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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 ( 4 , − 1 , 3 ) , ( 2 , 5 , 1 ) , ( − 1 , 2 , 1 ) .

The general form of the equation of the plane passing through the points ( 4 , − 1 , 3 ) , ( 2 , 5 , 1 ) , ( − 1 , 2 , 1 ) .

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 30E

To determine

To find: The general form of the equation of the plane passing through the points (4,−1,3) , (2,5,1) , (−1,2,1) .

Expert Solution & Answer

Answer to Problem 30E

The general form of the equation of the plane passing through the three is −x+y+4z−7=0 .

Explanation of Solution

Given information:

The points are (4,−1,3) , (2,5,1) , (−1,2,1) .

Calculation:

Find the vector u and v as follows.

u=〈2−4,5+1,1−3〉=〈−2,6,−2〉

and,

v=〈−1−4,2+1,1−3〉=〈−5,3,−2〉

Find the normal vector as follows.

u×v=|ijk−26−2−53−2|=−6i+6j+24k=〈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=〈−6,6,24〉 .

Substitute −6 for a , 6 for b , 24 for c , 4 for x1 , −1 for y1 and 3 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.

−6(x−4)+6(y+1)+24(z−3)=0−6x+24+6y+6+24z−72=0−6x+6y+24z−42=0−x+y+4z−7=0

Therefore, the general form of the equation of the plane passing through the three is −x+y+4z−7=0 .

Chapter 10.4, Problem 29E

Chapter 10.4, Problem 31E

Chapter 10 Solutions

Precalculus with Limits: A Graphing Approach

Ch. 10.1 - Prob. 1E Ch. 10.1 - Prob. 2E Ch. 10.1 - Prob. 3E Ch. 10.1 - Prob. 4E Ch. 10.1 - Prob. 5E Ch. 10.1 - Prob. 6E Ch. 10.1 - Prob. 7E Ch. 10.1 - Prob. 8E Ch. 10.1 - Prob. 9E Ch. 10.1 - Prob. 10E

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