Giải thích các bước giải:
1.Để $BN\perp CM$
$\to \vec{BN}\cdot \vec{CM}=0$
$\to (\vec{BA}+\vec{AN})\cdot (\vec{CA}+\vec{AM})=0$
$\to \vec{BA}\cdot \vec{CA}+\vec{BA}\cdot\vec{AM}+\vec{AN}\cdot\vec{CA}+\vec{AN}\cdot\vec{AM}=0$
$\to \vec{AB}\cdot \vec{AC}+\vec{AB}\cdot\dfrac13\vec{AB}-k\vec{AC}\cdot\vec{AC}+k\vec{AC}\cdot\dfrac13\vec{AB}=0$
$\to \vec{AB}\cdot \vec{AC}(1+\dfrac13-k+\dfrac13k)=0$
Vì $AB$ không vuông góc với $AC\to \vec{AB}\cdot\vec{AC}\ne 0$
$\to 1+\dfrac13-k+\dfrac13k=0$
$\to k=2$
2.Kẻ $DN//CM$
Để $\widehat{\vec{BN},\vec{CM}}=120^o$
$\to \widehat{\vec{BN},\vec{DN}}=120^o$
$\to \widehat{BND}=60^o$
$\to \widehat{BND}=\widehat{BAN}$
Mà $\widehat{DBN}=\widehat{ABN}$
$\to \Delta BND\sim\Delta BAN(g.g)$
$\to \dfrac{BN}{BA}=\dfrac{BD}{BN}$
$\to BN^2=BD\cdot BA$
Vì $\vec{AN}=k\vec{AC}\to \dfrac{AN}{AC}=k$
Ta có: $ND//CM$
$\to \dfrac{AD}{AM}=\dfrac{AN}{AC}=k$
$\to AD=kAM$
$\to AD=k\cdot \dfrac13AB$
$\to AD=\dfrac13kAB$
$\to BD=AB-AD=(1-\dfrac13k)AB$
$\to BN^2=(1-\dfrac13k)AB^2$
Xét $\Delta ABN$ có $\hat A=60^o$
$\to BN^2=AB^2+AN^2-2AB.AN.\cos60^o$
$\to BN^2=AB^2+AN^2-AB.AN$
$\to (1-\dfrac13k)AB^2=AB^2+k^2AC^2-AB.kAC$ vì $\vec{AN}=k\vec{AC}\to AN=kAC$
$\to (1-\dfrac13k)AB^2=AB^2+k^2AB^2-AB.kAB$ vì $AB=AC$
$\to (1-\dfrac13k)AB^2=AB^2(1+k^2-k)$
$\to 1-\dfrac13k=1+k^2-k$
$\to k=\dfrac23$ vì $k\ne 0$
