Giải thích các bước giải:
Ta có $BN=3NC$
$\to 4BN=3(BN+NC)$
$\to 4BN=3BC$
$\to BN=\dfrac34BC$
$\to\vec{BN}=\dfrac34\vec{BC}=\dfrac34(\vec{AC}-\vec{AB})$
Vì $M$ là trung điểm $AB$
$\to\vec{MB}=\dfrac12\vec{AB}$
$\to\vec{MN}=\vec{MB}+\vec{BN}$
$\to\vec{MN}=\dfrac12\vec{AB}+\dfrac34(\vec{AC}-\vec{AB})$
$\to\vec{MN}=-\dfrac14\vec{AB}+\dfrac34\vec{AC}$
Ta có:
$MN^2=(\vec{MN})^2=(-\dfrac14\vec{AB}+\dfrac34\vec{AC})^2$
$\to MN^2=\dfrac1{16}AB^2-\dfrac38\vec{AB}\cdot\vec{AC}+\dfrac9{16}AC^2$
$\to MN^2=\dfrac1{16}AB^2-\dfrac38\cdot AB\cdot AC\cdot \cos\widehat{BAC}+\dfrac9{16}AC^2$
$\to MN^2=\dfrac1{16}\cdot 2^2-\dfrac38\cdot 2\cdot 3\cdot \cos120^o+\dfrac9{16}\cdot 3^2$
$\to MN^2=\dfrac{103}{16}$
$\to MN=\dfrac{\sqrt{103}}{4}$
