M408m-Luecke_HW2 - romasko (qrr58) – Assignment 2 –...

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Unformatted text preview: romasko (qrr58) – Assignment 2 – luecke – (55035) This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Find the value of f (−2) when 1 1 f ( x) = 1 3 x2 + 1. D = −25 correct 2. D = −23 3. D = −27 10.0 points 001 1 13 11 4. D = −29 x. 5. D = −21 Explanation: For any 3 × 3 determinant 1. f (−2) = 12 correct 2. f (−2) = 16 A B C 3. f (−2) = 14 a1 b1 c1 a2 b2 c2 4. f (−2) = 8 5. f (−2) = 10 a b c d = ad − bc . 1 1 3 x2 + 13 11 + ((1) (1) − (1) (3)) x . Consequently, c2 −1 = 3 3 3 2 3 1 +C a1 b1 a2 b2 3 3 −1 3 −1 2 1 −2 3 2 −1 1 − = (3)(1) − (3)(2) − 2 ((3)(1) − (−1)(2)) f ( x) = 2 x 2 − 2 x , − ((3)(3) − (−1)(3)) . and so f (−2) = 12 . Consequently, keywords: determinant D = −25 . 10.0 points Find the value of the determinant 1 2 −1 D= c1 2 3 x = ((1) (3) − (1) (1)) x2 002 c2 a2 1 D= 1 b2 Thus Thus f ( x) = c1 a1 −B Explanation: For any 2 × 2 determinant b1 =A 3 3 2 −1 3 1 . keywords: determinant 003 10.0 points . romasko (qrr58) – Assignment 2 – luecke – (55035) 2 Find the cross product of the vectors a = −2i − 3j − k , b = −3i − j − 3k . 1. a × b = 9i − 4j − 8k 1. a × b = 8, 11, 2 2. a × b = 9, 7, 2 3. a × b = 2. a × b = 9i − 3j − 8k 9, 7, 3 4. a × b = 8, 11, 3 3. a × b = 8i − 5j − 8k 4. a × b = 8i − 3j − 7k correct 5. a × b = 9i − 3j − 7k 5. a × b = 9, 11, 2 6. a × b = 8, 7, 3 correct Explanation: By definition 6. a × b = 8i − 4j − 7k Explanation: One way of computing the cross product (−2i − 3j − k) × (−3i − j − 3k) is to use the fact that i × j = k, = j× k = i, k ×i = j, j ×j = 0, i −3 −2 a×b = k× k = 0. 3 1 jk 31 13 −3 1 i− −2 3 −3 1 j+ −2 3 3 k. 1 Consequently, while i×i = 0, a × b = 8, 7, 3 . For then a × b = 8i − 3j − 7k . keywords: vectors, cross product Alternatively, we can use the definition a×b = i −2 −3 −3 = −1 j −3 −1 k −1 −3 −2 −1 i− −3 −3 −2 + −3 10.0 points Which of the following expressions are welldefined for all vectors a, b, c, and d? −1 j −3 −3 k −1 to determine a × b. 004 005 I | a | × ( b × c) , II a · ( b × c) , III (a · b) × (c · d) . 1. I and III only 2. all of them 10.0 points Find the cross product of the vectors a = −3, 3, 1 , b = −2, 1, 3 . 3. none of them 4. II only correct romasko (qrr58) – Assignment 2 – luecke – (55035) 3 Explanation: The non-zero vectors orthogonal to a and b are all of the form 5. I and II only 6. II and III only v = λ(a × b) , 7. III only λ = 0, with λ a scalar. The only unit vectors orthogonal to a, b are thus 8. I only Explanation: 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. v=± But for the given vectors a and b, For the three given expressions, therefore, we see that I 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. = i 3 3 a×b = II is well-defined because it is the dot product of two vectors. a×b . |a × b| jk 14 26 3 14 i− 3 26 3 4 j+ 3 6 1 k 2 = −2 i − 6 j + 3 k . III is not well-defined because each term in the cross product is a dot product, hence a scalar. In this case, keywords: vectors, dot product, cross product, T/F, length, Consequently, 006 10.0 points |a × b|2 = 49 . v=± 6 3 2 i+ j− k 7 7 7 . Determine all unit vectors v orthogonal to a = 3 i + j + 4 k, b = 3i + 2j + 6k. 6 3 2 1. v = − i − j + k 7 7 7 2. v = ± 3. v = 2 3 6 i− j+ k 7 7 7 3 6 2 i− j+ k 7 7 7 4. v = −2 i − 6 j + 3 k 5. v = ± 6 3 2 i + j − k correct 7 7 7 6. v = 2 i − 3 j + 6 k keywords: vector product, cross product, unit vector, orthogonal, 007 10.0 points If a is a vector parallel to the xy -plane and b is a vector parallel to k, determine |a × b| when |a| = 2 and |b| = 4. 