13.1Problems 32-48

13.1Problems 32-48 - 302 C H A P T E R 13 V E C T O R G E O...

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302 CHAPTER 13 VECTOR GEOMETRY (ET CHAPTER 12) 32. A = ( 5 , 8 ) , B = ( 2 , 2 ) , P = ( 2 , 2 ) , Q = ( 3 , 8 ) SOLUTION Computing −→ AB and PQ gives: =h 2 5 , 2 8 i=h− 3 6 i =h− 3 2 , 8 2 5 , 6 i The vectors are not scalar multiples of each other, hence they are not parallel. In Exercises 33–36, let R = ( 2 , 7 ) . Calculate the following. 33. The length of OR Since 2 , 7 i , the length of the vector is k k= q ( 2 ) 2 + 7 2 = 53. 34. The components of u = PR ,where P = ( 1 , 2 ) We compute the components of the vector to obtain: u = 2 1 , 7 2 3 , 5 i 35. The point P such that has components h− 2 , 7 i Denoting P = ( x 0 , y 0 ) we have: 2 x 0 , 7 y 0 2 , 7 i Equating corresponding components yields: 2 x 0 =− 2 7 y 0 = 7 x 0 = 0 , y 0 = 0 P = ( 0 , 0 ) 36. The point Q such that RQ has components h 8 , 3 i We denote Q = ( x 0 , y 0 ) and have: x 0 ( 2 ), y 0 7 i=h x 0 + 2 , y 0 7 8 , 3 i Equating the corresponding components of the two vectors yields: x 0 + 2 = 8 y 0 7 3 x 0 = 6 , y 0 = 4 Q = ( 6 , 4 ) In Exercises 37–44, Fnd the given vector. 37. Unit vector e v where v = h 3 , 4 i The unit vector e v is the following vector: e v = 1 k v k v We Fnd the length of v 3 , 4 i : k v p 3 2 + 4 2 = 25 = 5 Thus e v = 1 5 h 3 , 4 i= ¿ 3 5 , 4 5 À . 38. Unit vector e w where w = h 21 , 20 i We compute the length of w 21 , 20 i : k w q 21 2 + ( 20 ) 2 = 841 = 29 The unit vector e w is, therefore: e w = 1 k w k · w w = 1 29 h 21 , 20 ¿ 21 29 , 20 29 À
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SECTION 13.1 Vectors in the Plane (ET Section 12.1) 303 39. Unit vector in the direction of u = h− 1 , 1 i
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This note was uploaded on 02/13/2010 for the course MATH MATH 32A taught by Professor Park during the Fall '09 term at UCLA.

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13.1Problems 32-48 - 302 C H A P T E R 13 V E C T O R G E O...

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