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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Chapter 10, Problem 15CT
To determine

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

Expert Solution & Answer
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Answer to Problem 15CT

The general equation of the plane passing through the points (−3,−4,2) , (−3,4,1) and (1,1,−2) is −27(x+3)−4(y+4)−32(z−2)=0 .

Explanation of Solution

Given information:

The passing points of the plane are (−3,−4,2) , (−3,4,1) and (1,1,−2) .

Calculation:

First find the component form of the vector u with initial point (−3,−4,2) and terminal point (−3,4,1) .

  u=〈−3−(−3),4−(−4),1−2〉=〈0,8,−1〉

Now find the component form of the vector v with initial point (−3,4,1) and terminal point (1,1,−2) .

  v=〈1−(−3),1−4,−2−1〉=〈4,−3,−3〉

Use the cross product of vectors u and v for the normal vector to the plane.

  u×v=|ijk08−14−3−3|=i(8×−3−(−3)×−1)−j(0×−3−4×−1)+k(0×−3−4×8)=i(−24−3)−j(0+4)+k(0−32)=−27i−4j−32k

Use the general equation of the plane.

  a(x−x1)+b(y−y1)+c(z−z1)=0

Substitute −27 , −4 , −32 for a , b , c and −3 , −4 , 2 for x1 , y1 , z1 respectively in the equation of plane.

  a(x−x1)+b(y−y1)+c(z−z1)=0−27(x−(−3))+(−4)(y−(−4))+(−32)(z−2)=0−27(x+3)−4(y+4)−32(z−2)=0

Therefore, the general equation of the plane passing through the points (−3,−4,2) , (−3,4,1) and (1,1,−2) is −27(x+3)−4(y+4)−32(z−2)=0 .

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Chapter 10, Problem 14CT
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Chapter 10, Problem 16CT

Chapter 10 Solutions

Precalculus with Limits: A Graphing Approach

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