If O and O' denote respectively the circumcentre and orthocentre of ΔABC, then $\vec{O'A} + \vec{O'B} + \vec{O'C} =$
Answer & explanation
Correct answer: option 4
Replacing S by O', we get
$\vec{O'A} + \vec{O'B} + \vec{O'C} =3\vec{O'G}$
$⇒\vec{O'A} + \vec{O'B} + \vec{O'C} =2\vec{O'G}+\vec{O'G}$
$⇒\vec{O'A} + \vec{O'B} + \vec{O'C} =2\vec{O'G}+\vec{GO}$ $[∵O'G=2GO]$
$⇒\vec{O'A} + \vec{O'B} + \vec{O'C} =\vec{O'O}$