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Theo các xác định điểm M, N ta có:
\(\overrightarrow{AM}=\dfrac{1}{2}\overrightarrow{AB};\overrightarrow{AN}=\dfrac{2}{3}\overrightarrow{AC}.\)
Theo tính chất trung điểm của MN ta có:
\(\overrightarrow{AK}=\dfrac{1}{2}\left(\overrightarrow{AM}+\overrightarrow{AN}\right)=\dfrac{1}{2}\left(\dfrac{1}{2}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\right)\)
\(=\dfrac{1}{4}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\).
\(\overrightarrow{CN}=2\overrightarrow{NA}\Leftrightarrow\overrightarrow{CA}+\overrightarrow{AN}=-2\overrightarrow{AN}\Leftrightarrow\overrightarrow{AN}=\frac{1}{3}\overrightarrow{AC}\)
\(\overrightarrow{AK}=\frac{1}{2}\overrightarrow{AM}+\frac{1}{2}\overrightarrow{AN}=\frac{1}{4}\overrightarrow{AB}+\frac{1}{6}\overrightarrow{AC}\Rightarrow\overrightarrow{KA}=-\frac{1}{4}\overrightarrow{AB}-\frac{1}{6}\overrightarrow{AC}\)
\(\overrightarrow{KD}=\overrightarrow{KA}+\overrightarrow{AD}=\left(-\frac{1}{4}\overrightarrow{AB}-\frac{1}{6}\overrightarrow{AC}\right)+\left(\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AC}\right)\)
\(=\frac{1}{4}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\Rightarrow\left\{{}\begin{matrix}m=\frac{1}{4}\\n=\frac{1}{3}\end{matrix}\right.\) \(\Rightarrow m-n=-\frac{1}{12}\)
\(\overrightarrow{CN}=2\overrightarrow{NA}\Leftrightarrow\overrightarrow{CA}+\overrightarrow{AN}=-2\overrightarrow{AN}\)
\(\Leftrightarrow-\overrightarrow{AC}=-3\overrightarrow{AN}\Rightarrow\overrightarrow{AN}=\frac{1}{3}\overrightarrow{AC}\)
\(\overrightarrow{AM}=\frac{1}{2}\overrightarrow{AB}\) (do M là trung điểm AB)
\(\overrightarrow{AK}=\frac{1}{2}\left(\overrightarrow{AM}+\overrightarrow{AN}\right)=\frac{1}{2}\left(\frac{1}{2}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\right)=\frac{1}{4}\overrightarrow{AB}+\frac{1}{6}\overrightarrow{AC}\)
\(\Rightarrow\left\{{}\begin{matrix}m=\frac{1}{4}\\n=\frac{1}{6}\end{matrix}\right.\)
MN là đường trung bình của tam giác ABC
\(\Rightarrow\overrightarrow{MN}=\dfrac{1}{2}\overrightarrow{BC}=\dfrac{1}{2}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=-\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\)
Từ giả thiết:
\(\overrightarrow{KM}=-2\overrightarrow{KN}=-2\left(\overrightarrow{KM}+\overrightarrow{MN}\right)\)
\(\Rightarrow3\overrightarrow{KM}=2\overrightarrow{NM}\Rightarrow\overrightarrow{KM}=\dfrac{2}{3}\overrightarrow{NM}\)
\(\Rightarrow\overrightarrow{MK}=\dfrac{2}{3}\overrightarrow{MN}=\dfrac{2}{3}\left(-\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\right)=-\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\)
M là trung điểm AB \(\Rightarrow\overrightarrow{AM}=\dfrac{1}{2}\overrightarrow{AB}\)
Do đó:
\(\overrightarrow{AK}=\overrightarrow{AM}+\overrightarrow{MK}=\dfrac{1}{2}\overrightarrow{AB}-\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}=\dfrac{1}{6}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{AC}\)
a) \(\overrightarrow {AB} .\overrightarrow {AC} = 2.3.\cos \widehat {BAC} = 6.\cos {60^o} = 3\)
