Let P = (0, 1, 5) and Q = (5, -1, 3) be the given points. The set of all points equidistant from P and Q lies on the perpendicular bisector of the line segment PQ.
To find the equation of the perpendicular bisector, we need to find its midpoint and its direction.
The midpoint M of PQ is given by the average of the coordinates of P and Q:
M = ((0+5)/2, (1+(-1))/2, (5+3)/2) = (2.5, 0, 4)
Now, let's find the direction vector of PQ. The direction vector is obtained by subtracting the coordinates of P from Q:
d = Q - P = (5, -1, 3) - (0, 1, 5) = (5, -2, -2)
To find the direction vector of the perpendicular bisector, we take the perpendicular vector to d. We can do this by swapping the x and y coordinates and changing the sign of one of them.
So, the direction vector of the perpendicular bisector is (-2, 5, -2).
Finally, using the coordinates of the midpoint M and the direction vector (-2, 5, -2), we can write the equation of the plane as:
-2(x - 2.5) + 5(y - 0) - 2(z - 4) = 0
which simplifies to:
-2x + 5y - 2z + 3 = 0
Therefore, the equation of the plane containing all the points equidistant from P and Q is -2x + 5y - 2z + 3 = 0.
The set of all points equidistant from (0, 1, 5) and 5, -1, 3) is a plane. Find the equation.
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