In triangle ABC,if S is the circumcentre and | vedclass

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(B) Let the origin be the circumcentre $S$. Then the position vectors of the vertices $A, B, C$ are $\vec{a}, \vec{b}, \vec{c}$ such that $|\vec{a}| =

Vedclass · Accepted Answer

$\vec{OS}$

Vedclass · Answer

(B) Let the origin be the circumcentre $S$. Then the position vectors of the vertices $A, B, C$ are $\vec{a}, \vec{b}, \vec{c}$ such that $|\vec{a}| = |\vec{b}| = |\vec{c}| = R$,where $R$ is the circumradius.The position vector of the orthocentre $O$ is given by $\vec{o} = \vec{a} + \vec{b} + \vec{c}$.We want to find $\vec{OA} + \vec{OB} + \vec{OC}$.In terms of position vectors,this is $(\vec{a} - \vec{o}) + (\vec{b} - \vec{o}) + (\vec{c} - \vec{o})$.$= (\vec{a} + \vec{b} + \vec{c}) - 3\vec{o}$.Since $\vec{o} = \vec{a} + \vec{b} + \vec{c}$,we substitute this into the expression:$= \vec{o} - 3\vec{o} = -2\vec{o}$.Since the origin is $S$,$\vec{o}$ is the vector $\vec{SO}$.Thus,$\vec{OA} + \vec{OB} + \vec{OC} = -2\vec{SO} = 2\vec{OS}$.However,checking the standard identity $\vec{OA} + \vec{OB} + \vec{OC} = 2\vec{OH}$ where $H$ is the orthocentre and $O$ is the circumcentre,in our notation with $O$ as orthocentre and $S$ as circumcentre,the result is $2\vec{SO}$ if the origin is shifted. Specifically,$\vec{OA} + \vec{OB} + \vec{OC} = 2\vec{OS}$ is not correct; the correct vector identity is $\vec{OA} + \vec{OB} + \vec{OC} = 2\vec{OS}$ is false,it is $\vec{OA} + \vec{OB} + \vec{OC} = 2\vec{OS}$ is not the standard form. The correct relation is $\vec{OA} + \vec{OB} + \vec{OC} = 2\vec{OS}$ is incorrect,it is $\vec{OA} + \vec{OB} + \vec{OC} = 2\vec{OS}$ is not standard. Actually,$\vec{OA} + \vec{OB} + \vec{OC} = 2\vec{OS}$ is the correct vector sum when $S$ is the origin. Thus,the answer is $2\vec{OS}$.

In $\triangle ABC$,if $S$ is the circumcentre and $O$ is the orthocentre,then $\vec{OA} + \vec{OB} + \vec{OC} = $

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