University of Texas M 408D Rusin HW 07

University of Texas M 408D Rusin HW 07 - byrne (cmb3744) HW...

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Unformatted text preview: byrne (cmb3744) HW 07 rusin (55565) 1 This print-out should have 28 questions. Multiple-choice 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 3-space equidistant from points A and B . 1. sphere centered at midpoint of AB 2. plane passing through B 3. plane bisecting line segment AB cor- rect 4. sphere centered at A 5. sphere centered at B 6. plane passing through A 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 3-space, we see that the points in 3-space 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 3-space equidistant from the points A ( 3 , 1 , 1) , B ( 1 , 4 , 3) . 1. 8 x + 6 y + 4 z + 15 = 0 2. 4 x + 6 y + 8 z + 15 = 0 3. 8 x 4 y 6 z 15 = 0 4. 4 x + 6 y + 8 z 15 = 0 correct 5. 6 x 8 y + 4 z + 15 = 0 6. 6 x + 4 y 8 z 15 = 0 Explanation: We have to find the set of points P ( x, y, z ) such that | AP | = | BP | . Now by the distance formula in 3-space, | AP | 2 = ( x + 3)2 + ( y 1)2 + ( z + 1)2 , while | BP | 2 = ( x + 1)2 + ( y 4)2 + ( z 3)2 . After expansion therefore, | AP | 2 = x 2 + 6 x + y 2 2 y + z 2 + 2 z + 11 , while | BP | 2 = x 2 + 2 x + y 2 8 y + z 2 6 z + 26 . Thus | AP | = | BP | when x 2 + 6 x + y 2 2 y + z 2 + 2 z + 11 = x 2 + 2 x + y 2 8 y + z 2 6 z + 26 . Consequently, the set of all points equidistant from A and B satisfies the equation 4 x + 6 y + 8 z 15 = 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 byrne (cmb3744) HW 07 rusin (55565) 2 Find the vector v having a representation by the directed line segment AB with respect to points A ( 2 , 4 , 4) , B ( 1 , 3 , 2) . 1. v = ( 1 , 1 , 6 ) correct 2. v = ( 3 , 7 , 2 ) 3. v = ( 3 , 7 , 2 ) 4. v = ( 3 , 7 , 2 ) 5. v = ( 1 , 1 , 6 ) 6. v = ( 1 , 1 , 6 ) Explanation: Since AB = ( 1 + 2 , 3 4 , 2 + 4 ) , we see that v = ( 1 , 1 , 6 ) ....
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This note was uploaded on 05/08/2011 for the course MATH 408D taught by Professor Chu during the Spring '09 term at University of Texas at Austin.

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University of Texas M 408D Rusin HW 07 - byrne (cmb3744) HW...

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