A larger sphere A, having a radius $R$ is snugly fitted in a cube (i.e. sphere A touches all six faces of the cube). Further, a small sphere B is snugly fitted in the corner of cube (i.e. sphere B touches sphere A & three orthogonal faces meeting at the same vertex). Further, a smaller sphere C, having a radius $r$, is snugly fitted in the same corner of the cube (i.e. sphere C touches sphere B & three orthogonal faces meeting at the same vertex).
How to find out the radius $r$ (of smaller sphere C) in terms of the radius $R$ (of larger sphere A)? I have tried this using vector analysis but I did not find the correct answer. Any help or pointing in the right direction is appreciated.
Note: None of the spheres touches any of 12 edges of the cube.
Unit vector equally inclined with three orthogonal axes X, Y & Z is given as $$ \hat{r}=\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$$
angle of inclination of the vector with three axes is $$=\cos^{-1}\left(\frac{1}{\sqrt{3}}\right)\approx 54.73^{o}$$ but further if angle of inclination of this vector with each of three orthogonal faces is $\alpha \space$ then I have $$\sin\alpha =\frac{R_{B}}{R\sqrt{3}-R-R_{B}}$$ Similarly, for radius $r$ $$\sin\alpha =\frac{r}{R_{B}\sqrt{3}-R_{B}-r}$$
Further, how I can determine angle of inclination $\alpha$ with the orthogonal planes?
1
Don't worry about the angles.2
Let's assume that the cube's centered on the origin & that the vertex of the cube surrounding the series of spheres is (1,1,1), so the radius of A is 1, the centers of the spheres are all on the line passing through (1,1,1), and A cuts that line at (a,a,a), where $a = \sqrt{1/3}$.3
Let $b$ be the radius of B. Find the center of B = (x,x,x) in terms of $a$ and $b$. Prove that for B to touch the cube, $x+b=1$4
Now use that info to find $b$. $\endgroup$