Let $ A $, $ B $ be two points lie on the sphere $$ x^2 + y^2 + (z-1)^2 = 25 $$ so that $ AB=6 $. Find the greateast value of $OA^2 - OB^2$.
I tried \begin{align*} OA^2- OB^2 & = (\overrightarrow{OA}^2 -\overrightarrow{OB}^2)\\ & = (\overrightarrow{OA} + \overrightarrow{OB}) (\overrightarrow{OA} - \overrightarrow{OB}) \\ & = 2\cdot \overrightarrow{OM} \cdot \overrightarrow{BA}\\ & \leqslant 2 \cdot OM . AB \\ &= 12 \cdot OM \end{align*} Where, $ M $ is midpoint of the segment $ AB $ and $O(0,0,0)$. From here, I can not solve the problem.
Let $P=(0,0,1)$ be the centre of the sphere. By Pythagoras, $PM^2=5^2-3^2$, so $M$ lies on $$ x^2+y^2+(z-1)^2=16 $$ and $OM$ is maximised when $M=(0,0,5)$. But the bound $OA^2-OB^2\leq 12 \max OM=60$ is too crude.
Instead, leave it as $2\overrightarrow{OM}\cdot\overrightarrow{BA}$ for now and further manipulate $$ 2\overrightarrow{OM}\cdot\overrightarrow{BA} =2\overrightarrow{OP}\cdot\overrightarrow{BA}+2\overrightarrow{PM}\cdot\overrightarrow{BA}. $$ But $\overrightarrow{PM}\perp\overrightarrow{BA}$, so it reduces to $$ 2\overrightarrow{OP}\cdot\overrightarrow{BA}. $$ Now the bound $OA^2-OB^2\leq 2\cdot OP\cdot BA=12$ is actually attainable with $\overrightarrow{BA}=6\,\mathbf{k}$, e.g., $B=(4,0,-2), A=(4,0,4)$.