13.2.Ex56

# 13.2.Ex56 - 324 C H A P T E R 13 V E C T O R G E O M E T...

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324 CHAPTER 13 VECTOR GEOMETRY (ET CHAPTER 12) We use the formula for the midpoint of a segment to Fnd the coordinates of the points P and Q .Thisgives P = µ 1 + 0 2 , 0 + 1 2 , 1 + 1 2 = µ 1 2 , 1 2 , 1 Q = µ 1 + 0 2 , 1 + 1 2 , 0 + 1 2 = µ 1 2 , 1 , 1 2 Substituting in (1) yields the following vector: v = −→ PQ = ¿ 1 2 1 2 , 1 1 2 , 1 2 1 À = ¿ 0 , 1 2 , 1 2 À . 54. ±ind the components of the vector w whose tail is C and head is the midpoint of AB in ±igure 18. B = (1, 1, 0) C = (0, 1, 1) A = (1, 0, 1) y x z FIGURE 18 SOLUTION We denote the midpoint of by M = ( a , b , c ) . To Fnd the coordinates of M we Frst parametrize the line through A = ( 1 , 0 , 1 ) and B = ( 1 , 1 , 0 ) : r ( t ) = ( 1 t ) h 1 , 0 , 1 i+ t h 1 , 1 , 0 i=h 1 t , 0 , 1 t i+h t , t , 0 1 , t , 1 t i The midpoint of occurs at t = 1 2 , hence the vector OM is h 1 , 1 2 , 1 2 i . B = (1, 1, 0) C = (0, 1, 1) A = (1, 0, 1) y x z The point M is the terminal point of ,thatis, M = ³ 1 , 1 2 , 1 2 ´ . We now Fnd the vector w = CM : w = ¿ 1 0 , 1 2 1 , 1 2 1 À = ¿ 1 , 1 2 , 1 2 À . Further Insights and Challenges In Exercises 55–59, we consider the equations of a line in symmetric form .

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13.2.Ex56 - 324 C H A P T E R 13 V E C T O R G E O M E T...

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