HW4 - Valencia (drv252) – assignment 4 – luecke –...

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Unformatted text preview: Valencia (drv252) – 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 (5, 1, −2), Q(0, 2, 1), R(1, −1, −1) . 1 keywords: vectors, dot product, angle between vectors 002 10.0 points Find the angle between the vectors √ √ a = 1, 2 3 , b = −5, 3 3 . 5π 1. angle = 6 π 2. angle = 4 2π 3. angle = 3 π 4. angle = correct 3 π 5. angle = 6 3π 6. angle = 4 Explanation: Since the dot product of vectors a and b can be written as a.b = |a| |b| cos θ , 0 ≤ θ ≤ π, Use vectors to decide which one of the following properties the triangle has. 1. right-angled at Q 2. right-angled at R correct 3. right-angled at P 4. not right-angled at P, Q, or R 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 (5, 1, −2), Q(0, 2, 1), R(1, −1, −1) we see that − − → − − → P Q = −5, 1, 3 , QR = 1, −3, −2 , while Thus − −− →→ QP · RP = 21, and − −− →− → P Q · RQ = 14, − → RP = 4, 2, −1 . where θ is the angle between the vectors, we see that cos θ = a.b , |a| |b| 0 ≤ θ ≤ π. But for the vertices But for the given vectors, √ √ a · b = (1)(−5) + (2 3)(3 3) = 13 , while |a| = Consequently, cos θ = √ 13 1 √ = 2 13 · 2 13 π . 3 √ 13 , |b| = √ 52 . −− →− → P R · QR = 0 . where 0 ≤ θ ≤ π . Thus angle = Consequently, ∆P QR is right-angled at R . Valencia (drv252) – assignment 4 – luecke – (58600) keywords: vectors, dot product, right triangle, perpendicular, 003 10.0 points 004 10.0 points 2 Find the vector projection of b onto a when b = i − 3j+ 2k, a = 2i+ j − 2k. Find the vector projection of b onto a when b = −2, 3 , a = 3, −1 . 5 1. vector projection = − ( i − 3 j + 2 k ) 9 4 2. vector projection = − ( i − 3 j + 2 k ) 9 3. vector projection = 4. vector projection = −1 (2i+ j−2k) 3 −1 ( i−3j+2k) 3 9 3, −1 cor1. vector projection = − 10 rect 2. vector projection = − 9 −2, 3 10 −8 3. vector projection = √ −2, 3 10 4. vector projection = − 4 3, −1 5 5 5. vector projection = − ( 2 i + j − 2 k ) 9 correct 4 6. vector projection = − ( 2 i + j − 2 k ) 9 Explanation: The vector projection of b onto a is given in terms of the dot product by proja b = Now when b = i − 3j+ 2k, we see that a · b = −5 , Consequently, a = 2i+ j − 2k. a·b a. |a|2 −8 3, −1 5. vector projection = √ 10 −9 6. vector projection = √ −2, 3 10 Explanation: The vector projection of b onto a is given in terms of the dot product by proja b = Now when b = −2, 3 , we see that a · b = −9 , Consequently, proja b = − keywords: 9 3, −1 10 |a| = (3)2 + (−1)2 . a = 3, −1 , a·b a. |a|2 |a|2 = (2)2 + (1)2 + (−2)2 . 5 proja b = − ( 2 i + j − 2 k ) . 9 keywords: . 005 The box shown in 10.0 points Valencia (drv252) – assignment 4 – luecke – (58600) z A 3 − → while C = (1, 1, 0). In this case CA is a directed line segment determining the vector u= −1, −1, 1 , −→ − while CB 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 θ between CA and CB . √ 3 1. cos θ = 2 1 2. cos θ = √ 2 3. cos θ = 0 4. cos θ = 1 2 A C y 0, −1, 0 . For these choices of u and v, √ u · v = 1 = 3 cos θ . Consequently, the cosine of the angle between CA and CB is given by cos θ = 1 u·v =√ . |u| |v | 3 keywords: vectors, dot product, unit cube, cosine, angle between vectors 006 The box shown in z 10.0 points 1 5. cos θ = √ correct 3 6. cos θ = 2 3 y D B x C Explanation: To use vectors we shall replace a line segment with the corresponding directed line segment. Now the angle θ between any pair of vectors u, v is given in terms of their dot product by u·v . cos θ = |u||v| On the other hand, since the unit cube has sidelength 1, A = (0, 0, 1), B = (1, 0, 0) , 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 = 2 (i + j − k) 3 Valencia (drv252) – assignment 4 – luecke – (58600) 1 2. vector projection = − (i − k) 2 2 3. vector projection = − (i + j − k) 3 1 4. vector projection = − (j − k) 2 5. vector projection = 6. vector projection = 1 (i − k) correct 2 1 (j − k) 2 Find the value of the determinant 1 D= −3 −1 1. D = −39 2. D = −33 correct 3. D = −35 4. D = −37 Explanation: For any 3 × 3 determinant A a1 a2 B b1 b2 C c1 c2 −B Thus 1 D= −3 −1 = 3 1 2 −3 3 2 −1 2 a1 a2 =A b1 b2 c1 c2 c1 c2 a1 a2 b1 b2 5. D = −41 3 2 −1 2 . 4 1 −3 On the other hand, since the unit cube has side-length 1, A = (0, 0, 1), B = (1, 0, 0) , Explanation: The vector projection of a vector b onto a vector a is given in terms of the dot product by a·b a. proja b = |a|2 +C . − − → while D = (0, 1, 0). 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. 