projection = - (i - k) 2 2 3. vector projection = (i+j-k) correct 3 2 4. vector projection = - (i + j - k) 3 5. vector projection = 1 (i - k) 2 Find the value of the determinant 1 D = 3 -1 2 -3 2 -1 2 2 . 4 1. D = -29 correct 2. D = -27 3. D = -21 4. D = -25 Explanation: For any 3 3 determinant A a1 a2 B b1 b2 C c1 c2 -B Thus 1 D = 3 -1 = -3 2 2 2 2 -3 2 -1 2 2 3 -1 2 2 - 3 -1 -3 2 a1 a2 = A b1 b2 c1 c2 c1 c2 a1 a2 b1 b2 5. D = -23 1 6. vector projection = - (j - k) 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 = |a|2 On the other hand, since the unit cube has side-length 1, A = (0, 0, 1), B = (1, 0, 0) , +C . - while C = (1, 1, 0). In this case AC is a directed line segment determining the vector a = 1, 1, -1 = i + j - k, , - - while AB determines the vector b = 1, 0, -1 = i - k. For these choices of a and b, ab = 2, |a|2 = 3 . -2 = (-3)(2) - (2)(2) - 2 ((3)(2) - (-1)(2)) - ((3)(2) - (-1)(-3)) . Consequently, D = -29 . keywords: determinant 008 10.0 points - - Consequently, the vector projection of AB - onto AC is given by proja b = 2 (i + j - k) . 3 keywords: vector projection, dot product, unit cube, component, 007 10.0 points Mirzoyeva (tim89) assignment 4 luecke (58600) Find the cross product of the vectors a = -i - 2j - 3k , b = 3i - j + k . 1. a b = - 5, 5, -4 2. a b = 3. a b = 4. a b = 5. a b = 6. a b = Explanation: By definition i 1 1 j 2 -1 k -3 -1 2 k. -1 - 4, -2, -4 - 5, 5, -3 -5, -2, -3 correct -4, -2, -3 - 4, 5, -4 5 1. a b = -5i - 8j + 7k correct 2. a b = -4i - 8j + 7k 3. a b = -5i + 5j + 7k 4. a b = -4i - 8j + 6k 5. a b = -4i + 5j + 6k Explanation: One way of computing the cross product (-i - 2j - 3k) (3i - j + k) is to use the fact that i j = k, while ii = 0, For then a b = -5i - 8j + 7k . Alternatively, we can use the definition ab = i -1 3 -2 -1 j -2 -1 k -3 1 -3 j 1 j j = 0, k k = 0. j k = i, k i = j, 6. a b = -5i + 4j + 6k ab = 2 -1 = 1 1 -3 -3 j+ i- 1 1 -1 -1 Consequently, a b = - 5, -2, -3 . keywords: vectors, cross product 010 10.0 points Find the value of f (-2) when f (x) = -3 1 2 -2 x2 + 2 -3 -2 1 x. = -1 -3 i- 3 1 + -1 3 1. f (-2) = 32 2. f (-2) = 26 3. f (-2) = 28 4. f (-2) = 24 correct 5. f (-2) = 30 Explanation: -2 k -1 to determine a b. 009 10.0 points Find the cross product of the vectors a = 1, 2, -3 , b = 1, -1, -1 . Mirzoyeva (tim89) assignment 4 luecke (58600) For any 2 2 determinant a c Thus f (x) = -3 1 2 -2 x2 + 2 -3 -2 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 well-defined because it is the cross product of two vectors. II is not well-defined because the second term in the cross product is a dot product, hence not a vector. III is not well-defined because each term in the cross product is a dot product, hence a scalar. keywords: vectors, dot product, cross product, T/F, length, 012 10.0 points = ((-3) (-2) - (1) (2)) x2 + ((2) (1) - (-3) (-2)) x . Consequently, f (x) = 4x2 - 4x , and so f (-2) = 24 . keywords: determinant 011 10.0 points Determine all unit vectors v orthogonal to a = 4 i + 3 j + k, b = 10 i9 j + 2 k . Which of the following expressions are welldefined for all vectors a, b, c, and d? I II a (b c) , a (b c) , 1. v = -3 i + 2 j + 6 k 2 6 3 2. v = - i + j + k 7 7 7 3. v = -6 i - 2 j + 3 k 6 2 3 4. v = - i - j + k 7 7 7 5. v = 6. v = 2 6 3 i - j - k correct 7 7 7 6 2 3 i+ j- k 7 7 7 III (a b) (c d) . 1. II and III only 2. II only 3. I only correct 4. none of them 5. III only 6. I and III only 7. all of them 8. I and II only Explanation: Explanation: The non-zero vectors orthogonal to a and b are all of the form v = (a b) , = 0, Mirzoyeva (tim89) assignment 4 luecke (58600) with a scalar. The only unit vectors orthogonal to a, b are thus ab v = . |a b| But for the given vectors a and b, ab = i j 4 3 10 9 k 1 2 For vectors a and b, |a b| = |a||b| sin 7 when the angle between them is . But = /2 in the case when a is parallel to the xy-plane and b is parallel to k becaus k is then perpendicular to the xy-plane. Consequently, for the given vectors, |a b| = 6 . keywords: = 4 3 4 1 3 1 k j+ i- 10 9 10 2 9 2 = -3 i + 2 j + 6 k . 10.0 points 014 In this case, |a b|2 = 49 . Consequently, 3 2 6 v = i- j- k 7 7 7 Find a vector v orthogonal to the plane through the points P (5, 0, 0), Q(0, 3, 0), R(0, 0, 4) . . 1. v = 3, 20, 15 2. v = 12, 20, 15 correct 3. v = 4, 20, 15 4. v = 12, 4, 15 5. v = 12, 5, 15 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 = -5, 3, 0 , Consequently, - - - v = P Q P R = 12, 20, 15 is othogonal to the plane through P, Q and R. - P R = -5, 0, 4 . keywords: vector product, cross product, unit vector, orthogonal, 013 10.0 points If a a vector parallel to the xy-plane and b is a vector parallel to k, determine |a b| when |a| = 2 and |b| = 3. 1. |a b| = -3 2. |a b| = 3 3. |a b| = -3 2 4. |a b| = -6 5. |a b| = 3 2 6. |a b| = 6 correct Explanation: 7. |a b| = 0 Mirzoyeva (tim89) assignment 4 luecke (58600) 015 10.0 points 8 Compute the volume of the parallelopiped with adjacent edges OP , OQ, and OR determined by vertices P (4, -4, 2) , Q(3, -1, 2) , R(3, 2, 2) , where O is the origin in 3-space. 1. volume = 3 2. volume = 6 correct 3. volume = 5 4. volume = 4 5. volume = 2 Explanation: The parallelopiped is determined by the vectors - - a = OP = 4, -4, 2 , - - b = OQ = - - c = OR = 3, -1, 2 , 3, 2, 2 . Thus its volume is given in terms of a scalar triple product by vol = |a (b c)| . But a (b c) = -1 2 2 2 4 -4 3 2 2 2 2 2 2 +2 3 3 -1 2 . 3 -1 3 3 = 4 +4 Consequently, the parallelopiped has volume = 6 . keywords: determinant, cross product scalar triple product, parallelopiped, volume, ...
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