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# Show that diameter of a sphere subtends a right angle at a point on the surface.

Toolbox:
• If $\overrightarrow a \perp \overrightarrow b$ then $\overrightarrow a.\overrightarrow b=0$ and for nonzero vectors if $\overrightarrow a.\overrightarrow b=0 \Rightarrow \overrightarrow a \perp \overrightarrow b.$
Let $AB$ be the diameter of a sphere with its centre at $C$ and radius $a$. Let P be any point on the surface of the sphere.
Now C is the midpoint of $AB. \therefore \overrightarrow {AC}=\overrightarrow {CB}=\large\frac{1}{2}\overrightarrow {AB}$
Consider $\overrightarrow {AP}.\overrightarrow {BP}=(\overrightarrow {AC}+\overrightarrow {CP}).(\overrightarrow {BC}+\overrightarrow {CP})$
$= (\overrightarrow {AC}+\overrightarrow {CD}).(-\overrightarrow {BC}+\overrightarrow {CD})$
$= (\overrightarrow {AC}+\overrightarrow {CP}).(\overrightarrow {CP}-\overrightarrow {AC})$
$=\overrightarrow {CP}^2-\overrightarrow {AC}^2=a^2-a^2=0$
$\overrightarrow {AP}.\overrightarrow {BP}=0 \Rightarrow <{APB} = 90^{\circ}$
Hence the diameter subtends a right angle at any point on the surface of the sphere.