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STD 12 - 10. vector algebra — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 10. vector algebra4 Marks
MCQ
The locus of  a point equidistant from two given points $ a$  and $ b $ is given by
  • ✓
    $[r - \frac{1}{2}(a + b)]\,.\,\,(a - b) = 0$
  • B
    $[r - \frac{1}{2}(a - b)]\,.\,\,(a + b) = 0$
  • C
    $[r - \frac{1}{2}(a + b)].(a + b) = 0$
  • D
    $[r - \frac{1}{2}(a - b)]\,.\,\,(a - b) = 0$
✓

Answer

Correct option: A.
$[r - \frac{1}{2}(a + b)]\,.\,\,(a - b) = 0$
a
(a) Let $P(r)$ be equidistant from $A\,\,(a)$ and $B\,\,(b)$ and $PM$ be perpendicular to $AB.$

Then $M$ is the mid point of $AB.$

$\therefore $ Position vector of $M$ is $\frac{1}{2}(a + b).$

$\overrightarrow {PM} \,.\,\overrightarrow {BA} = 0$ or $\left[ {r - \frac{1}{2}(a + b)} \right]\,.\,(a - b) = 0.$

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