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01spr-m2sols - EXAM II SOLUTIONS Math 51 Spring 2001 You...

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EXAM II SOLUTIONS Math 51, Spring 2001. You have 2 hours. No notes, no books. YOU MUST SHOW ALL WORK AND EXPLAIN ALL REASONING TO RECEIVE CREDIT Good luck! Name ID number 1. (/20 points) 2. (/20 points) 3. (/20 points) 4. (/20 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: Kuan Ju Liu (2 and 6) Robert Sussland (3 and 7) Hunter Tart (4 and 8) Alex Meadows (10) Dana Rowland (11) Circle your section meeting time: 11:00am 1:15pm 7pm 1
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1. (a) Use determinants to find the area of the triangle in R 2 with vertices located at 1 5 , 4 3 , - 2 1 Solution: The triangle determined by those three points is exactly half of the par- allelogram that is determined by any two sides. These sides can be represented as vectors, with tails on one of the given vectors and heads on the other two. For example, -→ v 1 = 1 5 - 4 3 = - 3 2 -→ v 2 = - 2 1 - 4 3 = - 6 - 2 The area of the parallelogram is then det - 3 - 6 2 - 2 = ( - 3)( - 2) - ( - 6)(2) = 18 So, the area of the triangle is then 9. (b) Use the cross product to determine the area of the triangle in R 3 with vertices located at 1 2 4 , 2 5 3 , 0 - 4 2 Solution: As above, the triangle here is exactly half of the parallelogram determined by -→ v 1 = 1 2 4 - 0 - 4 2 = 1 6 2 -→ v 2 = 2 5 3 - 0 - 4 2 = 2 9 1 This parallelogram has area equal to the length of the cross product of these two vectors. -→ v 1 × -→ v 2 = det -→ e 1 -→ e 2 -→ e 3 1 6 2 2 9 1 = - 12 3 - 3 So | -→ v 1 × -→ v 2 | = 12 2 + 3 2 + 3 2 = 162 = 9 2, and thus we conclude that the area of the triangle is half of that, or 9 / 2 2
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(c) Noticing that for vectors -→ v , -→ w , and -→ x , we have ( -→ v × -→ w ) · -→ x = det x 1 x 2 x 3 v 1 v 2 v 3 w 1 w 2 w 3 use the properties of the determinant to show that the cross product of two vectors is always perpendicular to each of those two vectors. Solution: To show that the cross product is perpendicular to -→ v , we simply compute the dot product of those two vectors, using the observed formula above. ( -→ v × -→ w ) · -→ v = det v 1 v 2 v 3 v 1 v 2 v 3 w 1 w 2 w 3 Since two of the rows of this matrix are identical, we immediately conclude that this determinant is zero, and so the dot product in question is zero. So, the cross product is perpendicular to -→ v .
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