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High School Math Calculus PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

The general form of the equation of the plane passes through ( 2 , 2 , 1 ) and ( − 1 , 1 , − 1 ) and perpendicular to the 2 x − 3 y + z = 3 .

The general form of the equation of the plane passes through ( 2 , 2 , 1 ) and ( − 1 , 1 , − 1 ) and perpendicular to the 2 x − 3 y + z = 3 .

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

8th Edition

ISBN: 9781337904285

Author: Larson

Publisher: CENGAGE L

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Question

Chapter 10.4, Problem 35E

To determine

To find: The general form of the equation of the plane passes through (2,2,1) and (−1,1,−1) and perpendicular to the 2x−3y+z=3 .

Expert Solution & Answer

Answer to Problem 35E

The general form of the equation of the plane passes through (2,2,1) and (−1,1,−1) and perpendicular to the 2x−3y+z=3 is −7x+y−11z−5=0 .

Explanation of Solution

Given information:

The plane passes through (2,2,1) and (−1,1,−1) and it is perpendicular to the 2x−3y+z=3 .

Calculation:

Since the plane is perpendicular to the 2x−3y+z=3 then the normal vector to the plane will be parallel to the plane that can be obtained by taking the coefficients of x , y and z .

v=〈2,−3,1〉

Calculate the second vector u as follows.

u=〈−1−2,1−2,−1−1〉=〈−3,−1,−2〉

Calculate the cross product of the vector u and v to find the normal vector on the plane.

u×v=|ijk2−31−3−1−2|=7i+j−11k=〈a,b,c〉

Substitute 7 for a , 1 for b , −11 for c , 2 for x1 , 2 for y1 and 1 for z1 in the equation a(x−x1)+b(y−y1)+c(z−z1)=0 to find the general form of the equation of the plane passes through (2,2,1) and (−1,1,−1) and perpendicular to the 2x−3y+z=3 .

7(x−2)+1(y−2)−11(z−1)=07x−14+y−2−11z+11=0−7x+y−11z−5=0

Therefore, the general form of the equation of the plane passes through (2,2,1) and (−1,1,−1) and perpendicular to the 2x−3y+z=3 is −7x+y−11z−5=0 .

Chapter 10.4, Problem 34E

Chapter 10.4, Problem 36E

Chapter 10 Solutions

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

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