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Question: If **r** be position vector of any point on a sphere and **a**, **b** are respectively position vect...

If r be position vector of any point on a sphere and a, b are respectively position vectors of the extremities of a diameter, then

A

r.(a−b)=0\mathbf { r } . ( \mathbf { a } - \mathbf { b } ) = 0

B

r.(r−a)=0\mathbf { r } . ( \mathbf { r } - \mathbf { a } ) = 0

C

(r+a)⋅(r+b)=0( \mathbf { r } + \mathbf { a } ) \cdot ( \mathbf { r } + \mathbf { b } ) = 0

D

(r−a)⋅(r−b)=0( \mathbf { r } - \mathbf { a } ) \cdot ( \mathbf { r } - \mathbf { b } ) = 0

Answer

(r−a)⋅(r−b)=0( \mathbf { r } - \mathbf { a } ) \cdot ( \mathbf { r } - \mathbf { b } ) = 0

Explanation

Solution

It is a fundamental property.