Question

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

विकल्प

• \[\frac{2}{3}\left( \vec{a} + \vec{b} \right)\]

• \[\frac{1}{6}\left( \vec{a} + \vec{b} \right)\]
• \[\frac{1}{3}\left( \vec{a} + \vec{b} \right)\]

   

• \[\frac{1}{2}\left( \vec{a} + \vec{b} \right)\]

Answer 1

\[\frac{1}{2}\left( \vec{a} + \vec{b} \right)\]
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, 
\[AG = \frac{\vec{a} + \vec{b}}{3} - \vec{0} = \frac{\vec{a} + \vec{b}}{3}\]