1 −3 −2 −3 −1 2 −3 − −3 −1 3 1 For these choices of a and b, a·b = 1, |a|2 = 2 . = (3)(−3) − (1)(2) − 2 ((−3)(−3) − (−1)(2)) − ((−3)(1) − (−1)(3)) . Consequently, D = −33 . keywords: determinant 008 10.0 points −→ − Consequently, the vector projection of AD − − → onto AB is given by proja b = 1 (i − k) . 2 keywords: vector projection, dot product, unit cube, component, 007 10.0 points Valencia (drv252) – assignment 4 – luecke – (58600) Find the cross product of the vectors a = 2i − 2j − 3k , b = −i + 3j − k . 1. a × b = 2. a × b = − 7, −7, 4 − 7, 4, 4 5 1. a × b = 11i − 6j + 4k 2. a × b = 11i + 5j + 4k correct 3. a × b = 12i − 6j + 3k 4. a × b = 12i + 5j + 4k 5. a × b = 12i + 5j + 3k Explanation: One way of computing the cross product (2i − 2j − 3k) × (−i + 3j − k) is to use the fact that i × j = k, while i×i = 0, For then a × b = 11i + 5j + 4k . Alternatively, we can use the definition a×b = i 2 −1 −2 3 j −2 3 k −3 −1 −3 j −1 j ×j = 0, k× k = 0. j× k = i, k ×i = j, 6. a × b = 11i − 7j + 3k 3. a × b = − 8, 4, 4 4. a × b = 5. a × b = 6. a × b = Explanation: By definition i j −3 2 2 −3 k −2 −1 −3 −2 j+ 2 −1 2 k. −3 − 8, 4, 5 −7, −7, 5 −8, −7, 5 correct a×b = 2 −3 = −3 −2 i− 2 −1 Consequently, a × b = − 8, −7, 5 . keywords: vectors, cross product 010 10.0 points Find the value of f (−2) when f ( x) = −3 2 3 −1 x2 + 3 −3 −1 2 x. = 2 −3 i− −1 −1 + 2 −1 1. f (−2) = −20 2. f (−2) = −24 3. f (−2) = −22 4. f (−2) = −16 5. f (−2) = −18 correct Explanation: −2 k 3 to determine a × b. 009 10.0 points Find the cross product of the vectors a = −3, 2, −2 , b = 2, −3, −1 . Valencia (drv252) – assignment 4 – luecke – (58600) For any 2 × 2 determinant a c Thus f ( x) = −3 2 3 −1 x2 + 3 −3 −1 2 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 well-defined because each term in the cross product is a dot product, hence a scalar. II is not well-defined because the first term in the cross product is a scalar because it is the length of a vector, not a vector. III is not well-defined because the second term in the cross product is a dot product, hence not a vector. keywords: vectors, dot product, cross product, T/F, length, 012 10.0 points = ((−3) (−1) − (2) (3)) x2 + ((3) (2) − (3) (−1)) 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 = i − 3 j + 4k, b = 2i − 3j + 6k. 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 1. v = − i + j + k 7 7 7 2. v = ± 2 6 3 i+ j− k 7 7 7 III a × (b · c) . 1. II and III only 2. I and III only 3. III only 4. none of them correct 5. all of them 6. I only 7. II only 8. I and II only Explanation: 3. v = −6 i + 2 j + 3 k 4. v = −3 i − 2 j + 6 k 5. v = ± 6 2 3 i − j − k correct 7 7 7 2 6 3 6. v = − i − j + k 7 7 7 Explanation: The non-zero vectors orthogonal to a and b are all of the form v = λ(a × b) , λ = 0, Valencia (drv252) – assignment 4 – luecke – (58600) with λ a scalar. The only unit vectors orthogonal to a, b are thus a×b v=± . |a × b| But for the given vectors a and b, a×b = −3 −3 i 1 2 j −3 −3 k 4 6 −3 k −3 For vectors a and b, |a × b| = |a||b| sin θ 7 when the angle between them is θ . But θ = π/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| = 4 . keywords: 014 10.0 points = 1 14 4 j+ i− 2 26 6 = −6 i + 2 j + 3 k . In this case, |a × b|2 = 49 . Consequently, 6 2 3 v = ± i− j− k 7 7 7 Find a vector v orthogonal to the plane through the points P (2, 0, 0), Q(0, 3, 0), R(0, 0, 5) . . 1. v = 3, 10, 6 2. v = 15, 10, 6 correct 3. v = 15, 5, 6 4. v = 15, 2, 6 5. v = 5, 10, 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 = −2, 3, 0 , Consequently, − − →− → v = P Q × P R = 15, 10, 6 is othogonal to the plane through P, Q and R. − → P R = −2, 0, 5 . 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| = 1 and |b| = 4. √ 1. |a × b| = −2 2 2. |a × b| = −4 3. |a × b| = 2 4. |a × b| = −2 √ 5. |a × b| = 2 2 6. |a × b| = 4 correct Explanation: 7. |a × b| = 0 Valencia (drv252) – 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, 2, −4) , 1. volume = 7 2. volume = 10 correct 3. volume = 9 4. volume = 8 5. volume = 6 Explanation: The parallelopiped is determined by the vectors − − → a = OP = 4, 2, −4 , −→ − b = OQ = − − → c = OR = 1, 3, 2 , 1, 3, 3 . Q(1, 3, 2) , R(1, 3, 3) , where O is the origin in 3-space. Thus its volume is given in terms of a scalar triple product by vol = |a · (b × c)| . But 4 a · ( b × c) = 3 3 2 3 1 1 −2 3 3 1 1 2 3 2 −4 2 3 −4 1 1 3 3 . =4 Consequently, the parallelopiped has volume = 10 . keywords: determinant, cross product scalar triple product, parallelopiped, volume, ...
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This note was uploaded on 03/26/2009 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas at Austin.

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