hw4solutions

hw4solutions - Math 215 HW #4 Solutions 1. Problem 2.1.6....

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Unformatted text preview: Math 215 HW #4 Solutions 1. Problem 2.1.6. Let P be the plane in 3-space with equation x + 2 y + z = 6. What is the equation of the plane P through the origin parallel to P ? Are P and P subspaces of R 3 ? Answer: For any real number r , the plane x + 2 y + z = r is parallel to P , since all such planes have a common normal vector i + 2 j + k = 1 2 1 . In particular, notice that the plane determined by the equation x + 2 y + z = 0 ( * ) is parallel to P and passes through the origin (since ( x, y, z ) = (0 , , 0) is a solution of the above equation). Hence, this is the equation which determines the plane P . Now, suppose x 1 y 1 z 1 , x 2 y 2 z 2 ∈ P ; i.e. the triples ( x 1 , y 1 , z 1 ) and ( x 2 , y 2 , z 2 ) both satisfy the equation ( * ). Then ( x 1 + x 2 ) + 2( y 1 + y 2 ) + ( z 1 + z 2 ) = ( x 1 + 2 y 1 + z 1 ) + ( x 2 + 2 y 2 + z 2 ) = 0 + 0 = 0 , so we have that x 1 y 1 z 1 + x 2 y 2 z 2 = x 1 + x 2 y 1 + y 2 z 1 + z 2 ∈ P . Also, if c ∈ R , then cx 1 + 2( cy 1 ) + cz 1 = c ( x 1 + 2 y 1 + z 1 ) = c (0) = 0 , so c x 1 y 1 z 1 = cx 1 cy 1 cz 1 ∈ P . Therefore, P is a subspace of R 3 . On the other hand, 6 and 3 are both in P , but 6 + 3 = 6 3 is not in P since 6 + 2(3) + 0 = 12 6 = 6 . Therefore, we see that P is not a subspace of R 3 . 1 2. Problem 2.1.12. The functions f ( x ) = x 2 and g ( x ) = 5 x are “vectors” in the vector space F of all real functions. The combination 3 f ( x )- 4 g ( x ) is the function h ( x ) = . Which rule is broken if multiplying f ( x ) by c gives the function f ( cx )? Answer: The combination 3 f ( x )- 4 g ( x ) is the function h ( x ) = 3 x 2- 20 x. If we tried to define scalar multiplication as cf ( x ) = f ( cx ) we would run into problems. Note that f (5 x ) = (5 x ) 2 = 25 x 2 , but f (2 x ) + f (3 x ) = (2 x ) 2 + (3 x ) 2 = 4 x 2 + 9 x 2 = 13 x 2 . Hence, this attempted definition of scalar multiplication would not satisfy rule 8 in the defi- nition of a vector space. 3. Problem 2.1.18. (a) The intersection of two planes through (0 , , 0) is probably a but it could be a . It can’t be the zero vector Z ! Answer: The intersection of two planes through the origin in R 3 is probably a line, but it could be a plane (if the two planes coincide). (b) The intersection of a plane through (0 , , 0) with a line through (0 , , 0) is probably a but it could be a . Answer: The intersection of a plane through the origin with a line through the origin in R 3 is probably just the single point (0 , , 0), but it could be a whole line (if the line lies in the plane)....
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hw4solutions - Math 215 HW #4 Solutions 1. Problem 2.1.6....

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