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Chapter 12.5, Problem 26E
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### Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

#### Solutions

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### Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

# Find an equation of the plane.26. The plane through the point (2, 0, 1) and perpendicular to the line x = 3t, y = 2 − t, z = 3 + 4t

To determine

To find: An equation of the plane that passes through the point (2,0,1) and perpendicular to the line x=3t,y=2t,z=3+4t .

Explanation

The parametric equations of the line are written as follows.

x=3t,â€‰y=2âˆ’t,â€‰z=3+4t

Rewrite the parametric equations of the line as follows.

x=0+3t,â€‰y=2âˆ’t,â€‰z=3+4t (1)

Formula:

Write the expression to find equation of the plane through the point P0(x0,â€‰y0,â€‰z0) and with the normal vector n=âŒ©a,â€‰b,â€‰câŒª as follows.

a(xâˆ’x0)+b(yâˆ’y0)+c(zâˆ’z0)=0 (2)

Write the expressions for the parametric equations for a line through the point (x0,â€‰y0,â€‰z0) and parallel to the direction vector âŒ©a,â€‰b,â€‰câŒª .

x=x0+at,â€‰y=y0+bt,â€‰z=z0+ct (3)

Compare equation (3) with (1) and writ the direction vector of the line (v) .

v=âŒ©3,â€‰âˆ’1,â€‰4âŒª

As the plane is perpendicular to the line, the direction of line âŒ©3,â€‰âˆ’1,â€‰4âŒª is considered as normal vector to the plane

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Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th