ข้อสอบ A-Level คณิตประยุกต์ 1 ปี 2563 (วิชาสามัญเก่า)

$$\begin{gathered} \mathrm{จากโจทย์} \\ \stackrel{\rightharpoonup }{i} = \left[\begin{array}{c}1\\ 0\\ 0\end{array}\right] , \stackrel{\rightharpoonup }{j} = \left[\begin{array}{c}0\\ 1\\ 0\end{array}\right] , \stackrel{\rightharpoonup }{k} = \left[\begin{array}{c}0\\ 0\\ 1\end{array}\right] \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} \left(\stackrel{\rightharpoonup }{v}\times \stackrel{\rightharpoonup }{i}\right)·\left(\stackrel{\rightharpoonup }{j}+\stackrel{\rightharpoonup }{k}\right) \\ = \left(\left[\begin{array}{c}2\\ 3\\ 1\end{array}\right]\times \left[\begin{array}{c}1\\ 0\\ 0\end{array}\right]\right) · \left(\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]+\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]\right) \\ = \left[\begin{array}{c}\left(3\right)\left(0\right)-\left(1\right)\left(0\right)\\ \left(1\right)\left(1\right)-\left(2\right)\left(0\right)\\ \left(2\right)\left(0\right)-\left(3\right)\left(1\right)\end{array}\right] · \left[\begin{array}{c}0\\ 1\\ 1\end{array}\right] \end{gathered}$$

$$\begin{gathered} = \left[\begin{array}{c}0\\ 1\\ -3\end{array}\right] · \left[\begin{array}{c}0\\ 1\\ 1\end{array}\right] \\ = \left(0\right)\left(0\right)+\left(1\right)\left(1\right)+\left(-3\right)\left(1\right) \\ = -2 \\ \mathrm{ดังนั้น} \mathrm{ตอบ} 2 \end{gathered}$$