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. |
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. Statement-1 is True, Statement-2 is True; Statement-2 is not a correct explanation for Statement-1. Statement-1 is True, Statement-2 is False. Statement-1 is False, Statement-2 is True. |
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-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. |