1. |a × b| = −4 √ 2. |a × b| = 4 2 3. |a × b| = 4 4. |a × b| = 0 romasko (qrr58) – Assignment 2 – luecke – (55035) 4 Consequently, 5. |a × b| = −8 − − →− → v = P Q × P R = 20, 15, 12 √ 6. |a × b| = −4 2 7. |a × b| = 8 correct is othogonal to the plane through P, Q and R. Explanation: For vectors a and b, 10.0 points 009 |a × b| = |a||b| sin θ when the angle between them is θ , 0 ≤ θ < π . But θ = π/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| = 8 . Compute the volume of the parallelopiped with adjacent edges OP , OQ, and OR determined by vertices P (3, −4, 2) , Q(2, −3, 2) , R(2, 4, 4) , where O is the origin in 3-space. 1. volume = 17 2. volume = 14 keywords: cross product, length, angle, 008 10.0 points Find a vector v orthogonal to the plane through the points P (3, 0, 0), Q(0, 4, 0), R(0, 0, 5) . 1. v = 20, 5, 12 3. volume = 16 correct 4. volume = 15 5. volume = 18 Explanation: The parallelopiped is determined by the vectors − − → a = OP = 3, −4, 2 , − − → b = OQ = 2. v = 4, 15, 12 − − → c = OR = 3. v = 20, 3, 12 2, 4, 4 . Thus its volume is given in terms of a scalar triple product by 4. v = 5, 15, 12 vol = |a · (b × c)| . 5. v = 20, 15, 12 correct 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. But a · ( b × c) = Here − − → P Q = −3, 4, 0 , 2, −3, 2 , − → P R = −3, 0, 5 . =3 −3 4 2 4 3 −4 2 2 4 2 −3 +4 4 2 22 24 +2 2 −3 2 4 . romasko (qrr58) – Assignment 2 – luecke – (55035) 5 Consequently, the parallelopiped has 10.0 points 011 volume = 16 . Which of the following surfaces is the graph of keywords: determinant, cross product scalar triple product, parallelopiped, volume, 010 10.0 points 3x + 6y + 4z = 12 Which of the following statements are true for all lines and planes in 3-space? I. two lines perpendicular to a plane are parallel, II. two planes parallel to a third plane are parallel, III. two lines perpendicular to a third line are parallel. in the first octant? z correct 1. 1. II and III only 2. I only 3. I and II only correct y 4. III only x 5. none of them z 6. I and III only 2. 7. all of them 8. II only Explanation: I. TRUE: the two lines will have direction vectors parallel to the normal vector of the plane, and so be parallel, hence the two lines are parallel. y x z 3. II. TRUE: each of the two planes has a normal vector parallel to the normal vector of the third plane, and so are parallel, hence the planes are parallel. III. FALSE: the x-axis and y -axis are perpendicular, not parallel, yet both are perpendicular to the z -axis. y x romasko (qrr58) – Assignment 2 – luecke – (55035) z 6 z 4. y y x x 012 z 10.0 points Which equation has the surface 5. z z y x x y y x z 6. as its graph? 1. y + z = 1 x y 2. y − x = 1 3. x + z = 1 correct Explanation: Since the equation is linear, it’s graph will be a plane. To determine which plane, we have only to compute the intercepts of 4. x + y = 1 5. z − y = 1 6. z − x = 1 3x + 6y + 4z = 12 . Now the x-intercept occurs at y = z = 0, i.e. at x = 4; similarly, the y -intercept is at y = 2, while the z -intercept is at z = 3. By inspection, therefore, the graph is Explanation: As the surface is a plane, the equation must be linear; in addition, for fixed y and varying x, the corresponding value of z stays constant. So the equation must depend only on y and z . To determine the equation precisely, we thus look at the trace of the surface on the yz -plane as shown in romasko (qrr58) – Assignment 2 – luecke – (55035) 7 Consequently, the linear equation is 014 10.0 points x+z = 1 . 