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M25008SelSol8

# M25008SelSol8 - Mathematics 250 Selected Solutions Fall...

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Mathematics 250 Fall 2008 Selected Solutions Assignment # 8 AS8.1: What is the relation between 7.5.8 and 7.5.19 a)? Response: In problem 7.5.8, it is shown that 1 2 det 1 1 1 a 1 b 1 c 1 a 2 b 2 c 2 is the area of the triangle with vertices A = a 1 a 2 , B = b 1 b 2 , and C = c 1 c 2 . If we think of A and B as fixed, and C as a variable, then the area of Δ ABC is 1 2 times the distance | A - B | (the base) times the height, which is the perpendicular distance from C to the line through A and B . Thus, the area will be zero exactly when C is on the line through A and B , and this is the assertion of problem 7.5.19. AS8.2: a) List the standard basis for Λ 3 ( R 5 ). b) Compute the exterior product of the following triples of vectors in R 5 . Express the result in terms of the basis of part a). i ) 1 1 1 0 0 , 0 1 - 1 3 2 , and 1 0 - 1 1 - 2 ii ) 1 1 1 0 0 , 0 1 1 1 0 , and 0 0 1 1 1 c) Compute the Gram matrix of each of the triples of vectors in part b). What is the volume of the parallelopiped spanned by each? What are the areas of the faces? d) Compute the lengths of the wedge products of part b). Compare them to the parallelopiped volumes computed in part c). Response: a) The standard basis for Λ k ( R n ) consists of k -fold wedge products of distinct standard basis elements e j of R n . (The factors must all be distinct, or the wedge product will be zero.) Thus, the standard basis of Λ k ( R n )) is labeled by k element subsets of the whole numbers from 1 to n . In particular, the basis for Λ 3 ( R 5 ) is given by e { 1 , 2 , 3 } = e 1 e 2 e 3 . e { 1 , 2 , 4 } = e 1 e 2 e 4 , e { 1 , 2 , 5 } = e 1 e 2 e 5 , e { 1 , 3 , 4 } = e 1 e 3 e 4 . e { 1 , 3 , 5 } = e 1 e 3 e 5 , e { 1 , 4 , 5 } = e 1 e 4 e 5 , e { 2 , 3 , 4 } = e 2 e 3 e 4 . e { 2 , 3 , 5 } = e 2 e 3 e 5 , e { 2 , 4 , 5 } = e 2 e 4 e 5 , and e { 3 , 4 , 5 } = e 3 e 4 e 5 .

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