54-60in13.2

54-60in13.2 - 324 C H A P T E R 13 V E C T O R G E O M E T...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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 .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

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.

Page1 / 3

54-60in13.2 - 324 C H A P T E R 13 V E C T O R G E O M E T...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online