Determine the point in Z-X plane which is equidistant from the points (1,-1,0) , (2,1,2), (3,2,-1)
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A general approach to such a problem would be as follows:
Let us find the plane whose points are equidistant from these two points and whose normal is the vector connecting these two points. (This plane is the locus of all points equidistant from A and B). The plane has as its normal:
page 1
and the plane passes through the midpoint of line connecting A and B. The point is given by
page 2
Thus the equation of plane is: 6x−2y=66x−2y=6
Let us find the plane whose points are equidistant from these two points and whose normal is the vector connecting these two points. (This plane is the locus of all points equidistant from A and B). The plane has as its normal:
page 1
and the plane passes through the midpoint of line connecting A and B. The point is given by
page 2
Thus the equation of plane is: 6x−2y=66x−2y=6
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