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 2 of 7: Identify the orthogonal edge sum with PQ
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In plain words

The space diagonal PQ of the box is literally the sum of its three perpendicular edges from P.

PA⃗+PB⃗+PC⃗=Q−P\vec{PA}+\vec{PB}+\vec{PC} = Q-P
Detailed analysis

From the previous step, A+B+C−3P=Q−PA+B+C-3P = Q-P, and the left side is exactly PA⃗+PB⃗+PC⃗\vec{PA}+\vec{PB}+\vec{PC} by definition, so this sum of three mutually perpendicular vectors equals PQ⃗\vec{PQ}.