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Unformatted text preview: lawrence (cdl678) – Homework 7 – ODELL – (56280) 1 This printout should have 23 questions. Multiplechoice 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 welldefined because it is the product of two scalars. II. is not welldefined because the second term in the dot product is not a vector. III. is not welldefined 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.
 Spring '10
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