Q. The locus of a point equidistant from two points whose position vectors are $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$, is
A) $\left\{ \overrightarrow{\mathbf{r}} - \frac{1}{2}\left( \ \overrightarrow{\mathbf{a}}\mathbf{+}\overrightarrow{\mathbf{b}} \right) \right\}\left( \mathbf{\ }\overrightarrow{\mathbf{a}}\mathbf{-}\overrightarrow{\mathbf{b}} \right) = 0$
B) $\left\{ \overrightarrow{\mathbf{r}} - \left( \ \overrightarrow{\mathbf{a}}\mathbf{+}\overrightarrow{\mathbf{b}} \right) \right\} \bullet \overrightarrow{\mathbf{b}} = 0$
C) $\left\{ \overrightarrow{\mathbf{r}} - \frac{1}{2}\left( \ \overrightarrow{\mathbf{a}}\mathbf{+}\overrightarrow{\mathbf{b}} \right) \right\} \bullet \overrightarrow{\mathbf{a}} = 0$
D) $\left\{ \overrightarrow{\mathbf{r}} - \frac{1}{2}\left( \ \overrightarrow{\mathbf{a}}\mathbf{-}\overrightarrow{\mathbf{b}} \right) \right\} \bullet \left( \ \overrightarrow{\mathbf{a}}\mathbf{+}\overrightarrow{\mathbf{b}} \right) = 0$
Explanation
The correct answer is Option A: $\left\{ \overrightarrow{\mathbf{r}} - \frac{1}{2}\left( \ \overrightarrow{\mathbf{a}}\mathbf{+}\overrightarrow{\mathbf{b}} \right) \right\}\left( \mathbf{\ }\overrightarrow{\mathbf{a}}\mathbf{-}\overrightarrow{\mathbf{b}} \right) = 0$.
A detailed explanation for this question is being prepared.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.