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Part (b)(ii)

What it means by "Vec(SR) dot Vec(QR) = 0": (You can rotate or zoom the view.) Move the red point R. This is when Vec(SR)dotVec(QR)=0, giving a right angle. Hence R is on the circle with QS as diameter. BUT THIS IS 3D SPACE! Hence you can see how the light-blue circle can 'rotate'. This is summarised as R now lies on the SPHERE with QS as diameter. (The sphere is on the LHS of the applet. Click the bubble next to the 'Sphere' 'd'. Since the question has said that R lies on the unit sphere (bubble on LHS next to 'Sphere' 'a'), this additional constraint forces R to be on the intersection of these two spheres, and thus concyclic with those other points. (The solution gave an algebraic approach for this last issue, which argue for the coplanarity (something vectors have a good advantage).)