42-54in13.3

# 42-54in13.3 - 334 C H A P T E R 13 V E C T O R G E O M E T...

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334 CHAPTER 13 VECTOR GEOMETRY (ET CHAPTER 12) 42. u = h 2 , 0 i , v = h 4 , 3 i SOLUTION We compute the dot products: u · v =h 2 , 0 i·h 4 , 3 i= 2 · 4 + 0 · 3 = 8 v · v =k v k 2 = 4 2 + 3 2 = 25 The projection of u along v is the following vector: proj v ( u ) = ³ u · v v · v ´ v = 8 25 h 4 , 3 ¿ 32 25 , 24 25 À . 43. u = h 1 , 1 , 1 i , v = h 1 , 1 , 0 i We Frst compute the following dot products: u · v 1 , 1 , 1 1 , 1 , 0 1 · 1 + 1 · 1 + 1 · 0 = 2 v · v v k 2 = 1 2 + 1 2 + 0 2 = 2 The projection of u along v is the following vector: proj v ( u ) = ³ u · v v · v ´ v = 2 2 v = v 1 , 1 , 0 i 44. u = h 3 , 2 , 1 i , v = h 6 , 4 , 2 i To Fnd the projection proj v ( u ) = ( u · v v · v ) v , we need to compute the dot products: u · v 3 , 2 , 1 6 , 4 , 2 3 · 6 + 2 · 4 + 1 · 2 = 28 v · v v k 2 = 6 2 + 4 2 + 2 2 = 56 Hence, proj v ( u ) = 28 56 h 6 , 4 , 2 i=h 3 , 2 , 1 i . 45. u = 5 i + 7 j 4 k , v = k The projection of u along v is the following vector: proj v ( u ) = ³ u · v v · v ´ v We compute the dot products: u · v = ( 5 i + 7 j 4 k ) · k =− 4 k · k 4 v · v v k 2 k k 2 = 1 Hence, proj v ( u ) = 4 1 k 4 k 46. u = i + 29 k , v = j Since u · v = ( i + 29 k ) · j = i · j + 29 k · j = 0, u is orthogonal to v , the projection of u along v is the zero vector proj v ( u ) = 0 47. u = h a , b , c i , v = i The component of u along v is a

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## This homework help was uploaded on 04/22/2008 for the course MATH 32A taught by Professor Gangliu during the Winter '08 term at UCLA.

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42-54in13.3 - 334 C H A P T E R 13 V E C T O R G E O M E T...

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