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Find the equation for the line passing through P = (2, -2, 1) and perpendicular to the plane -x + y – 3z = 3. Note that the correct answer below can be either in parametric or symmetric form. О А. (2t, — 2t, t) о в. (2 + t, -2 - t,1 + 3t) x +1 y - 1 z+3 -2 1 y+2 %3D z - 1 -3 O E. -x + y – 3z = -6

Find the equation for the line passing through P = (2, -2, 1) and perpendicular to the plane -x + y – 3z = 3. Note that the correct answer below can be either in parametric or symmetric form. О А. (2t, — 2t, t) о в. (2 + t, -2 - t,1 + 3t) x +1 y - 1 z+3 -2 1 y+2 %3D z - 1 -3 O E. -x + y – 3z = -6

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 the equation for the line passing through P= (2, –2,1) and perpendicular to the plane -x + y – 3z = 3. Note that the correct answer below can be either in parametric or symmetric form. О А. (2t, — 2t, t) Free-form Snip о в. (2 + t, —2 — t,1+ 3t) x +1 с. 2 у — 1 z +3 -2 1 2 y +2 z – 1 D. -1 -3 O E. -x + y – 3z = -6
Transcribed Image Text:Find the equation for the line passing through P= (2, –2,1) and perpendicular to the plane -x + y – 3z = 3. Note that the correct answer below can be either in parametric or symmetric form. О А. (2t, — 2t, t) Free-form Snip о в. (2 + t, —2 — t,1+ 3t) x +1 с. 2 у — 1 z +3 -2 1 2 y +2 z – 1 D. -1 -3 O E. -x + y – 3z = -6
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