Calculus Notes 13.5

# Calculus Notes 13.5 - Calculus-Stewart Dr. Berg Summer 09...

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Calculus- Stewart Dr. Berg Summer ‘09 Page 1 13.5 13.5 Equations of Lines and Planes Vector Parameterization of a Line A line is determined by two points, say P and Q. Let r 0 = OP and d = PQ ; and let r ( t ) = OX where X is an arbitrary point on the line. Notice that r ( t ) r 0 = t d for some scalar t , since r ( t ) r 0 and d are parallel. Definition The line through r 0 = x 0 i + y 0 j + z 0 k in the direction of d = d 1 i + d 2 j + d 3 k is parameterized by r ( t ) = t d + r 0 = t ( d 1 i + d 2 j + d 3 k ) + ( x 0 i + y 0 j + z 0 k ) = ( td 1 + x 0 ) i + ( td 2 + y 0 ) j + ( td 3 + z 0 ) k . The scalar parametric equations are x ( t ) = td 1 + x 0 , y ( t ) = td 2 + y 0 , and z ( t ) = td 3 + z 0 . The symmetric form is x x 0 d 1 = y y 0 d 2 = z z 0 d 3 for d i 0 . Note: 1) This is merely the simplest parameterization. Any real valued function can replace t to parameterize all or part of the line. 2) To get the symmetric form, solve each scalar parametric equation for t . Example A The line through the points P (1,0, 2) and Q (1,2,3) is the line through r 0 = OP = i 2 k in the direction of d = PQ = ( i + 2 j + 3 k ) ( i 2 k ) = 2 j + 5 k . Its parameterization is r ( t ) = t d + r 0 = t (2 j + 5 k ) + ( i 2 k ) . The scalar parametric equations are x ( t ) = 1 , y ( t ) = 2 t , and z ( t ) = 5 t

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## This note was uploaded on 06/06/2010 for the course M 408 taught by Professor Hodges during the Spring '08 term at University of Texas.

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Calculus Notes 13.5 - Calculus-Stewart Dr. Berg Summer 09...

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