` vec(OA) + vec(OB) +vec(OC)` is equal to

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Q 2212612530.     Let `O, N, G`, and `O'` are the circumcentre, nine point centre, centroid and orthocentre of a `Delta ABC` respectively. `AL` and `BM` are
perpendiculars from `A` and `B` on sides `BC` and `CA` respectively. Let `AD` be the median and `OD` is perpendicular to side `BC`. Let `R`
be the circum radius of `Delta ABC`, then `OA = OB = OC = R`.

Now, in `Delta OBD, OD = R cos A`, in `Delta ABM, AO' =AM sec (90^(circ) -C) ( :. angle O' AM = 90^(circ) -C )`

` = AM cosec C= (C cos A)/(sin C)`

`= 2 R cos A`

` :. AO' =2 OD`


If `S` be any point in the plane of `Delta ABC` and `AP` is the diameter of the circum circle.
` vec(OA) + vec(OB) +vec(OC)` is equal to
A

`vec(OO')`

B

`2 vec(O'O)`

C

`2 vec(AO)`

D

`vec(ON)`

HINT


Solution
(Provided By a Student and Checked/Corrected by EXXAMM.com Team)

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