105指考數學甲選填第9題詳解

1. 先計算 $\overset{\large\rightharpoonup}{u}\times\overset{\large\rightharpoonup}{v}$： $$\overset{\large\rightharpoonup}{u}\times\overset{\large\rightharpoonup}{v} = \begin{vmatrix} \overset{\large\rightharpoonup}{i} & \overset{\large\rightharpoonup}{j} & \overset{\large\rightharpoonup}{k} \\ 1 & 2 & 3 \\ 1 & 0 & -1 \end{vmatrix} = (2\cdot(-1)-3\cdot 0,\; 3\cdot 1-1\cdot(-1),\; 1\cdot 0-2\cdot 1) = (-2,\; 4,\; -2)$$ 2. 因為 $\overset{\large\rightharpoonup}{w}$ 與 $\overset{\large\rightharpoonup}{u}\times\overset{\large\rightharpoonup}{v}$ 平行，所以可設 $\overset{\large\rightharpoonup}{w}=t(-2,4,-2)=(-2t,4t,-2t)$，其中 $t$ 為實數。 3. 將 $\overset{\large\rightharpoonup}{w}$ 代入行列式條件： $$\begin{vmatrix} 1 & 2 & 3 \\ 1 & 0 & -1 \\ -2t & 4t & -2t \end{vmatrix} = -12$$ 因為三階行列式的值等於 $\overset{\large\rightharpoonup}{w}\cdot(\overset{\large\rightharpoonup}{u}\times\overset{\large\rightharpoonup}{v})$，且 $\overset{\large\rightharpoonup}{w}\parallel\overset{\large\rightharpoonup}{u}\times\overset{\large\rightharpoonup}{v}$，所以此行列式值 $=t|\overset{\large\rightharpoonup}{u}\times\overset{\large\rightharpoonup}{v}|^2$： $$|\overset{\large\rightharpoonup}{u}\times\overset{\large\rightharpoonup}{v}|^2 = (-2)^2+4^2+(-2)^2 = 4+16+4 = 24$$ 所以 $t\cdot 24 = -12$，得 $t=-\frac{1}{2}$。 4. 因此 $\overset{\large\rightharpoonup}{w} = -\frac{1}{2}(-2,4,-2) = (1,-2,1)$。 故填 $(1,-2,1)$。