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 \[AB.\]You need to login to perform this action.
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