To find the equation of the plane consisting of all points that are equidistant from the points A(1, 0, 2) and B(3, 4, 0), we start by recalling that a plane can be determined by a point and a normal vector or can be defined as all the points that satisfy a specific distance condition.
1. **Understanding the distance**: All points P(x, y, z) that are equidistant from points A and B satisfy the equation:
Distance(P, A) = Distance(P, B)
We can express this using the distance formula:
Distance(P, A) = √[ (x – 1)² + (y – 0)² + (z – 2)² ]
Distance(P, B) = √[ (x – 3)² + (y – 4)² + (z – 0)² ]
2. **Setting the distances equal**:
√[ (x – 1)² + (y – 0)² + (z – 2)² ] = √[ (x – 3)² + (y – 4)² + (z – 0)² ]
3. **Squaring both sides**: To eliminate the square roots, we will square both sides of the equation:
(x – 1)² + y² + (z – 2)² = (x – 3)² + (y – 4)² + z²
4. **Expanding both sides**:
Left side:
(x² – 2x + 1) + y² + (z² – 4z + 4) = x² + y² + z² – 2x – 4z + 5
Right side:
(x² – 6x + 9) + (y² – 8y + 16) + z² = x² + y² + z² – 6x – 8y + 25
5. **Setting the equations equal**:
x² + y² + z² – 2x – 4z + 5 = x² + y² + z² – 6x – 8y + 25
Canceling identical terms and rearranging gives:
4x – 8y – 4z + 20 = 0
6. **Simplifying the equation**:
Divide the entire equation by 4:
x – 2y – z + 5 = 0
7. **Final equation**:
The equation of the plane consisting of all points that are equidistant from the points A(1, 0, 2) and B(3, 4, 0) is:
x – 2y – z + 5 = 0
This plane represents all the points that maintain equal distance to both given points.