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(A) Let the points be $A$ and $B$ with position vectors $a$ and $b$ respectively. The midpoint $M$ of the line segment $AB$ has the position vector $\

Vedclass · Accepted Answer

$(2r - a - b) \cdot (a - b) = 0$

Vedclass · Answer

(A) Let the points be $A$ and $B$ with position vectors $a$ and $b$ respectively. The midpoint $M$ of the line segment $AB$ has the position vector $\frac{a+b}{2}$.The perpendicular bisector passes through $M$ and is perpendicular to the vector $\vec{AB} = b - a$ (or $a - b$).Let $P$ be any point on the perpendicular bisector with position vector $r$. Then the vector $\vec{MP} = r - \frac{a+b}{2}$ must be perpendicular to the vector $\vec{AB} = a - b$.Since the dot product of two perpendicular vectors is zero,we have:$\left(r - \frac{a+b}{2}\right) \cdot (a - b) = 0$Multiplying by $2$,we get:$(2r - (a + b)) \cdot (a - b) = 0$Thus,the equation is $(2r - a - b) \cdot (a - b) = 0$.

The equation of the perpendicular bisector of the line segment joining the points whose position vectors are $a$ and $b$ respectively is

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