If S is the circumcentre, O is the orthocentr | vedclass

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(D) Let the origin be at the circumcentre $S$. Then $\vec{SA} = \vec{a}, \vec{SB} = \vec{b}, \vec{SC} = \vec{c}$, where $|\vec{a}| = |\vec{b}| = |\vec

Vedclass · Accepted Answer

$i \rightarrow D, ii \rightarrow C, iii \rightarrow A, iv \rightarrow B$

Vedclass · Answer

(D) Let the origin be at the circumcentre $S$. Then $\vec{SA} = \vec{a}, \vec{SB} = \vec{b}, \vec{SC} = \vec{c}$, where $|\vec{a}| = |\vec{b}| = |\vec{c}| = R$.$(i)$ The orthocentre $O$ is given by $\vec{SO} = \vec{a} + \vec{b} + \vec{c}$. Thus, $\vec{SA} + \vec{SB} + \vec{SC} = \vec{SO}$. So, $(i) \rightarrow D$.(ii) Let $G$ be the centroid. Then $\vec{G} = \frac{\vec{A} + \vec{B} + \vec{C}}{3}$. With $S$ as origin, $\vec{GA} + \vec{GB} + \vec{GC} = (\vec{a} - \vec{g}) + (\vec{b} - \vec{g}) + (\vec{c} - \vec{g}) = (\vec{a} + \vec{b} + \vec{c}) - 3\vec{g} = 3\vec{g} - 3\vec{g} = \vec{0}$. So, $(ii) \rightarrow C$.(iii) Since $\vec{OA} = \vec{a} - \vec{o}$, $\vec{OB} = \vec{b} - \vec{o}$, and $\vec{OC} = \vec{c} - \vec{o}$, we have $\vec{OA} + \vec{OB} + \vec{OC} = (\vec{a} + \vec{b} + \vec{c}) - 3\vec{o} = \vec{o} - 3\vec{o} = -2\vec{o} = 2\vec{SO} = 2\vec{OS}$ (since $\vec{SO} = \vec{o}$). So, $(iii) \rightarrow A$.(iv) The centroid $G$ divides the line segment joining the orthocentre $O$ and circumcentre $S$ in the ratio $2:1$. Thus, $\vec{OG} = \frac{2}{3}\vec{OS}$. So, $(iv) \rightarrow B$.Therefore, the correct match is $(i) \rightarrow D, (ii) \rightarrow C, (iii) \rightarrow A, (iv) \rightarrow B$.

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