Find an equation of the set of all points equidistant from the points $A(-2, 5, 3)$ and $B(5, 2, -3)$.

Describe the set.

1. A line perpendicular to $AB$
2. A sphere with diameter $AB$
3. A plane perpendicular to $AB$
4. A cube with diagonal $AB$

Neetesh Kumar | December 19, 2024

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This is the solution to Math 1C
Assignment: 12.1 Question Number 22
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Step-by-step solution:

Step 1: Understanding the problem:

The set of all points equidistant from two points $A$ and $B$ lies on the perpendicular bisector of the line segment $AB$. In three-dimensional space, this perpendicular bisector is a plane.

Step 2: Equation of the plane:

To find the equation of the plane:

1. Midpoint of $AB$:
   The midpoint of $AB$ is given by: $M = \left(\frac{-2 + 5}{2}, \frac{5 + 2}{2}, \frac{3 - 3}{2}\right) = \left(\frac{3}{2}, \frac{7}{2}, 0\right)$

2. Direction vector of $AB$:
   The direction vector from $A$ to $B$ is: $\vec{AB} = (5 - (-2), 2 - 5, -3 - 3) = (7, -3, -6)$

3. Normal vector to the plane:
   The normal vector of the plane is the same as the direction vector $\vec{AB} = (7, -3, -6)$.

4. Equation of the plane:
   The general equation of a plane is: $7(x - \frac{3}{2}) - 3(y - \frac{7}{2}) - 6(z - 0) = 0$

   Simplifying: $7x - 3y - 6z - \frac{21}{2} + \frac{21}{2} = 0$

   $7x - 3y - 6z = 0$

Thus, the equation of the plane is: $7x - 3y - 6z = 0$

Step 3: Verifying the description of the set:

The set of all points equidistant from $A$ and $B$ forms a plane perpendicular to $AB$.

1. Option 1 (A line perpendicular to $AB$):
   This is incorrect as the solution is a plane, not a line.

2. Option 2 (A sphere with diameter $AB$):
   This is incorrect as a sphere would describe points equidistant from the midpoint of $AB$, not from both points $A$ and $B$.

3. Option 3 (A plane perpendicular to $AB$):
   This is correct as the perpendicular bisector is a plane.

4. Option 4 (A cube with diagonal $AB$):
   This is incorrect as there is no cube involved in the problem.

Final Answer:

1. The equation of the set is:

$\boxed{7x - 3y - 6z = 0}$

2. The set is described as:

3. $\boxed{\text{A plane perpendicular to} \space AB}$

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