This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Math 215 HW #4 Solutions 1. Problem 2.1.6. Let P be the plane in 3space 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)....
View
Full
Document
This document was uploaded on 08/20/2011.
 Spring '09
 Math

Click to edit the document details