If P is any arbitary point on the circumcurcle of the equilateral triangle of side length l units, then `|vec(PA)|^(2)+|vec(PB)|^(2)+|vec(PC)|^(2)` is always equal to

Options

A. `2l^(2)`
B. `2sqrt3l^(2)`
C. `l^(2)`
D. `3l^(2)`

Answer

a

Explanation

Let P.V. of A,B and C be `vecp, veca, vecb and vecc`
respectively, and `O(vec0)` be the circumcentre of equilateral traingle ABC. Then
`|vecP| = |vecb| = |veca|= |vecc| = l/ sqrt3`
Now `|vec(PA)|^(2) = |veca - vecp|^(2)= |veca|^(2) + |vecp|^(2) - 2vecp`
` and |vecPC|^(2) = |vecc|^(2) + |vecp|^(2) - 2vecp. vecc`
` Rightarrow sum|vec(PA)|^(2) = 6. l^(2)/3 - 2vecp . (veca + vecb + vecc)`
` 2l^(2) (as veca + vecb + vecc//3 = vec0)`