b)
Ta có: \(\overrightarrow {AB} + \overrightarrow {AC} = 2\overrightarrow {AM} \)(do M là trung điểm của BC)
\( \Leftrightarrow \overrightarrow {AM} = \frac{1}{2}\overrightarrow {AB} + \frac{1}{2}\overrightarrow {AC} \)
+) \(\overrightarrow {BD} = \overrightarrow {AD} - \overrightarrow {AB} = \frac{7}{{12}}\overrightarrow {AC} - \overrightarrow {AB} \)
c) Ta có:
\(\begin{array}{l}\overrightarrow {AM} .\overrightarrow {BD} = \left( {\frac{1}{2}\overrightarrow {AB} + \frac{1}{2}\overrightarrow {AC} } \right)\left( {\frac{7}{{12}}\overrightarrow {AC} - \overrightarrow {AB} } \right)\\ = \frac{7}{{24}}\overrightarrow {AB} .\overrightarrow {AC} - \frac{1}{2}{\overrightarrow {AB} ^2} + \frac{7}{{24}}{\overrightarrow {AC} ^2} - \frac{1}{2}\overrightarrow {AC} .\overrightarrow {AB} \\ = - \frac{1}{2}A{B^2} + \frac{7}{{24}}A{C^2} - \frac{5}{{24}}\overrightarrow {AB} .\overrightarrow {AC} \\ = - \frac{1}{2}{.2^2} + \frac{7}{{24}}{.3^2} - \frac{5}{{24}}.3\\ = 0\end{array}\)
\( \Rightarrow AM \bot BD\)
a) \(\overrightarrow {AC} + \overrightarrow {BD} = \overrightarrow {AM} + \overrightarrow {MN} + \overrightarrow {NC} + \overrightarrow {BM} + \overrightarrow {MN} + \overrightarrow {ND} \\= \left( {\overrightarrow {AM} + \overrightarrow {BM} } \right) + \left( {\overrightarrow {MN} + \overrightarrow {MN} } \right) + \left( {\overrightarrow {NC} + \overrightarrow {ND} } \right) \\= \overrightarrow 0 + 2\overrightarrow {MN} + \overrightarrow 0 = 2\overrightarrow {MN} \) (đpcm)
b) \(\overrightarrow {AC} + \overrightarrow {BD} = \overrightarrow {BC} + \overrightarrow {AD} \)
\(\)\(\overrightarrow {BC} + \overrightarrow {AD} = \overrightarrow {BM} + \overrightarrow {MN} + \overrightarrow {NC} + \overrightarrow {AM} + \overrightarrow {MN} + \overrightarrow {ND} \)
\(\left( {\overrightarrow {BM} + \overrightarrow {AM} } \right) + \left( {\overrightarrow {MN} + \overrightarrow {MN} } \right) + \left( {\overrightarrow {NC} + \overrightarrow {ND} } \right) = 2\overrightarrow {MN} \)
Mặt khác ta có: \(\overrightarrow {AC} + \overrightarrow {BD} = 2\overrightarrow {MN} \)
Suy ra \(\overrightarrow {AC} + \overrightarrow {BD} = \overrightarrow {BC} + \overrightarrow {AD} \)
Cách 2:
\(\begin{array}{l}
\overrightarrow {AC} + \overrightarrow {BD} = \overrightarrow {BC} + \overrightarrow {AD} \\
\Leftrightarrow \overrightarrow {AC} - \overrightarrow {AD} = \overrightarrow {BC} - \overrightarrow {BD} \\
\Leftrightarrow \overrightarrow {DC} = \overrightarrow {DC} (đpcm)
\end{array}\)
\(\overrightarrow{KD}=\overrightarrow{KA}+\overrightarrow{AD}=-\overrightarrow{AK}+\overrightarrow{AD}\)
\(=-\frac{1}{2}\left(\overrightarrow{AM}+\overrightarrow{AN}\right)+\frac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)
\(=-\frac{1}{2}\left(\frac{1}{2}\overrightarrow{AB}+\frac{2}{3}\overrightarrow{AC}\right)+\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AC}\)
\(=\frac{1}{4}\overrightarrow{AB}+\frac{1}{6}\overrightarrow{AC}\)