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Unformatted text preview: HW06a gilbert (55035) 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. Consequently, length = 38 . 1 10.0 points A rectangular box is constructed in 3 space with one corner at the origin and other vertices at (5, 0, 0), (0, 2, 0), (0, 0, 3) . Find the length of the diagonal of the box. 1. length = 17 length = 2. 33 3. length = 33 4. length = 38 correct 5. length = 17 6. 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 yzcoordinate plane. 1. x 2 + y 2 + z 2 6x 8y + 4z + 25 = 2. x 2 + y 2 + z 2 + 6x + 8y 4z + 25 = 3. x 2 + y 2 + z 2 6x 8y + 4z + 20 = correct 4. x 2 + y 2 + z 2 6x 8y + 4z + 13 = 5. x 2 + y 2 + z 2 + 6x + 8y 4z + 13 = Explanation: We have to find the length of BD in the figure D G E F O C A B given that OA = 5 , OC = 2 , OD = 3 . Now by Pythagoras theorem, 6. x 2 + y 2 + z 2 + 6x + 8y 4z + 20 = Explanation: Since the sphere touches the yzplane, its radius, r, is the distance from its center, (3, 4, 2) to the yzplane; thus r = 3. Consequently (x 3) 2 + (y 4) 2 + (z + 2) 2 = 9 is an equation for the sphere. After expansion this becomes x 2 + y 2 + z 2 6x 8y + 4z + 20 = 0 . 003 10.0 points Find the trace on the xyplane of the sphere having center at (1, 3, 1) and radius 3. length OB = length AC = 29 . 1. x 2 + y 2 2x 6y + 2 = 0 correct But then, again by Pythagoras, 2. y 2 + z 2 6y 2z + 2 = length BD = 38 ....
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 Spring '07
 Sadler

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