"Find an equation for the set of points P(x, y, z) that are equidistant from the points A(1, 2, 3) and B(4, 0, 1)."
Can some one help?
Wouldn't that be the plane passing through the midpoint of AB and having as its normal the direction vector of AB ??
midpoint of AB = (5/2, 1, 2)
direction vector AB = (3,-2,-2)
the the plane has equation
3x - 2y - 2z + C = 0
but (5/2,1,2) lies on it, so
15/2 - 2 - 4 + C = 0
C = -3/2
so 3x - 2y - 2z - 3/2 = 0 or
6x - 4y - 4z - 3 = 0
To find an equation for the set of points that are equidistant from points A(1, 2, 3) and B(4, 0, 1), we can use the concept of the midpoint formula.
The midpoint formula states that the midpoint M of two points can be found by taking the average of their x, y, and z coordinates individually.
Using this formula, we can find the coordinates of the midpoint M between points A and B:
Midpoint M(x, y, z) = ((x1 + x2)/2, (y1 + y2)/2, (z1 + z2)/2)
In this case, A(1, 2, 3) and B(4, 0, 1):
M(x, y, z) = ((1 + 4)/2, (2 + 0)/2, (3 + 1)/2)
= ((5/2), (2/2), (4/2))
= (2.5, 1, 2)
Now, we know that the set of points that are equidistant from A and B will be the locus of points that are equidistant from the midpoint M(2.5, 1, 2).
To represent this mathematically, we can use the distance formula between a general point P(x, y, z) and the midpoint M(2.5, 1, 2):
Distance Formula:
√((x - x₁)² + (y - y₁)² + (z - z₁)²) = √((x - 2.5)² + (y - 1)² + (z - 2)²)
Simplifying this equation will give us the equation for the set of points P(x, y, z) that are equidistant from points A and B.
√((x - 1)² + (y - 2)² + (z - 3)²) = √((x - 2.5)² + (y - 1)² + (z - 2)²)
This equation represents the set of points that are equidistant from points A(1, 2, 3) and B(4, 0, 1).