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Vector Algebra — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsVector Algebra1 Mark
MCQ
Let $G$ be the centroid of $\triangle{\text{ABC}}$. if $\overrightarrow{\text{AB}}=\vec{\text{a}},\overrightarrow{\text{AC}}=\vec{\text{b}}$, then the bisector $\overrightarrow{\text{AG}}$, in terms of $\vec{\text{a}}$ and $\vec{\text{b}}$ is,
  • A
    $\frac{2}3\big(\vec{\text{a}}+\vec{\text{b}}\big)$
  • B
    $\frac{1}6\big(\vec{\text{a}}+\vec{\text{b}}\big)$
  • ✓
    $\frac{1}3\big(\vec{\text{a}}+\vec{\text{b}}\big)$
  • D
    $\frac{1}2\big(\vec{\text{a}}+\vec{\text{b}}\big)$
✓

Answer

Correct option: C.
$\frac{1}3\big(\vec{\text{a}}+\vec{\text{b}}\big)$
Taking $A$ as origin. Then, position vector of $A, B$ and $C$ are $\vec0,\vec{\text{a}}$ and $\vec{\text{b}}$ respectively.
Then, Centroid $G$ has position vector $\frac{\vec0+\vec{\text{a}}+\vec{\text{b}}}3=\frac{\vec{\text{a}}+\vec{\text{b}}}3$
Therefore, $\text{AG}=\frac{\vec{\text{a}}+\vec{\text{b}}}3-\vec0=\frac{\vec{\text{a}}+\vec{\text{b}}}3$

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