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Question
If O and O' are circumcentre and orthocentre of ∆ ABC, then \[\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC}\] equals
Options
2\[\overrightarrow{OO}\]
- \[O \overrightarrow{O'}\]
- \[\overrightarrow{OO'}\]
- \[2 \overrightarrow{O'O}\]
Solution
Given: O be the circumcentre and \[O'\] be the orthocentre of \[\bigtriangleup ABC\].
Let G be the centroid of the triangle.
We know that O, G and H are collinear and by geometry \[\overrightarrow{O'G} = 2 \overrightarrow{OG} .\]
This yields, \[\overrightarrow{O'O} = \overrightarrow{O'G} + \overrightarrow{GO} = 2 \overrightarrow{GO} + \overrightarrow{GO} = 3 \overrightarrow{GO} . \]
In other words \[\overrightarrow{OO'} = 3 \overrightarrow{OG} .\]
Since, \[\overrightarrow{OG} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}\]
∴ \[\overrightarrow{OO'} = 3 \times \frac{\vec{a} + \vec{b} + \vec{c}}{3} = \vec{a} + \vec{b} + \vec{c} = \overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} .\]
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