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What is the equation to the plane through P(4,-2,3) and parallel to 3x-7z=12
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I guess just a quick run though of any ideas how to get there in english would help... Any help helps!
I get the normal vector <3,0,7> and know it runs perpendicular to the plane...
I can use the point to make a parametric equation... x=4+3t y=-2 z=3+7t
parallel to a y=0 plane eh ...
3x - 7z = 12 <3,0,-7> is the normal to this "plane" so its good for the normal for any plane parallel to it right?
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given a normal and a point, all we do is put them together to get the plane equation. 3(x-4)+0(y--2)-7(z-3)=0 3x -12 -7z +21 = 0 3x -7z +9 = 0 should do it
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