Engineering Calculus Notes 50

Engineering Calculus Notes 50 - P 1 and P 2 is given by the...

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38 CHAPTER 1. COORDINATES AND VECTORS The vector −−−→ P 1 P 2 joining them lies along , so we can use it as a direction vector: −→ v = −→ p 2 −→ p 1 = x −→ ı + y −→ + z −→ k where x = x 2 x 1 , y = y 2 y 1 , and z = z 2 z 1 . Using P 1 as basepoint, this leads to the parametrization −→ p ( t ) = −→ p 1 + t −→ v = −→ p 1 + t ( −→ p 2 −→ p 1 ) = (1 t ) −→ p 1 + t −→ p 2 . Note that we have set this up so that −→ p (0) = −→ p 1 −→ p (1) = −→ p 2 . The full line through these points corresponds to allowing the parameter to take on all real values. However, if we restrict t to the interval 0 t 1, the corresponding points Fll out the line segment P 1 P 2 . Since the point −→ p ( t ) is travelling with constant velocity, we have the following observations: Remark 1.3.1 (Two-Point ±ormula) . Suppose P 1 ( x 1 ,y 1 ,z 1 ) and P 2 ( x 2 ,y 2 ,z 2 ) are distinct points. The line through
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Unformatted text preview: P 1 and P 2 is given by the parametrization 7 p ( t ) = (1 t ) p 1 + t p 2 with coordinates x = (1 t ) x 1 + tx 2 y = (1 t ) y 1 + ty 2 z = (1 t ) z 1 + tz 2 . The line segment P 1 P 2 consists of the points for which t 1 . The value of t gives the portion of P 1 P 2 represented by the segment P 1 p ( t ) ; in particular, the midpoint of P 1 P 2 has position vector 1 2 ( p 1 + p 2 ) = p 1 2 ( x 1 + x 2 ) , 1 2 ( y 1 + y 2 ) , 1 2 ( z 1 + z 2 ) P . 7 This is the parametrized form of the two-point formula for a line in the plane determined by a pair of points....
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This note was uploaded on 10/20/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

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