A smooth curve is normal to a surface f(x,y,z) = c at a point of intersection if the curve's velocity vector is a nonzero scalar multiple of Vf at the point. Show that the 7 curve r(t) = /ti+ vtj-7(t+3)k is normal to the surface x + y - 7z = 51 when t= 1.

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter11: Topics From Analytic Geometry
Section: Chapter Questions
Problem 18T
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A smooth curve is normal to a surface f(x,y,z) = c at a point of intersection if the curve's velocity vector is a nonzero scalar multiple of Vf at the point. Show that the
curve r(t) = Vti+ Vtj-
7
7(t + 3)k is normal to the surface x +y - 7z = 51 when t= 1.
Transcribed Image Text:A smooth curve is normal to a surface f(x,y,z) = c at a point of intersection if the curve's velocity vector is a nonzero scalar multiple of Vf at the point. Show that the curve r(t) = Vti+ Vtj- 7 7(t + 3)k is normal to the surface x +y - 7z = 51 when t= 1.
Expert Solution
Step 1

Given:

A smooth curve is normal to a surface fx,y,z=c at a point of intersectionif the curve's velocity vector is a non-zero scalar multiple of f at the point.

To Find:

The curve rt=ti+tj-74t+3k is normal to the surface x2+y2-7z=51when t=1.

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