Question

Let $P$ is any arbitrary point on the circumcircle of a given equilateral triangle $ABC$ of side length $\ell$ units, then $|\overrightarrow{PA}{|}^2+|\overrightarrow{PB}{|}^2+|\overrightarrow{PC}{|}^2$ is always equal to

• A. $2{\ell }^2$
• B. $2\sqrt{3}{\ell }^2$
• C. ${\ell }^2$
• D. $3{\ell }^2$

Answer 1

Answer: (A)

Let position vector of $P,A,B,C$ are $\vec{p},\vec{a},\vec{b}$ and $\vec{c}\&O(\vec{O})$ be the circumcentre of equilateral
triangle $ABC\Rightarrow |\vec{p}|=|\vec{a}|=|\vec{b}|=|\vec{c}|=\frac{\ell }{\sqrt{3}}$
$|\overrightarrow{PA}{|}^2=|\vec{a}-\vec{p}{|}^2=|\vec{a}{|}^2+|\vec{p}{|}^2-2\vec{a}\cdot \vec{p}$
Now, $\sum \mid \overrightarrow{PA}{\mid }^2=6\cdot \frac{{\ell }^2}{3}-2\vec{p}(\vec{a}+\vec{b}+\vec{c})$ as
$\frac{\vec{a}+\vec{b}+\vec{c}}{3}=0$ hence $\sum |\overrightarrow{PA}{|}^2=2{\ell }^2$