Problem 2

We consider a fixed point

P

in the interior of a fixed sphere. We construct three segments

PA, PB, PC

, perpendicular two by two, with the vertices

A, B, C

on the sphere. We consider the vertex

Q

which is opposite to

P

in the parallelepiped (with right angles) with

PA, PB, PC

as edges. Find the locus of the point

Q

when

A, B, C

take all the positions compatible with our problem.

Step 4 of 7: Expand each edge length using the sphere condition

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In plain words

Expanding each squared distance turns the geometric sum into pure algebra in the position vectors.

|X-P|^2 = R^2 - 2X\cdot P + d^2 \text{ for } X\in\{A,B,C\},\quad A+B+C = Q+2P

Detailed analysis

Since $|A|=|B|=|C|=R$ and $|P|=d$ , each term $|\vec{PA}|^2=|A-P|^2$ expands to $R^2-2A\cdot P+d^2$ , and similarly for $B,C$ ; summing uses $A+B+C=Q+2P$ from the first step.