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प्रश्न
Let G be the centroid of ∆ ABC. If \[\overrightarrow{AB} = \vec{a,} \overrightarrow{AC} = \vec{b,}\] then the bisector \[\overrightarrow{AG} ,\] in terms of \[\vec{a}\text{ and }\vec{b}\] is
पर्याय
\[\frac{2}{3}\left( \vec{a} + \vec{b} \right)\]
- \[\frac{1}{6}\left( \vec{a} + \vec{b} \right)\]
- \[\frac{1}{3}\left( \vec{a} + \vec{b} \right)\]
- \[\frac{1}{2}\left( \vec{a} + \vec{b} \right)\]
उत्तर
Taking A as origin.
Then, position vector of A, B and C are \[\vec{0} , \vec{a}\] and \[\vec{b}\] respectively.
Then, Centroid G has position vector \[\frac{\vec{0} + \vec{a} + \vec{b}}{3} = \frac{\vec{a} + \vec{b}}{3}\]
Therefore,
\[AG = \frac{\vec{a} + \vec{b}}{3} - \vec{0} = \frac{\vec{a} + \vec{b}}{3}\]
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