013 Find parametric equations for the line passing through the point P (1, −2, 4) and perpendicular to the plane 10.0 points A line ℓ passes through the point P (1, 4, 1) and is parallel to the vector 3, 1, 1 . At what point Q does ℓ intersect the xy plane? 4x + y − 2z = 4 . 1. x = 1 − 4t, y = 2 − t, z = 4 − 2t 2. x = 4 + t, y = 1 − 2t, z = −2 + 4t 1. Q(−2, 3, 0) correct 3. x = 1 + 4t, y = −2 + t, z = 4 − 2t correct 2. Q(0, 4, 3) 3. Q(0, −2, 5) 4. x = −1 + 4t, y = 2 + t, z = −4 − 2t 4. Q(0, 5, 4) 5. x = 4 − t, y = −1 + 2t, z = −2 + 4t 5. Q(3, −2, 0) 6. x = 4 + t, y = 1 + 2t, z = 2 − 4t 6. Q(4, 5, 0) Explanation: Since the xy -plane is given by z = 0, we have to find an equation for ℓ and then set z = 0. Now a line passing through a point P (a, b, c) and having direction vector v is given parametrically by r(t) = a + tv , a= a, b, c . Explanation: A line passing through a point P (a, b, c) and having direction vector v is given parametrically by r(t) = a + tv , Now for the given line, its direction vector will be parallel to the normal to the plane 4x + y − 2z = 4 . But for ℓ, a = 1, 4, 1 , v = 3, 1, 1 . Thus Thus a = 1, 2, 4, , r(t) = a = a, b, c . 1 + 3t, 4 + t, 1 + t , so z = 0 when t = −1. Consequently, the line ℓ intersects the xy -plane at Q(−2, 3, 0) . keywords: line, parametric equations, direction vector, point on line, intercept, coordinate plane v = 4, 1, −2 , and so r(t) = 1 + 4t, 2 + t, 4 + 2t . Consequently, x = 1 + 4t, y = 2 + t, z = 4 + 2t are parametric equations for the line. romasko (qrr58) – Assignment 2 – luecke – (55035) 1. 2x − 4y − 3z + 6 = 0 correct 10.0 points 015 8 Find parametric equations for the line passing through the points P (4, 1, 1) and Q(3, 5, 5). 2. 2x − 3y + 4z + 6 = 0 3. 2x + 3y + 4z − 6 = 0 1. x = −1 − 4t, y = 4 − t, z = 4 + t 4. 2x − 3y + 4z − 6 = 0 2. x = −1 − 4t, y = 4 + t, z = 4 + t 5. 2x + 4y + 3z + 6 = 0 3. x = −1 + 4t, y = 4 + t, z = 4 − t 6. 2x − 4y − 3z − 6 = 0 4. x = 4 − t, y = 1 − 4t, z = 1 + 4t Explanation: Since the points Q, R, and S lie in the plane, the displacement vectors 5. x = 4 + t, y = 1 + 4t, z = 1 − 4t 6. x = 4 − t, correct y = 1 + 4t, Now − − → PQ = a= is a direction vector for the given line, so a = 4, 1, 1 , v = −1, 4, 4 . ijk − − →− → n = QR × QS = 2 1 0 30 4 − t, 1 + 4t, 1 + 4t . 2 is normal to the plane. On the other hand, if P (x, y, z ) is an arbitrary point on the plane, then the displacement vector − − → v = PQ = Thus r(t) = 3, 0, 2 , lie in the plane. Thus the cross product a, b, c . −1, 4, 4 2, 1, 0 , − → QS = z = 1 + 4t Explanation: A line passing through a point P (a, b, c) and having direction vector v is given parametrically by r(t) = a + tv , − − → QR = x + 2, y + 1, z − 2 lies in the plane, so Consequently, v ·n = 0. x = 4 − t, y = 1 + 4t, z = 1 + 4t Now i 016 Find an equation for the plane passing through the points Q(−2, −1, 2) , R(0, 0, 2) , S (1, −1, 4) . k 0 30 10.0 points j 21 are parametric equations for the line. 2 = 2, −4, −3 . But then v · n = 2(x + 2) − 4(y + 1) − 3(z − 2) = 2x − 4y − 3z + 6 = 0 . romasko (qrr58) – Assignment 2 – luecke – (55035) Consequently, the plane 2x − 4y − 3z + 6 = 0 9 will be parallel to the plane and the point Q(4, 2, 4) on the line obtained by setting t = 0 lies in the plane also. Thus − − → b = QP = 3, −2, −7 passes through Q, R and S . too is parallel to the plane. In this case, keywords: plane, cross product, plane determined by three points, dot product 017 a × b = −28 − (−8), −21 + 12, 6 − 12 = 10.0 points Find an equation for the plane passing through the point P (7, 0, −3) and containing the line x = 4 − 3t, y = 2 + 4t, z = 4 + 4t . 