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Unformatted text preview: suleimenov (bs26835) HW 07 rusin (55565) 1 This printout should have 28 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. Welcome back from Spring Break! Now, get to work! :) . I have made the assignment a little longer following some requests for more ongoing practice. Let me know how you like it should I go back to about 20 problems per week or do you like having more? 001 (part 1 of 2) 10.0 points 1. Describe the set of all points in 3space equidistant from points A and B . 1. sphere centered at B 2. sphere centered at A 3. plane passing through B 4. plane passing through A 5. sphere centered at midpoint of AB 6. plane bisecting line segment AB cor rect Explanation: In the plane the set of all points equidistant from points C and D is a straight line bisect ing the line segment CD . But if we now apply this to every plane passing through points A and B in 3space, we see that the points in 3space equidistant from A and B will be the plane bisecting the line segment AB . 002 (part 2 of 2) 10.0 points 2. Confirm your answer to part 1 by finding an equation for the set of all points in 3space equidistant from the points A (1 , 3 , 1) , B (4 , 1 , 2) . 1. 3 x + 2 y + z 5 = 0 correct 2. x + 2 y + 3 z + 5 = 0 3. x 3 y 2 z 5 = 0 4. 2 x y + 3 z + 5 = 0 5. 3 x + 2 y + z + 5 = 0 6. 2 x + 3 y z 5 = 0 Explanation: We have to find the set of points P ( x, y, z ) such that  AP  =  BP  . Now by the distance formula in 3space,  AP  2 = ( x 1)2 + ( y 3)2 + ( z 1)2 , while  BP  2 = ( x 4)2 + ( y + 1)2 + ( z 2)2 . After expansion therefore,  AP  2 = x 2 2 x + y 2 + 6 y + z 2 2 z + 11 , while  BP  2 = x 2 8 x + y 2 + 2 y + z 2 4 z + 21 . Thus  AP  =  BP  when x 2 2 x + y 2 + 6 y + z 2 2 z + 11 = x 2 8 x + y 2 + 2 y + z 2 4 z + 21 . Consequently, the set of all points equidistant from A and B satisfies the equation 3 x + 2 y + z 5 = 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 003 10.0 points suleimenov (bs26835) HW 07 rusin (55565) 2 Find the vector v having a representation by the directed line segment AB with respect to points A ( 2 , 3 , 2) , B ( 1 , 5 , 3) . 1. v = ( 3 , 8 , 1 ) 2. v = ( 3 , 8 , 1 ) 3. v = ( 3 , 8 , 1 ) 4. v = ( 1 , 2 , 5 ) correct 5. v = ( 1 , 2 , 5 ) 6. v = ( 1 , 2 , 5 ) Explanation: Since AB = ( 1 + 2 , 5 + 3 , 3 2 ) , we see that v = ( 1 , 2 , 5 ) ....
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 Spring '11
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