108指考數乙的平面向量與行列式計算面積問題

=問題= 

考慮座標平面上相異五點$O, A, B, C, D$。已知向量$\overrightarrow{OC} = 3\overrightarrow{OA}, \overrightarrow{OD} = 3\overrightarrow{OB}$，且向量$\overrightarrow{AB}$的座標表示為$\overrightarrow{AB} = \left[ \begin{array}{c} 3 \\ -4 \end{array} \right]$。試回答下列問題：

(1) 試以座標表示$\overrightarrow{DC}$。

(2) 若$\overrightarrow{OA} = \left[ \begin{array}{c} 1 \\ 2 \end{array} \right]$，試利用二階行列式與面積的關係，求$\triangle OCD$的面積。

=解答=

(1) 利用向量分解，得

$$\overrightarrow{DC} = \overrightarrow{DO} + \overrightarrow{OC} = \overrightarrow{OC} - \overrightarrow{OD} = 3\overrightarrow{OA} - 3\overrightarrow{OB} = 3(\overrightarrow{OA} - \overrightarrow{OB}) = 3(\overrightarrow{OA}+\overrightarrow{BO}) = 3\overrightarrow{BA} = -3\overrightarrow{AB} = \left[ \begin{array}{c} -9 \\ 12 \end{array} \right].$$

(2) 先求出$\overrightarrow{OB}$，

$$\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{AB} = \left[ \begin{array}{c} 1 \\ 2 \end{array} \right] + \left[ \begin{array}{c} 3 \\ -4 \end{array} \right] = \left[ \begin{array}{c} 4 \\ -2 \end{array} \right].$$

於是

$$\overrightarrow{OC} = 3\overrightarrow{OA} = 3\left[ \begin{array}{c} 1 \\ 2 \end{array} \right] = \left[ \begin{array}{c} 3 \\ 6 \end{array} \right],$$

$$\overrightarrow{OD} = 3\overrightarrow{OB} = 3 \left[ \begin{array}{c} 4 \\ -2 \end{array} \right] =  \left[ \begin{array}{c} 12 \\ -6 \end{array} \right].$$

因此

$$\triangle OCD = \frac{1}{2} \left| \det (\overrightarrow{OC}, \overrightarrow{OD}) \right| = \frac{1}{2} |\left| \begin{array}{cc} 3 & 12 \\ 6 & -6 \end{array} \right|| = 45.$$