ข้อสอบ A-Level คณิตประยุกต์ 1 ปี 2566 (หลักสูตรใหม่)

$$\begin{gathered} \mathrm{จากรูป} A \mathrm{อยู่บนแกน} Y (\mathrm{แกน} y \mathrm{จะได้} x, z=0), \overline{AB} \mathrm{ขนานกับแกน} X \\ \mathrm{จะได้} \mathrm{จุด} A \mathrm{คือ} A\left(0, 1, 0\right) \\ \mathrm{จากรูป} D \mathrm{อยู่บนแกน} Y (\mathrm{แกน} y \mathrm{จะได้} x, z=0), \overline{DC} \mathrm{ขนานกับแกน} X \\ \mathrm{จะได้} \mathrm{จุด} D \mathrm{คือ} D\left(0, 2, 0\right) \\ \mathrm{จากรูป} E \mathrm{อยู่บนระนาบ} YZ, \overline{ED} \mathrm{ขนานกับ} \overline{FA}, \overline{EF} \mathrm{ขนานกับ} \overline{DA} \\ \mathrm{จะได้} \mathrm{จุด} E \mathrm{คือ} E\left(0, 2, 2\right) \\ \mathrm{จากสูตร} \overline{u}·\overline{v} = \left|\overline{u}\right|\left|\overline{v}\right|\cos \theta \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} A\hat{C}E = \theta \\ \mathrm{จะได้} \overline{CA}·\overline{CE} = \left|\overline{CA}\right|\left|\overline{CE}\right|\cos \theta \to \boxed{1} \end{gathered}$$

$- \mathrm{หาค่า} \overline{CA}, \left|\overline{CA}\right|$
$$\begin{gathered} \mathrm{จะได้} \overline{CA} = A\left(0, 1, 0\right)-C\left(1, 2, 0\right) \\ = \left[\begin{array}{c}0-1\\ 1-2\\ 0-0\end{array}\right] = \left[\begin{array}{c}-1\\ -1\\ 0\end{array}\right] \\ \mathrm{แสดงว่า} \left|\overline{CA}\right| = \sqrt{{\left(-1\right)}^2+{\left(-1\right)}^2+0^2} \\ = \sqrt{2} \end{gathered}$$

$$\begin{gathered} - \mathrm{หาค่า} \overline{CE}, \left|\overline{CE}\right| \\ \mathrm{จะได้} \overline{CE} = E\left(0, 2, 2\right)-C\left(1, 2, 0\right) \\ = \left[\begin{array}{c}0-1\\ 2-2\\ 2-0\end{array}\right]=\left[\begin{array}{c}-1\\ 0\\ 2\end{array}\right] \\ \mathrm{แสดงว่า} \left|\overline{CE}\right| = \sqrt{{\left(-1\right)}^2+0^2+2^2} \\ = \sqrt{5} \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \boxed{1} \overline{CA}·\overline{CE} = \left|\overline{CA}\right|\left|\overline{CE}\right|\cos \theta \\ \mathrm{จะได้} \left[\begin{array}{c}-1\\ -1\\ 0\end{array}\right]·\left[\begin{array}{c}-1\\ 0\\ 2\end{array}\right] = \sqrt{2}\sqrt{5}\cos \theta \end{gathered}$$

$$\begin{gathered} \left(-1\right)\left(-1\right)+\left(-1\right)\left(0\right)+0\left(2\right) = \sqrt{10}\cos \theta \\ 1+0+0 = \sqrt{10}\cos \theta \\ 1 = \sqrt{10}\cos \theta \\ \cos \theta = \frac{1}{\sqrt{10}} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} \mathrm{ค่าของ} sec\theta \mathrm{เท่ากับเท่าใด} \\ \mathrm{ดังนั้น} sec\theta = \frac{1}{\cos \theta } \\ sec\theta = \frac{1}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\sqrt{10}$}\right.}} = \sqrt{10} \end{gathered}$$