1. 20x + 9y − 6z = 122 20x + 9y + 6z = 122 is an equation for the plane. keywords: plane, normal, cross product, line, 018 3. 3x + 4y − 4z = 10 4. 20x + 9y + 6z = 122 correct Explanation: If we can find vectors a, b parallel to the plane, their cross product A, B, C will be a vector normal to the plane. In this case A(x − 7) + By + C (z + 3) = 0 will be an equation for the plane passing through the point P (7, 0, −3) and containing the line x + 3y + 2z = 3 . 1. x + 3y + 2z = 5 correct 2. 2x + y + 3z = −2 3. 2x + y + 3z = 1 4. x + 3y + 2z = 2 5. 3x + 2y + z = 3 6. 3x + 2y + z = 6 Explanation: The scalar equation for the plane through P (a, b, c) with normal vector x = 4 − 3t, y = 2 + 4t, z = 4 + 4t . To determine a and b, note first that the given line lies in the plane, so its direction vector a = −3, 4, 4 10.0 points Find an equation for the plane passing through the point P (1, 2, −1) and parallel to the plane 5. 20x − 9y + 6z = 122 6. 3x − 4y − 4z = 10 = n. Consequently, 2. 3x + 4y + 4z = 10 n = a×b = −20, −9, −6 n = Ai + B j + C k is A(x − a) + B (y − b) + C (z − c) = 0 . romasko (qrr58) – Assignment 2 – luecke – (55035) In this question P (a, b, c) = (1, 2, −1) , while n = i + 3j + 2k if the plane is parallel to 10 Explanation: The graph is a circular cylinder whose axis of symmetry is parallel to the x-axis, so it will be the graph of an equation containing no x-term. This already eliminates the equations x2 + y 2 + 2 y = 0 , x2 + y 2 + 4 y = 0 , x + 3y + 2z = 3 x2 + y 2 − 2 x = 0 , x2 + y 2 − 4 x = 0 . since parallel planes have parallel normal vectors. Consequently, the plane has equation On the other hand, the intersection of the graph with the yz -plane, i.e. the x = 0 plane, is a circle centered on the y -axis and passing through the origin as shown in x + 3y + 2z = 5 . z keywords: plane, normal vector, point on plane, scalar equation y 019 10.0 points Which one of the following equations has graph But this circle has radius 2 because the cylinder has radius 2, and so its equation is (y − 2)2 + z 2 = 4 as a circle in the yz -plane. Consequently, after expansion we see that the cylinder is the graph of the equation when the circular cylinder has radius 2. y 2 + z 2 − 4y = 0 . 1. x2 + y 2 + 4y = 0 2. x2 + y 2 − 2x = 0 3. y 2 + z 2 − 4y = 0 correct 2 keywords: quadric surface, graph of equation, cylinder, 3D graph, circular cylinder, trace 020 10.0 points 2 4. x + y + 2y = 0 5. y 2 + z 2 − 2y = 0 Which one of the following is the graph of the equation 6. x2 + y 2 − 4x = 0 z − y2 = 1 ? romasko (qrr58) – Assignment 2 – luecke – (55035) 11 z z 5. 1. y x y x z z 6. 2. x y y x Explanation: The graph must be a cylinder with axis parallel to the x-axis. In addition, its trace on the yz -plane must be a parabola opening in the positive z -direction and having positive z -intercept. Consequently, the graph of z 3. x y z − y2 = 1 is z z 4. cory rect y x x keywords: quadric surface, cylinder, parabolic cylinder, 3D-graph, surface, Surfaces, ...
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This note was uploaded on 06/15/2011 for the course M 408M taught by Professor Gilbert during the Fall '07 term at University of Texas at Austin.

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