For the triangle ABC : |overrightarrowAB|=2…

For the triangle $ABC$: $|\overrightarrow{AB}|=2$, $|\overrightarrow{AC}|=3$, $\angle(\overrightarrow{AB},\overrightarrow{AC})=\dfrac{\pi}{3}$. Determine the angle between the altitude and the median drawn from the vertex $A$.

Step 1: Notation

\[ \vec b=\overrightarrow{AB},\quad \vec c=\overrightarrow{AC},\qquad |\vec b|=2,\quad |\vec c|=3,\quad \vec b\cdot\vec c=2\cdot3\cdot\tfrac12=3. \]

Step 2: The median

\[ \overrightarrow{AM}=\frac{\vec b+\vec c}{2},\qquad |\overrightarrow{AM}|^2=\frac{4+6+9}{4}=\frac{19}{4},\qquad |\overrightarrow{AM}|=\frac{\sqrt{19}}{2}. \]

Step 3: The altitude direction

Write $\vec h=\lambda\vec b+\mu\vec c$ and require $\vec h\perp\overrightarrow{BC}=\vec c-\vec b$: \[\lambda(\vec b\cdot\vec c-\vec b^{\,2})+\mu(\vec c^{\,2}-\vec b\cdot\vec c)=-\lambda+6\mu=0\ \Longrightarrow\ \lambda=6\mu.\]

\[ \vec h=6\vec b+\vec c,\qquad \vec h\cdot(\vec c-\vec b)=18-24+9-3=0\ \checkmark \]

\[ |\vec h|^2=36\cdot4+12\cdot3+9=189,\qquad |\vec h|=3\sqrt{21}. \]

Indeed $\overrightarrow{AH}=\tfrac17(6\vec b+\vec c)$: the foot $H$ divides $BC$ in the ratio $BH:HC=1:6$.

Step 4: The angle

\[ \overrightarrow{AM}\cdot\vec h=\tfrac12\left(6\vec b^{\,2}+7\,\vec b\cdot\vec c+\vec c^{\,2}\right)=\tfrac12(24+21+9)=27, \]

\[ \cos\varphi=\frac{27}{\frac{\sqrt{19}}{2}\cdot3\sqrt{21}}=\frac{18}{\sqrt{399}}=\frac{6\sqrt{399}}{133}\approx0.9011. \]

\[\varphi=\arccos\frac{18}{\sqrt{399}}\approx25.7^\circ\]

Step 5: Independent check

In the right triangle $AHM$ (right angle at $H$), $\cos\varphi=\dfrac{AH}{AM}$. Here $|BC|=\sqrt{4-6+9}=\sqrt7$, the area is $S=\tfrac12\cdot2\cdot3\sin60^\circ=\tfrac{3\sqrt3}{2}$, and

\[ AH=\frac{2S}{|BC|}=\frac{3\sqrt3}{\sqrt7},\qquad \cos\varphi=\frac{3\sqrt3/\sqrt7}{\sqrt{19}/2}=\frac{6\sqrt3}{\sqrt{133}}=\frac{18}{\sqrt{399}}.\ \checkmark \]