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ector equation for this line is r(t) = (5, -4, 3) + t1, 4, - ½)

ector equation for this line is r(t) = (5, -4, 3) + t1, 4, - ½)

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Find a vector equation and parametric equations for the line. (Use the parameter t.) The line through the point (5, -4, 3) and parallel to the vector Step 1 Recall that if a line is parallel to the vector v and passes through the point Po, which is the tip of the position vector ro, then the vector equation of the line is given by r(t) = ro + tv. For the given line, we have ro = (5, -4, 3) and v = -3). So the vector equation for this line is ---) 4, - r(t) = (5,-4, 3) + 1, 4, -
Transcribed Image Text:Find a vector equation and parametric equations for the line. (Use the parameter t.) The line through the point (5, -4, 3) and parallel to the vector Step 1 Recall that if a line is parallel to the vector v and passes through the point Po, which is the tip of the position vector ro, then the vector equation of the line is given by r(t) = ro + tv. For the given line, we have ro = (5, -4, 3) and v = -3). So the vector equation for this line is ---) 4, - r(t) = (5,-4, 3) + 1, 4, -
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