（1）根據向量減法： $$\overset{\large\rightharpoonup}{DC} = \overset{\large\rightharpoonup}{OC} - \overset{\large\rightharpoonup}{OD}$$ 代入已知條件 $\overset{\large\rightharpoonup}{OC} = 3\overset{\large\rightharpoonup}{OA}$ 與 $\overset{\large\rightharpoonup}{OD} = 3\overset{\large\rightharpoonup}{OB}$： $$\overset{\large\rightharpoonup}{DC} = 3\overset{\large\rightharpoonup}{OA} - 3\overset{\large\rightharpoonup}{OB} = 3(\overset{\large\rightharpoonup}{OA} - \overset{\large\rightharpoonup}{OB}) = 3\overset{\large\rightharpoonup}{BA} = -3\overset{\large\rightharpoonup}{AB}$$ 已知 $\overset{\large\rightharpoonup}{AB} = (3, -4)$，故： $$\overset{\large\rightharpoonup}{DC} = -3(3, -4) = (-9, 12)$$ （2）已知 $\overset{\large\rightharpoonup}{OA} = (1, 2)$ 且 $\overset{\large\rightharpoonup}{AB} = \overset{\large\rightharpoonup}{OB} - \overset{\large\rightharpoonup}{OA} = (3, -4)$： $$\overset{\large\rightharpoonup}{OB} = \overset{\large\rightharpoonup}{OA} + \overset{\large\rightharpoonup}{AB} = (1, 2) + (3, -4) = (4, -2)$$ 由於 $\overset{\large\rightharpoonup}{OC} = 3\overset{\large\rightharpoonup}{OA} = (3, 6)$ 且 $\overset{\large\rightharpoonup}{OD} = 3\overset{\large\rightharpoonup}{OB} = (12, -6)$，$\triangle OCD$ 的面積為： $$\text{Area} = \dfrac{1}{2} \left| \det \begin{bmatrix} 3 & 6 \\ 12 & -6 \end{bmatrix} \right| = \dfrac{1}{2} |3(-6) - 6(12)| = \dfrac{1}{2} |-18 - 72| = \dfrac{90}{2} = 45$$ 故 $\triangle OCD$ 的面積為 $45$。