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Course: M 408D 408D, Spring 2012
School: University of Texas
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gilbert HW06a (55035) This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 1 10.0 points A rectangular box is constructed in 3space with one corner at the origin and other ver-tices at (5, 0, 0), (0, 2, 0), (0, 0, 3) . Find the length of the diagonal of the box. 1. length = 17 2. length = 33 Consequently, length...

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gilbert HW06a (55035) This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 1 10.0 points A rectangular box is constructed in 3space with one corner at the origin and other ver-tices at (5, 0, 0), (0, 2, 0), (0, 0, 3) . Find the length of the diagonal of the box. 1. length = 17 2. length = 33 Consequently, length = 38 . keywords: length diagonal, rectangular solid, Pythagoras theorem, ThreeDimSys, 002 10.0 points Find an equation for the sphere centered at (3, 4, 2) that is tangent to the yz-coordinate plane. 2 2 2 1. x + y + z 6x 8y + 4z + 25 = 0 2 2 2 2. x + y + z + 6x + 8y 4z + 25 = 0 2 2 2 3. x + y + z 6x 8y + 4z + 20 = 0 correct 3. length = 33 4. length = 38 correct 5. length = 17 2 2 2 4. x + y + z 6x 8y + 4z + 13 6. length = 38 Explanation: We have to find the length of BD in the figure D 1 G E F A 2 2 2 5. x + y + z + 6x + 8y 4z + 13 = 0 2 2 2 6. x + y + z + 6x + 8y 4z + 20 = 0 Explanation: Since the sphere touches the yz-plane, its radius, r, is the distance from its center, (3, 4, 2) to the yz-plane; thus r = 3. Con-sequently 2 2 2 (x 3) + (y 4) + (z + 2) = 9 C O =0 is an equation for the sphere. After expansion this becomes B 2 2 2 x + y + z 6x 8y + 4z + 20 = 0 given that OA = 5 , OC = 2 , OD = 3 . But then, again by Pythagoras, length BD = 38 . 10.0 points Find the trace on the xy-plane of the sphere having center at (1, 3, 1) and radius 3. Now by Pythagoras theorem, length OB = length AC = 003 29 . 2 2 1. x + y 2x 6y + 2 = 0 correct 2 2 2 . y + z 6 y 2z + 2 = 0 2 2 3. y + z + 6y + 2z = 2 . HW06a gilbert (55035) 2 Now by the distance formula in 3-space, 2 2 4. x + y + 2x + 6y = 2 2 2 2 2 |AP | = (x 2) + (y + 3) + (z 1) , 2 2 5. x + z 2x 2z + 2 = 0 while 2 2 6. x + z + 2x + 2z = 2 Explanation: The sphere consists of all points P (x, y, z) 2 2 2 2 |BP | = (x 4) + (y + 2) + (z 4) . After expansion therefore, such that dist ((x, y, z) , (1, 3, 1)) = 3 . 2 2 2 2 |AP | = x 4x + y + 6y + z 2z + 14 , while Thus, by the distance formula in 3-space, 2 2 2 |BP | (x 1) + (y 3) + (z 1) = 9, which after expansion becomes 2 2 2 x + y + z 2x 6y 2 z + 2 = 0 . But the xy-plane is the plane z = 0. Consequently, the trace of the sphere on the xy plane is obtained by setting z = 0, so its equation is 2 2 x + y 2x 6y + 2 = 0 004 . 10.0 points Find an equation for the set of all points in 3-space equidistant from the points A(2, 3, 1) , 2 2 2 2 = x 8x + y + 4y + z 8z + 36 . Thus |AP |= |BP | when 2 2 2 x 4x + y + 6y + z 2z + 14 2 2. x 3y + 2z + 11 = 0 2 = x 8x + y + 4y + z 8z + 36 . Consequently, the set of all points equidistant from A and B satisfies the equation 2x + y + 3z 11 = 0 . Notice that this is a plane perpendicular to the line segment joining A and B (since it must contain the perpendicular bisector of the line segment AB). keywords: plane, locus points, equidistant two points B(4, 2, 4) . 1. 3x 2y z 11 = 0 2 005 10.0 points The parallelopiped in 3 -space shown in A C 3. 3x + y + 2z + 11 = 0 4. x + 2y 3z 11 = 0 R B 5. 2x + y + 3z + 11 = 0 6. 2x + y + 3z 11 = 0 correct Explanation: We have to find the set of points P (x, y, z) such that |AP | = |BP | . S P D Q HW06a gilbert (55035) 3 z is determined by its vertices P (1, 4), 4, Q(4, 4, 2) , R(1, A 0, 1), S(2, 1, 3) . C Find the vector v represented by the directed line segment P A. R O 1. v = h 6, 7, 6 O correct 2. v = h 0, 2, 1 O y Q P x 3. v = h 7, 4, 5 O is the cube of side length 2 units having one corner at the origin O and the coordinate planes for three of its faces. 