question_answer

The locus of a point equidistant from two given points a and b is given by                 

A)            \[[\mathbf{r}-\frac{1}{2}(\mathbf{a}+\mathbf{b})]\,.\,\,(\mathbf{a}-\mathbf{b})=0\]  

B)            \[[\mathbf{r}-\frac{1}{2}(\mathbf{a}-\mathbf{b})]\,.\,\,(\mathbf{a}+\mathbf{b})=0\]

C)            \[[\mathbf{r}-\frac{1}{2}(\mathbf{a}+\mathbf{b})].(\mathbf{a}+\mathbf{b})=0\]

D)            \[[\mathbf{r}-\frac{1}{2}(\mathbf{a}-\mathbf{b})]\,.\,\,(\mathbf{a}-\mathbf{b})=0\]

Correct Answer: A

Solution :

                   Let \[P(\mathbf{r})\] be equidistant from \[A\,\,(\mathbf{a})\] and \[B\,\,(\mathbf{b})\] and \[PM\] be perpendicular to \[AB.\]                    Then \[M\] is the mid point of \[AB.\]                    Position vector of \[M\] is \[\frac{1}{2}(\mathbf{a}+\mathbf{b}).\]                    \[\overrightarrow{PM}\,.\,\overrightarrow{BA}=0\] or \[\left[ \mathbf{r}-\frac{1}{2}(\mathbf{a}+\mathbf{b}) \right]\,.\,(\mathbf{a}-\mathbf{b})=0.\]