If O and O' denote respectively the circum- centre and orthocentre of ΔABC, then $\vec{OA} + \vec{OB} + \vec{OC} =$
Answer & explanation
Correct answer: option 2
Let G be the centroid of ΔABC. Then O', G, O are collinear and G divides O' O in the ratio 2: 1 i.e. $\frac{O'G}{OG}=\frac{2}{1}$
Replacing S by O, we have
$\vec{OA} + \vec{OB} + \vec{OC} =3\vec{OG}$
$⇒\vec{OA} + \vec{OB} + \vec{OC} =2\vec{OG}+\vec{OG}$
$⇒\vec{OA} + \vec{OB} + \vec{OC} =\vec{GO}+\vec{OG}$ $[∵2OG=GO']$
$⇒\vec{OA} + \vec{OB} + \vec{OC} =\vec{OO'}$