Vector Algebra Question 1
Question: The locus of a point equidistant from two given points a and b is given by
Options:
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 $
Show Answer
Answer:
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 $ [ \mathbf{r}-\frac{1}{2}(\mathbf{a}+\mathbf{b}) ],.,(\mathbf{a}-\mathbf{b})=0. $
BETA
