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(D) Let $P$ be any point on the sphere with position vector $r$. Let $A$ and $B$ be the extremities of a diameter with position vectors $a$ and $b$ re

Vedclass · Accepted Answer

$(r - a) \cdot (r - b) = 0$

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(D) Let $P$ be any point on the sphere with position vector $r$. Let $A$ and $B$ be the extremities of a diameter with position vectors $a$ and $b$ respectively.Since $AB$ is a diameter,the angle subtended by the diameter at any point $P$ on the sphere is a right angle $(90^{\circ})$.Therefore,the vector $\vec{AP}$ is perpendicular to the vector $\vec{BP}$.The vector $\vec{AP} = r - a$ and the vector $\vec{BP} = r - b$.Since they are perpendicular,their dot product must be zero:$(r - a) \cdot (r - b) = 0$.

If $r$ is the position vector of any point on a sphere and $a$ and $b$ are the position vectors of the extremities of a diameter,then which of the following is true?

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