I\'m having trouble working through a problem involving planes and a point. The

Question

I'm having trouble working through a problem involving planes and a point. The question is as follows:

Find the equation of a plane through (1,2,-1) Perpendicular to each of the planes:

2x-y+z = 0
x+3y-4z=1

The best I can figure is finding the normal vector of the two planes that both contain the point given and then calculate the equation of the plane created by those two normal vectors. Am I on the right track? I'm still not sure how to execute this even if I am, but I'm working on it...

Any help is greatly appreciated!

Explanation / Answer

You are on the right track.

Cross product: normal vector for plane1 = <2,-1,1> X normal vector for plane2 <1,3,-4>.

 i    j    k
2  -1   1
1   3  -4                  (4-3)i -(-8-1)j + (6-(-1))k = <1,9,7> is a vector orthogonal to both planes.


Now you have a direction vector <1,9,7> and a point (1,2,-1).

a(x-xi) + b(y-yi) + c(z-zi) = 0

a = 1, b=9, c=7 & xi=1, yi=2, zi=3.

1(x-1) + 9(y-2) + 7(z-3) = 0

simplifing give you:

x-1+9y-18+7z-21=0

answer  x+9y+7z= 40

This is an equation of the plane which is perpendicular to the planes and containing the point.

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