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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 ( 0 , 0 , 0 ) , ( 1 , 2 , 3 ) , ( − 2 , 3 , 3 ) .

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

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 29E

To determine

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

Expert Solution & Answer

Answer to Problem 29E

The general form of the equation of the plane passing through the three is −3x+9y+7z=0 .

Explanation of Solution

Given information:

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

Calculation:

Find the vector u and v as follows.

u=〈1−0,2−0,3−0〉=〈1,2,3〉

and,

v=〈−2−0,3−0,3−0〉=〈−2,3,3〉

Find the normal vector as follows.

u×v=|ijk123−233|=−3i+9j+7k=〈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=〈−3,9,7〉 .

Substitute −3 for a , 9 for b , 7 for c , 0 for x1 , 0 for y1 and 0 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.

−3(x−0)+9(y−0)+7(z−0)=0−3x+9y+7z=0

Therefore, the general form of the equation of the plane passing through the three is −3x+9y+7z=0 .

Chapter 10.4, Problem 28E

Chapter 10.4, Problem 30E

Chapter 10 Solutions

Precalculus with Limits: A Graphing Approach

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