2002Exam1Solutions

# 2002Exam1Solutions - EXAM I SOLUTIONS Math 51 Spring 2002...

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EXAM I SOLUTIONS Math 51, Spring 2002. You have 2 hours. No notes, no books, no calculators. YOU MUST SHOW ALL WORK AND EXPLAIN ALL REASONING TO RECEIVE CREDIT Good luck! Name ID number 1. (/20 points) 2. (/30 points) 3. (/20 points) 4. (/10 points) 5. (/20 points) Bonus (/10 points) Total (/100 points) “On my honor, I have neither given nor received any aid on this examination. I have furthermore abided by all other aspects of the honor code with respect to this examination.” Signature: Circle your TA’s name: Tarn Adams (2 and 6) Mariel Saez (3 and 7) Yevgeniy Kovchegov (4 and 8) Heaseung Kwon (A02) Alex Meadows (A03) Circle your section meeting time: 11:00am 1:15pm 7pm 1

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1. Let -→ u = 0 3 1 -→ v = 3 2 2 -→ w = 2 3 4 (a) Compute the length of each of the given vectors, and the dot product for each pair. Solution: k -→ u k = 0 2 + 3 2 + 1 2 = 10 k -→ v k = 3 2 + 2 2 + 2 2 = 17 k -→ w k = 2 2 + 3 2 + 4 2 = 29 -→ u · -→ v = (0)(3) + (3)(2) + (1)(2) = 8 -→ u · -→ w = (0)(2) + (3)(3) + (1)(4) = 13 -→ v · -→ w = (3)(2) + (2)(3) + (2)(4) = 20 (b) Let θ -→ u , -→ v be the angle formed at the origin between the vectors -→ u and -→ v . Evaluate cos( θ -→ u , -→ v ) Solution: -→ u · -→ v = k -→ u kk -→ v k cos( θ -→ u , -→ v ) 8 = 10 17cos( θ -→ u , -→ v ) 8 170 = cos( θ -→ u , -→ v ) 2
(c) Find a linear dependence of the three vectors given, or prove that they are indepen- dent. Solution: A linear dependence of the vectors means that there are constants c 1 ,c 2 ,c 3 (not all zero) such that c 1 -→ u + c 2 -→ v + c 3 -→ w = -→ 0 Interpreting the matrix-vector product as a linear combination of column vectors, this is equivalent to having a nontrivial solution to the equation A -→ x = -→ 0 , with A = 0 3 2 3 2 3 1 2 4 To determine if this is the case, we row reduce the matrix: 0 3 2 3 2 3 1 2 4 1 2 4 3 2 3 0 3 2 r 3 r 2 r 1 1 2 4 0 - 4 - 9 0 3 2 r 1 r 2 - 3 r 1 r 3 2 0 - 1 0 - 4 - 9 0 0 - 19 2 r 1 + r 2 r 2 4 r 3 + 3 r 2 38 0 0 0 - 76 0 0 0 - 19 19 r 1 - r 3 19 r 2 - 9 r 3 r 3 At this point, we see that the RREF of A will have a pivot in every column, and thus that there will not be any free variables. Thus we conclude that solutions are

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2002Exam1Solutions - EXAM I SOLUTIONS Math 51 Spring 2002...

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