54-60in13.2 - 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 C H A P T E R 13 VECTOR GEOMETRY (ET CHAPTER 12) We use the formula for the midpoint of a segment to find the coordinates of the points P and Q . This gives P = 1 + 0 2 , 0 + 1 2 , 1 + 1 2 = 1 2 , 1 2 , 1 Q = 1 + 0 2 , 1 + 1 2 , 0 + 1 2 = 1 2 , 1 , 1 2 Substituting in (1) yields the following vector: v = −→ P Q = 1 2 1 2 , 1 1 2 , 1 2 1 = 0 , 1 2 , 1 2 . 54. Find the components of the vector w whose tail is C and head is the midpoint of AB in Figure 18. B = (1, 1, 0) C = (0, 1, 1) A = (1, 0, 1) y x z FIGURE 18 SOLUTION We denote the midpoint of AB by M = ( a , b , c ) . To find the coordinates of M we first parametrize the line through A = ( 1 , 0 , 1 ) and B = ( 1 , 1 , 0 ) : r ( t ) = ( 1 t ) 1 , 0 , 1 + t 1 , 1 , 0 = 1 t , 0 , 1 t + t , t , 0 = 1 , t , 1 t The midpoint of AB occurs at t = 1 2 , hence the vector OM is 1 , 1 2 , 1 2 . B = (1, 1, 0) C = (0, 1, 1) A = (1, 0, 1) y x z The point M is the terminal point of OM , that is, M = 1 , 1 2 , 1 2 . We now find the vector w = CM : w = 1 0 , 1 2 1 , 1 2 1 = 1 , 1 2 , 1 2 . Further Insights and Challenges In Exercises 55–59, we consider the equations of a line in symmetric form .
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