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9. Find a parametric equation for the line that is orthogonal to the surface 2 at the point (1, 1, 2). Xyz

9. Find a parametric equation for the line that is orthogonal to the surface 2 at the point (1, 1, 2). Xyz

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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Can you please explain each step of the solution especially the part I circled. Confused on how to get that answer. 

9. Find a parametric equation for the line that is orthogonal to the surface xyz = 2 at the point (1, 1, 2). Solution. The normal vector to the surface f(x, y, z) = 2 with f (x, y, z) = xyz is Vf(7, y, 2) yzi + α2j + xyk. Thus, the normal vector at (x, y, z) = (1, 1,2) is i = Vf(1,1, 2) = 2i + 2j + k. The normal vector n (2, 2, 1) is a direction vector of the line, so a parametric equation of the line is (x, y, z) = (1, 1, 2) + t(2,2, 1), or x = 1+ 2t, y = 1+ 2t, z = 2+t.
Transcribed Image Text:9. Find a parametric equation for the line that is orthogonal to the surface xyz = 2 at the point (1, 1, 2). Solution. The normal vector to the surface f(x, y, z) = 2 with f (x, y, z) = xyz is Vf(7, y, 2) yzi + α2j + xyk. Thus, the normal vector at (x, y, z) = (1, 1,2) is i = Vf(1,1, 2) = 2i + 2j + k. The normal vector n (2, 2, 1) is a direction vector of the line, so a parametric equation of the line is (x, y, z) = (1, 1, 2) + t(2,2, 1), or x = 1+ 2t, y = 1+ 2t, z = 2+t.
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