MathLabs

Problem 2

We consider a fixed point PP in the interior of a fixed sphere. We construct three segments PA,PB,PCPA, PB, PC, perpendicular two by two, with the vertices A,B,CA, B, C on the sphere. We consider the vertex QQ which is opposite to PP in the parallelepiped (with right angles) with PA,PB,PCPA, PB, PC as edges. Find the locus of the point QQ when A,B,CA, B, C take all the positions compatible with our problem.
Step 1 of 7: Express Q using the parallelepiped's vertex structure
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In plain words

In a box, the opposite vertex is reached by walking along all three edges from P.

Q=A+B+C−2PQ = A+B+C-2P
The fixed sphere on which A, B, C lie, with interior point P.
A sphere rendered in 3D, representing the fixed sphere of the problem that contains points A, B, C on its surface.
Detailed analysis

With OO the sphere's center as origin of position vectors, QQ is the vertex opposite PP in the box built on the perpendicular edges PA,PB,PCPA,PB,PC, so Q=P+PA⃗+PB⃗+PC⃗=A+B+C−2PQ = P+\vec{PA}+\vec{PB}+\vec{PC} = A+B+C-2P.