Let O and O' be respectively the circumcentre and the orthocentre of a triangle ABC and, S be any point in the plane of the triangle.
Statement-1: $\vec{OA}+\vec{OB}+\vec{OC}=\vec{OO'}$
Statement-2: $\vec{SA} + \vec{SB} + \vec{SC} = 3 \vec{SG}$, where G is the centroid of ΔABC.
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
$⇒\vec{OA} +\vec{OB}+ \vec{OC} = 3\vec{OG}$
$⇒\vec{OA} +\vec{OB}+ \vec{OC} =2\vec{OG}+\vec{OG}$
$⇒\vec{OA} +\vec{OB}+ \vec{OC} =\vec{GO'} + \vec{OG}$ $[∵ 2OG = GO']$
$⇒\vec{OA} +\vec{OB}+ \vec{OC} =\vec{OG}+\vec{GO'}$
$⇒\vec{OA} +\vec{OB}+ \vec{OC} =\vec{OO'}$
So, statement-1 is true and statement-2 is a correct explanation for statement-1.