平行四辺形ABCDにおいて、以下の内積を求める問題です。 (1) $\overrightarrow{AB} \cdot \overrightarrow{AF}$ (2) $\overrightarrow{EA} \cdot \overrightarrow{EF}$ (3) $\overrightarrow{FE} \cdot \overrightarrow{FD}$ (4) $\overrightarrow{EB} \cdot \overrightarrow{EC}$

1. 問題の内容

平行四辺形ABCDにおいて、以下の内積を求める問題です。

(1) $\overrightarrow{AB} \cdot \overrightarrow{AF}$

(2) $\overrightarrow{EA} \cdot \overrightarrow{EF}$

(3) $\overrightarrow{FE} \cdot \overrightarrow{FD}$

(4) $\overrightarrow{EB} \cdot \overrightarrow{EC}$

2. 解き方の手順

(1) $\overrightarrow{AB} \cdot \overrightarrow{AF}$

$\overrightarrow{AB} \cdot \overrightarrow{AF} = |\overrightarrow{AB}| |\overrightarrow{AF}| \cos{\angle BAF}$

$|\overrightarrow{AB}| = 2$

$\angle BAF = 90^{\circ} - 60^{\circ} = 30^{\circ}$

$AF = AB \sin{60^{\circ}} = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}$

$|\overrightarrow{AF}| = \sqrt{3}$

$\overrightarrow{AB} \cdot \overrightarrow{AF} = 2 \times \sqrt{3} \times \cos{30^{\circ}} = 2 \times \sqrt{3} \times \frac{\sqrt{3}}{2} = 3$

(2) $\overrightarrow{EA} \cdot \overrightarrow{EF}$

$\overrightarrow{EA} \cdot \overrightarrow{EF} = |\overrightarrow{EA}| |\overrightarrow{EF}| \cos{\angle AEF}$

$AE = AB \cos{60^{\circ}} = 2 \times \frac{1}{2} = 1$

$|\overrightarrow{EA}| = 1$

$\angle AEF = 180^{\circ} - 90^{\circ} - 60^{\circ} = 30^{\circ}$

$EF = 2EC = 2\sqrt{3}$

$|\overrightarrow{EF}| = 2 \sqrt{3}$

$\overrightarrow{EA} \cdot \overrightarrow{EF} = 1 \times 2 \sqrt{3} \times \cos{30^{\circ}} = 2 \sqrt{3} \times \frac{\sqrt{3}}{2} = 3$

(3) $\overrightarrow{FE} \cdot \overrightarrow{FD}$

$\overrightarrow{FE} \cdot \overrightarrow{FD} = |\overrightarrow{FE}| |\overrightarrow{FD}| \cos{\angle EFD}$

$|\overrightarrow{FE}| = 2 \sqrt{3}$

$FD = \sqrt{3}$

$|\overrightarrow{FD}| = \sqrt{3}$

$\angle FDC = 60^{\circ}$

$\angle DFC = 90^{\circ}$

$\angle FCD = 30^{\circ}$

よって$\angle EFD = \angle AFC = 90^{\circ}-60^{\circ} = 30^{\circ}$

$\overrightarrow{FE} \cdot \overrightarrow{FD} = 2 \sqrt{3} \times \sqrt{3} \times \cos{150^{\circ}} = 2 \sqrt{3} \times \sqrt{3} \times (-\frac{\sqrt{3}}{2}) = 6 \times (-\frac{\sqrt{3}}{2}) = -3 \sqrt{3}$

$\angle AFD=180-(90+60) = 30$

$\angle EFD = 60 + \angle AFD = 60 +30 =90$

$\overrightarrow{FE} \cdot \overrightarrow{FD} = |\overrightarrow{FE}| |\overrightarrow{FD}| \cos{\angle EFD}$

$\overrightarrow{FE} \cdot \overrightarrow{FD} = 2\sqrt{3}*\sqrt{3} * \cos(90)$

$\overrightarrow{FE} \cdot \overrightarrow{FD} = 2\sqrt{3}*\sqrt{3}*0$

$\overrightarrow{FE} \cdot \overrightarrow{FD} = 0$

(4) $\overrightarrow{EB} \cdot \overrightarrow{EC}$

$\overrightarrow{EB} \cdot \overrightarrow{EC} = |\overrightarrow{EB}| |\overrightarrow{EC}| \cos{\angle BEC}$

$|\overrightarrow{EB}| = \sqrt{3}$

$|\overrightarrow{EC}| = \sqrt{3}$

$\angle BEC = 180^{\circ} - 60^{\circ} = 120^{\circ}$

$\overrightarrow{EB} \cdot \overrightarrow{EC} = \sqrt{3} \times \sqrt{3} \times \cos{120^{\circ}} = 3 \times (-\frac{1}{2}) = -\frac{3}{2}$

3. 最終的な答え

(1) 3

(2) 3

(3) 0

(4) $-\frac{3}{2}$