Final answer:
The distance between points P(4, 3, -1) and Q(6, -1, 3) is 6 units, calculated using the three-dimensional distance formula. This distance is invariant under rotations of the coordinate system due to the nature of orthogonal rotation matrices. The equation of a sphere centered at P with a radius equal to the distance between P and Q is (x-4)² + (y-3)² + (z+1)² = 36.
Step-by-step explanation:
The question asks to find the distance between two points in three-dimensional space and to prove that this distance is invariant under rotations of the coordinate system. The two points are P(4, 3, -1) and Q(6, -1, 3). To find the distance, we use the distance formula for three dimensions, which is √((x2-x1)² + (y2-y1)² + (z2-z1)²). Applying this formula, the distance between P and Q is:
√((6-4)² + (-1-3)² + (3-(-1))²) = √(2² + (-4)² + 4²) = √(4 + 16 + 16) = √36 = 6 units.
Regarding invariance under rotation, we consider that the lengths of vectors (and therefore the distances between points) are preserved under rotation transformations because rotation matrices are orthogonal. This means any rotation can be represented by a matrix R that satisfies R²=R^{-1}, ensuring that the inner product (and thus the lengths) remains unchanged. Consequently, the distance is invariant under rotations as it depends on the scalar product of the vectors representing the points.
For the sphere, we need a point and a radius. Assuming the question asks for a sphere centered at either P or Q with radius equal to the distance between P and Q, if centered at P, the equation is (x-4)² + (y-3)² + (z+1)² = 36.