4. v = h 4, 3 , 3 O 5. v = h 1, 7 , 4 O Find the vector v represented by the di rected line segment CB where B is the mid-point of the line segment AR. 6. v = h 7, 8 , 6 O Explanation: As a vector sum, But PA PQ = = 5, 0, 2 , h while 1. v = h 2, 1, 1 O PQ + PS . 2, 4, 3 , = O h 4. v = h 1, 1, 1 O 5. v = h 1, 2, 1 O PS = 1, 3, 1 . h 6. v = h 2, 1, 1 O correct Explanation: Since B is the point O Consequently, v= 2. v = h 1, 2, 1 O 3. v = h 1, 1, 1 O PR O PR + PA = . 6, 7, 6 h O B= 1 2 (0, 0, 2) + (0, 2, 0) = (0, 1, 1) , while C = (2, 2, 2) , keywords: parallelopiped, 3-space, coordi-nates, vertex, directed line segment, vector sum 006 we see that v= 2, CB = h 1, 1 . O 10.0 points The box shown in keywords: cube, 3-space, coordinates, vertex, directed line segment, mid-point, HW06a gilbert (55035) 007 10.0 points Determine the length of the vector 2a b when a = i + 2j + 3k , b = i + 2j + k . 4 4. a b = 6 5. a b = 10 Explanation: The dot product, a a = a1i + a2j + a3k , 1. length = 6 2. length = 4 2 3. length = 38 correct b, of vectors b = b1i + b2j + b3k is defined by a b = a1 b 1 + a2 b 2 + a3 b 3 . 4. length = 34 Consequently, when a = 3i 2j + 3k, 5. length = 30 b = 2 i j 2k , we see that Explanation: The length, |c|, of the vector c = c1i + c2j + c3k ab = 2 is defined by 009 . 10.0 points q 2 2 2 c 1+c 2+c 3. Consequently, when |c| = a = i + 2j + 3k , b = i + 2j + k , and Find the angle between the vectors a = h 2 3 , 1 O , b = h 7, 1. angle = c = 2a b = 3i + 2j + 5k , 3O. 3 4 2. angle = 6 correct we see that |2a b| = 38 . 008 10.0 points Determine the dot product of the vectors a = 3i 2j + 3k, 1. a b = 4 2. a b = 2 correct 3. a b = 8 b = 2i j 2k . 2 3 4. angle = 4 5. angle = 3 3. angle = 6. angle = 5 6 Explanation: Since the dot product of vectors a and b can be written as a.b = kak kbk cos , 0 , HW06a gilbert (55035) where is the angle between the vectors, we see that we see that a.b = 5 k ak kbk cos But for the given vectors, , 0. PQ while h cos = QP RP 52 . and Consequently, =2 13 3 3 13 2 13 1, = h 4, 2, 1 , 2 3, O . O Thus kbk = 13 , QR O = h kak = Consequently, RP a b = (2 3)(7) + (1)( 3) = 13 3 , while 5, 1, 3 , = = 21, PQ RQ = 14, PR QR = 0. P QR is right-angled at R . where 0 . Thus angle = 10 A triangle 6 keywords: vectors, dot product, right trian-gle, perpendicular, . 10.0 points P QR in 3-space has vertices P (2, 4, 0), Q(3, 5, 3), R(2, 2, 1) . Use vectors to decide which one of the follow-ing 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. But for the vertices P (2, 4, 0), Q(3, 5, 3), R(2, 2, 1)
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University of Texas - M 408D - 408D
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University of Texas - M 408D - 408D
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University of Texas - M 408D - 408D
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University of Texas - M 408D - 408D
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University of Texas - M 408D - 408D
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University of Texas - M 408D - 408D
DiscGQ02 gilbert (55035)This print-out should have 4 questions.Multiple-choice questions may continueon the next column or page find allchoices before answering.00110.0 points1and the series11B=F (N) .N=61. A &lt; BDetermine whether the series
University of Texas - M 408D - 408D
EXAM 02 gilbert (55035)This print-out should have 14 questions.Multiple-choice questions may continueon the next column or page find allchoices before answering.0011I.E., y = x. Thus by the constraint equation,g (x, x) = 2x + x 6 = 0 ,I.E., x = 2
University of Texas - M 408D - 408D
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University of Texas - M 408D - 408D
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University of Texas - M 408D - 408D
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University of Texas - M 408D - 408D
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University of Texas - M 408D - 408D
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University of Texas - M 408D - 408D
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