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 3 of 7: Apply the Pythagorean identity for orthogonal vectors
Previous stepNext step
In plain words

Perpendicular edges add their squared lengths, just like the 2D Pythagorean theorem extended to three dimensions.

∣PQ⃗∣2=∣PA⃗∣2+∣PB⃗∣2+∣PC⃗∣2|\vec{PQ}|^2 = |\vec{PA}|^2+|\vec{PB}|^2+|\vec{PC}|^2
Detailed analysis

Because PA⃗,PB⃗,PC⃗\vec{PA},\vec{PB},\vec{PC} are pairwise orthogonal, squaring their sum kills every cross term, so ∣PQ⃗∣2|\vec{PQ}|^2 equals the sum of the three squared edge lengths — a 3D Pythagorean theorem.