homework1_solutions - Solutions MATH 211 (Siefken):...

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Unformatted text preview: Solutions MATH 211 (Siefken): Homework #1 Due May 11, 2011 Question #1 Given three non-zero vectors in R 3 , the set of all linear combinations, or span , of these vectors is either a line, a plane or all of R 3 . For d R , define u = [3 , 4 , 1], v = [4 ,- 4 ,- 4], and w = [14 , ,d ]. (a) For what value(s) of d does span { u , v , w } (that is, the set of all linear combinations a u + b v + c w ) consist of only a plane in R 3 ? Since u and v are not multiples of eachother (and neither is the zero vector), span { u , v } is a plane. Since span { u , v } span { u , v , w } , the only way for span { u , v , w } to be a plane is if w is a linear combination of u and v . We must now find constants a,b,d so that a u + b v = w , and since the first two coordinates of w are 14 , 0, we know that the first two coordinates of a u + b v must be 14 , 0. By inspection, we notice that 2 u + 2 v = [14 , ,- 6], and so if d =- 6, w span { u , v } giving that span { u , v , w } = span { u , v } is a plane. d =- 6 is the only value that works since [14 , , * ] is linearly independent from u and v if * 6 =- 6. Thus, if d 6 =- 6, span { u , v , w } = R 3 is not a plane. (b) Is there a value of d for which span { u , v , w } is only a line? (Either give such a d or show that there cannot exist such a d .) No. span { u , v } is a plane and span { u , v } span { u , v , w } , so span { u , v , w } must be at least a plane. Question #2 Let v = [- 1 , 1 , 1 ,- 1]. (a) Find three nonzero vectors in R 4 that are orthogonal to v and orthogonal to eachother. By guess and check, we notice [1 , 1 , 1 , 1], [- 1 ,- 1 , 1 , 1], and [1 ,- 1 , 1 ,- 1] are such vectors....
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homework1_solutions - Solutions MATH 211 (Siefken):...

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