Exam 3 Practice-solutions

Exam 3 Practice-solutions - rodriguez (lar2636) Exam 3...

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Unformatted text preview: rodriguez (lar2636) Exam 3 Practice rodin (54565) 1 This print-out should have 45 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 0.0 points Find the minimum length of u v when u = 2 j and v is a position vector of length 8 in the yz-plane. 1. minimum length = 18 2. minimum length = 0 correct 3. minimum length = 17 4. minimum length = 19 5. minimum length = 16 6. minimum length = 15 Explanation: The length of the cross product of u and v is given by | u v | = | u || v | sin = 16sin where 0 is the angle between u and v . Now j lies in the yz-plane, so the angle between j and v varies from 0 to . Consequently, u v has minimum length = 0 . 002 0.0 points Compute the volume of the parallelopiped determined by the vectors a = ( 4 , 3 , 1 ) , b = ( 3 , 3 , 3 ) , and c = ( 1 , 1 , 3 ) . 1. volume = 8 2. volume = 7 3. volume = 5 4. volume = 9 5. volume = 6 correct Explanation: For the parallelopiped determined by vec- tors a , b , and c its volume = | a ( b c ) | . But a ( b c ) = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 4 3 1 3 3 3 1 1 3 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = 4 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 3 3 1 3 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle + 3 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 3 3 1 3 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 3 3 1 1 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle . Consequently, volume = 6 . keywords: determinant, cross product, scalar triple product, parallelopiped, volume, 003 0.0 points Determine all unit vectors v orthogonal to a = i + 4 j + 3 k , b = 2 i + 6 j + 3 k . 1. v = parenleftBig 3 7 i 6 7 j 2 7 k parenrightBig 2. v = 6 i + 3 j 2 k 3. v = 3 7 i + 6 7 j 2 7 k 4. v = 6 7 i + 3 7 j 2 7 k 5. v = parenleftBig 6 7 i 3 7 j + 2 7 k parenrightBig correct rodriguez (lar2636) Exam 3 Practice rodin (54565) 2 6. v = 3 i + 6 j + 2 k Explanation: The non-zero vectors orthogonal to a and b are all of the form v = ( a b ) , negationslash = 0 , with a scalar. The only unit vectors orthog- onal to a , b are thus v = a b | a b | . But for the given vectors a and b , a b = vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle i j k 1 4 3 2 6 3 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle vextendsingle 4 3 6 3 vextendsingle vextendsingle vextendsingle vextendsingle i vextendsingle vextendsingle vextendsingle vextendsingle...
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Exam 3 Practice-solutions - rodriguez (lar2636) Exam 3...

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