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Question: If O be the circumcentre and O' be the orthocentre of the triangle ABC, then \(\overset{\rightarrow}...

If O be the circumcentre and O' be the orthocentre of the triangle ABC, then O′A→+O′B→+O′C→=\overset{\rightarrow}{O'A} + \overset{\rightarrow}{O'B} + \overset{\rightarrow}{O'C} =

A

OO→′\overset{\rightarrow}{OO}'

B

2O′O→2\overset{\rightarrow}{O'O}

C

2OO′→2\overset{\rightarrow}{OO'}

D

0

Answer

2O′O→2\overset{\rightarrow}{O'O}

Explanation

Solution

O′A→=O′O→+OA→\overset{\rightarrow}{O^{'}A} = \overset{\rightarrow}{O^{'}O} + \overset{\rightarrow}{OA}

O′B→=O′O→+OB→\overset{\rightarrow}{O^{'}B} = \overset{\rightarrow}{O^{'}O} + \overset{\rightarrow}{OB}

O′C→=O′O→+OC→\overset{\rightarrow}{O^{'}C} = \overset{\rightarrow}{O^{'}O} + \overset{\rightarrow}{OC}

⇒O′A→+O′B→+O′C→\Rightarrow \overset{\rightarrow}{O^{'}A} + \overset{\rightarrow}{O^{'}B} + \overset{\rightarrow}{O^{'}C}

=3O′O→+OA→+OB→+OC→= 3\overset{\rightarrow}{O^{'}O} + \overset{\rightarrow}{OA} + \overset{\rightarrow}{OB} + \overset{\rightarrow}{OC}

Since OA→+OB→+OC→=OO′→=−O′O→\overset{\rightarrow}{OA} + \overset{\rightarrow}{OB} + \overset{\rightarrow}{OC} = \overset{\rightarrow}{OO^{'}} = - \overset{\rightarrow}{O^{'}O}

∴\therefore O′A→+O′B→+O′C→=2O′O→\overset{\rightarrow}{O^{'}A} + \overset{\rightarrow}{O^{'}B} + \overset{\rightarrow}{O^{'}C} = 2\overset{\rightarrow}{O^{'}O}.