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High School Math Calculus Precalculus with Limits: A Graphing Approach

The general form of the equation of the plane passing through the points ( 2 , 3 , − 2 ) , ( 3 , 4 , 2 ) , ( 1 , − 1 , 0 ) .

The general form of the equation of the plane passing through the points ( 2 , 3 , − 2 ) , ( 3 , 4 , 2 ) , ( 1 , − 1 , 0 ) .

Precalculus with Limits: A Graphing Approach

Precalculus with Limits: A Graphing Approach

7th Edition

ISBN: 9781305071711

Author: Ron Larson

Publisher: Brooks/Cole

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Question

Chapter 10.4, Problem 31E

To determine

To find: The general form of the equation of the plane passing through the points (2,3,−2) , (3,4,2) , (1,−1,0) .

Expert Solution & Answer

Answer to Problem 31E

The general form of the equation of the plane passing through the three is 6x−2y−z−8=0 .

Explanation of Solution

Given information:

The points are (2,3,−2) , (3,4,2) , (1,−1,0) .

Calculation:

Find the vector u and v as follows.

u=〈3−2,4−3,2+2〉=〈1,1,4〉

and,

v=〈1−2,−1−3,0+2〉=〈−1,−4,2〉

Find the normal vector as follows.

u×v=|ijk114−1−42|=18i−6j−3k=〈a,b,c〉

The standard equation of the plane is given by a(x−x1)+b(y−y1)+c(z−z1)=0 , where n=〈a,b,c〉 is the nonzero normal vector. So, the normal vector is n=〈18,−6,−3〉 .

Substitute 18 for a , −6 for b , −3 for c , 2 for x1 , 3 for y1 and −2 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 passing through the point and perpendicular the specified vector.

18(x−2)−6(y−3)−3(z+2)=018x−36−6y+18−3z−6=06x−12−2y+6−z−2=06x−2y−z−8=0

Therefore, the general form of the equation of the plane passing through the three is 6x−2y−z−8=0 .

Chapter 10.4, Problem 30E

Chapter 10.4, Problem 32E

Chapter 10 Solutions

Precalculus with Limits: A Graphing Approach

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