Exam 3 Practice-solutions

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 θ = 16sin θ where 0 ≤ θ ≤ π is the angle between u and v . Now j lies in the yz-plane, 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 non-zero 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...
View Full Document

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.

Page1 / 25

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online