202601292

3.基底化

$在平行四边形 A B C D 中, A B = 4, A D = 3, 点 E 是 A B 的中点, 点 F 满足$
$\vec{B F} = 2 \vec{F C}, 且 D F = \sqrt{13}, 求 \vec{B F} \cdot \vec{D F}$

[解析]

用基底统一向量，由条件选择 $\vec{A B}, \vec{A D}$
$\vec{D F} = \vec{D C} + \vec{C F} = \vec{A B} - \frac{1}{3} \vec{A D}$
$\vec{E F} = \frac{1}{2} \vec{A B} + \frac{2}{3} \vec{A D}$
${\vec{D F}}^{2} = (\vec{A B} - \frac{1}{3} \vec{A D})^{2} = {\vec{A B}}^{2} - \frac{2}{3} \vec{A B} \cdot \vec{A D} + \frac{1}{9} {\vec{A D}}^{2} = 13$
$∴ \vec{A B} \cdot \vec{A D} = 6$

$\vec{B F} \cdot \vec{D F} = (\vec{A B} - \frac{1}{3} \vec{A D}) \cdot (\frac{1}{2} \vec{A B} + \frac{2}{3} \vec{A D}) = 9$