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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 C H A P T E R 13 VECTOR GEOMETRY (ET CHAPTER 12) 32. A = ( 5 , 8 ) , B = ( 2 , 2 ) , P = ( 2 , 2 ) , Q = ( 3 , 8 ) SOLUTION Computing −→ AB and −→ P Q gives: −→ AB = 2 5 , 2 8 = − 3 6 −→ P Q = − 3 2 , 8 2 = − 5 , 6 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 −→ O R SOLUTION Since −→ O R = − 2 , 7 , the length of the vector is −→ O R = ( 2 ) 2 + 7 2 = 53. 34. The components of u = −→ P R , where P = ( 1 , 2 ) SOLUTION We compute the components of the vector to obtain: u = −→ P R = 2 1 , 7 2 = − 3 , 5 35. The point P such that −→ P R has components 2 , 7 SOLUTION Denoting P = ( x 0 , y 0 ) we have: −→ P R = − 2 x 0 , 7 y 0 = − 2 , 7 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 8 , 3 SOLUTION We denote Q = ( x 0 , y 0 ) and have: −→ RQ = x 0 ( 2 ), y 0 7 = x 0 + 2 , y 0 7 = 8 , 3 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, find the given vector. 37. Unit vector e v where v = 3 , 4 SOLUTION The unit vector e v is the following vector: e v = 1 v v We find the length of v = 3 , 4 : v = 3 2 + 4 2 = 25 = 5 Thus e v = 1 5 3 , 4 = 3 5 , 4 5 .
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