Let O, O' and G be the circumcentre, orthocentre and centroid of a ΔABC and S be any point in the plane of the triangle.
Statement-1: $\vec{O'A}+\vec{O'B}+\vec{O'C} = 2\vec{O'O}$
Statement-2: $\vec{SA} + \vec{SB} + \vec{SC} = 3 \vec{SG}$
Answer & explanation
Correct answer: option 1
We have,
$\vec{SA} + \vec{SB} + \vec{SC} =\vec{SA} + (\vec{SB} + \vec{SC})$
$⇒\vec{SA} + \vec{SB} + \vec{SC} = \vec{SA} + 2 \vec{SD}$ [∵ D is the mid-point of BC]
$⇒\vec{SA} + \vec{SB} + \vec{SC} =(1+2) \vec{SG} = 3 \vec{SG}$
So, statement-2 is true.
Replacing S by O' in statement-2, we get
$\vec{O'A}+\vec{O'B}+\vec{O'C} = 3\vec{O'G}$
$⇒\vec{O'A}+\vec{O'B}+\vec{O'C} =2\vec{O'G}+\vec{O'G}$
$⇒\vec{O'A}+\vec{O'B}+\vec{O'C} =2\vec{O'G}+2\vec{GO}$ $[∵ 2OG = GO']$
$⇒\vec{O'A}+\vec{O'B}+\vec{O'C} =2(\vec{O'G}+\vec{GO})$
$⇒\vec{O'A}+\vec{O'B}+\vec{O'C} =2\vec{O'O}$
So, statement-1 is true and statement-2 is a correct explanation for statement-1.