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Unformatted text preview: rodriguez (lar2636) – Exam 3 Practice – rodin – (54565) 1 This printout should have 45 questions. Multiplechoice 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 yzplane. 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 θ = 16sin θ where 0 ≤ θ ≤ π is the angle between u and v . Now j lies in the yzplane, so the angle θ between j and v varies from 0 to π . 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 nonzero vectors orthogonal to a and b are all of the form v = λ ( a × b ) , λ negationslash = 0 , with λ 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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This note was uploaded on 11/17/2011 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas.
 Spring '07
 Holcombe

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