Planes 1 & 2 Intersect: Find Scalar Equations

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SUMMARY

The discussion focuses on finding the scalar equations of two intersecting planes given their symmetric equation and specific points. The symmetric equation of the line of intersection is (x-1)/2 = (y-2)/3 = (z+4)/1. Plane 1 contains the point A(2,1,1), while Plane 2 contains the point B(1,2,-1). To derive the scalar equations, three points in each plane must be identified, starting with the given points and using the symmetric equation to find additional points.

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Two planes, plane 1 and plane 2, intersect in the line with symmetric equation (x-1)/2 = (y-2)/3 = (z+4)/1. Plane 1 contains the point A(2,1,1) and plane 2 contains the point B(1,2,-1). Find the scalar equations of planes plane 1 and plane 2.

I have no idea how to do it, all help will be appreciated.
 
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jessicajx22 said:
Two planes, plane 1 and plane 2, intersect in the line with symmetric equation (x-1)/2 = (y-2)/3 = (z+4)/1. Plane 1 contains the point A(2,1,1) and plane 2 contains the point B(1,2,-1). Find the scalar equations of planes plane 1 and plane 2.

I have no idea how to do it, all help will be appreciated.

Hi Jessica, welcome to MHB!

A plane can be determined by 3 points that are in the plane.
So let's see if we can find 3 such points.

Obviously plane 1 contains point A(2,1,1).
So we need to use (x-1)/2 = (y-2)/3 = (z+4)/1 to find 2 more points.
Suppose each of them is 0. Then we must have x=1, y=2, z=-4. That is because for instance (1-1)/2=0.
Alternatively, if each of them is 1, then we must have x=3, y=5, z=-3, don't we?

Now we have 3 points in plane 1.
Do you already know what a scalar equation of a plane is?
And perhaps how to find it based on 3 points?
 

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