HW 07-solutions

# HW 07-solutions - suleimenov(bs26835 HW 07 rusin(55565 This...

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suleimenov (bs26835) – HW 07 – rusin – (55565) 1 This print-out should have 28 questions. Multiple-choice questions may continue on the next column or page – fnd 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 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 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. Confrm your answer to part 1 by fnding an equation For the set oF all points in 3-space 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 fnd the set oF points P ( x, y, z ) such that | AP | = | BP | . Now by the distance Formula in 3-space, | AP | 2 = ( x 1)2 + ( y 3)2 + ( z 1)2 , while | | 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 | | 2 = x 2 8 x + y 2 + 2 y + z 2 4 z + 21 . Thus | AP | = | | 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 satisfes 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

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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 = a− 3 , 8 , 1 A 2. v = a 3 , 8 , 1 A 3. v = a− 3 , 8 , 1 A 4. v = a 1 , 2 , 5 A correct 5. v = a 1 , 2 , 5 A 6. v = a− 1 , 2 , 5 A Explanation: Since AB = a− 1 + 2 , 5 + 3 , 3 2 A , we see that v = a 1 , 2 , 5 A . 004 10.0 points When u , v are the displacement vectors u = AB , v = −→ AP , determined by the parallelogram A B C D P Q R S O express AR in terms of u and v . 1. AR = 2 v 2. AR = 2( u + v ) 3. AR = 2( u v ) 4. AR = 2 u 5. AR = 2 v u 6. AR = u + 2 v correct Explanation: By the parallelogram law for the addition of vectors we see that AR = u + 2 v .
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HW 07-solutions - suleimenov(bs26835 HW 07 rusin(55565 This...

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