Let G be the centroid of ΔABC and S be any point in the plane of a ΔABC.
Statement-1: $\vec{SA} + \vec{SB} + \vec{SC} = 3 \vec{SG}$
Statement-2: If O, A and B are three points in a plane, then $m\vec{OA}+n\vec{OB} = (m + n)\vec{OC}$, where C is a point dividing AB in the ratio $n : m$.
Answer & explanation
Correct answer: option 1
Let O be the origin.
Using section formula, we have
$\vec{OC}=\frac{n\vec{OB}+m\vec{OA}}{n+m}⇒ (m+n)\vec{OC}=m\vec{OA}+n\vec{OB}$
So, statement-2 is true.
Now,
$\vec{SA} + \vec{SB} + \vec{SC} =\vec{SA}+(\vec{SB} + \vec{SC})$
$⇒\vec{SA} + \vec{SB} + \vec{SC} =\vec{SA}+2\vec{SD}$ [Using statement-2 for m = n = 1]
$⇒\vec{SA} + \vec{SB} + \vec{SC} =(1+2)\vec{SG}$ [Using statement-2]
$⇒\vec{SA} + \vec{SB} + \vec{SC} =3\vec{SG}$
So, both the statements are true and statement-2 is a correct explanation for statement-1.