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If `P` is any arbitrary point on the circumcirlce of the equllateral trangle of side length `l` units, then `| vec P A|^2+| vec P B|^2+| vec P C|^2` is always equal to `2l^2` b. `2sqrt(3)l^2` c. `l^2` d. `3l^2`

A

`2l^(2)`

B

`2sqrt3l^(2)`

C

`l^(2)`

D

`3l^(2)`

Text Solution

Verified by Experts

The correct Answer is:
a
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)`
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