Find the equation of the plane satisfying the given condition: Passing trough (1,1,1) and 2,0,3) and perpendicular to the plane x+2y-3z=0
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To get the equation of a plane, you want the vector normal (perpendicular to) the plane. So that is the first thing we want to try to find: a vector normal to our plane.
If we had two vectors that were in our plane (and were not multiples of each other), then we could take the cross product of the vectors, which would give us a vector normal to our plane. So that is what we are looking for: two vectors (not multiples of each other) that lie in our plane.
The points (1,1,1) and (2,0,3) are in the plane, so the vector that "connects" them must lie in the plane. Thus, the vector <2-1, 0-1, 3-1> =<1,-1,2> is in the plane.
The plane x+2y-3z=0 has normal vector <1,2,-3>. This vector must also lie on our plane (since the two planes are perpendicular)
Thus, we have our two vectors:
<1,2,-3> and <1,-1,2> lie in our plane, so their cross-product:
<1,-5,-3> is normal to the plane.
Thus, we want the plane that is normal to <1,-5,-3> and goes through the point (1,1,1).
The formula for the plane is:
1(x-1)-5(y-1)-3(z-1)=0
which simplifies to:
x-5y-3z=-7
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