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 10.4, Problem 28E
To determine

To find: The general form of the equation of the plane passing through the point and perpendicular the specified vector.

Expert Solution & Answer
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Answer to Problem 28E

The general form of the equation of the plane passing through the point and perpendicular the specified vector is −x+y−2z+12=0 .

Explanation of Solution

Given information:

The point is (0,0,6) which perpendicular to x=1−t , y=2+t , z=4−2t .

Calculation:

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=〈−1,1,−2〉 .

Substitute −1 for a , 1 for b , −2 for c , 0 for x1 , 0 for y1 and 6 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.

  −1(x−0)+1(y−0)−2(z−6)=0−x+y−2z+12=0

Therefore, the general form of the equation of the plane passing through the point and perpendicular the specified vector is −x+y−2z+12=0 .

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Chapter 10.4, Problem 27E
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Chapter 10.4, Problem 29E

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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