calc hw 7 - lawrence (cdl678) – Homework 7 – ODELL –...

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Unformatted text preview: lawrence (cdl678) – Homework 7 – ODELL – (56280) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Which of the following expressions are well- defined for all vectors a , b , and c ? I. | a | ( b · c ) , II. ( a + b ) · | c | , III. ( a · b ) · c . 1. II only 2. none of them 3. II and III only 4. I only correct 5. I and II only 6. I and III only 7. all of them 8. III only Explanation: The dot product is defined only for two vectors, and its value is a scalar. On the other hand, the length of any vector is always a scalar. For the three given expressions, therefore, we see that I. is well-defined because it is the product of two scalars. II. is not well-defined because the second term in the dot product is not a vector. III. is not well-defined because the first term in the dot product with c is not a vector. keywords: vectors, dot product, T/F, length, TrueFalse, 002 10.0 points Which of the following statements are true for all vectors a , b ? A. | a · b | = | a || b | = ⇒ a bardbl b , B. a · b = 0 = ⇒ a = 0 or b = 0, C. | a + b | 2 = | a | 2 + 2 a · b + | b | 2 . 1. all of them 2. A and B only 3. A only 4. none of them 5. B only 6. B and C only 7. A and C only correct 8. C only Explanation: If θ is the angle between a and b , then a · b = | a || b | cos θ . A. TRUE: since | a · b | = | a || b | = ⇒ | cos θ | = 1 , it follows that θ = 0 or π , in which case a bardbl b . B. FALSE: if a ⊥ b , then θ = π/ 2. But then cos θ = 0. So a · b = 0 when a ⊥ b , as well as when a = 0 or b = 0. C. TRUE: since | a | 2 = a · a , | a + b | 2 = ( a + b ) · ( a + b ) = | a | 2 + a · b + b · a + | b | 2 = | a | 2 + 2 a · b + | b | 2 because a · b = b · a . keywords: lawrence (cdl678) – Homework 7 – ODELL – (56280) 2 003 10.0 points For which positive value of x are the vectors (− 24 x, 2 , 1 ) , ( 2 , 5 x 2 , − 10 ) orthogonal? Correct answer: 5. Explanation: Since (− 24 x, 2 , 1 ) · ( 2 , 5 x 2 , − 10 ) = 10 x 2 − 48 x − 10 , the vectors will be orthogonal when 2(5 x 2 − 24 x − 5) = 2(5 x + 1)( x − 5) = 0 . Consequently, the only positive value of x for which this occurs is x = 5 . 004 10.0 points Find the scalar projection of b onto a when b = 2 i + 2 j − k , a = 2 i − j − 2 k . 1. scalar projection = 2 3 2. scalar projection = 1 3 3. scalar projection = 0 4. scalar projection = 1 5. scalar projection = 4 3 correct Explanation: The scalar projection of b onto a is given in terms of the dot product by comp a b = a · b | a | . Now when b = 2 i + 2 j − k , a = 2 i − j − 2 k , we see that a · b = 4 , | a | = radicalBig (2) 2 + ( − 1) 2 + ( − 2) 2 ....
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This note was uploaded on 04/09/2010 for the course M 56280 taught by Professor Odell during the Spring '10 term at University of Texas.

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calc hw 7 - lawrence (cdl678) – Homework 7 – ODELL –...

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