ข้อสอบ A-Level คณิตประยุกต์ 1 ปี 2565 (หลักสูตรใหม่)

$$\begin{gathered} \mathrm{จากโจทย์} A\left(0, 3, 2\right) B\left(1, -1, 0\right) \\ \mathrm{จะได้} \overline{AB} = \left(1, -1, 0\right)-\left(0, 3, 2\right) \\ = 1\stackrel{\rightharpoonup }{i}-4\stackrel{\rightharpoonup }{j}-2\stackrel{\rightharpoonup }{k} \\ \mathrm{จากโจทย์} C\left(2, 1, 3\right) D\left(x, 5, 1\right) \\ \mathrm{จะได้} \overline{CD} = \left(x, 5, 1\right)-\left(2, 1, 3\right) \\ = \left(x-2\right)\stackrel{\rightharpoonup }{i}+4\stackrel{\rightharpoonup }{j}-2\stackrel{\rightharpoonup }{k} \end{gathered}$$

$$\begin{gathered} \mathrm{และจากโจทย์เราทราบว่า} \overline{AB} \mathrm{ตั้งฉากกับ} \overline{CD} \mathrm{หมายความว่า} \overline{AB} ·\overline{CD}=0 \\ \overline{AB} ·\overline{CD} = \left(1\stackrel{\rightharpoonup }{i}-4\stackrel{\rightharpoonup }{j}-2\stackrel{\rightharpoonup }{k}\right)·\left(\left(x-2\right)\stackrel{\rightharpoonup }{i}+4\stackrel{\rightharpoonup }{j}-2\stackrel{\rightharpoonup }{k}\right) \\ 0 = \left(x-2\right)-16+4 \\ 0 = x-2-12 \\ x = 14 \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้ว่า} \mathrm{จุด} D\left(14, 5, 1\right) \mathrm{และเราทราบอยู่แล้วว่า} A\left(0, 3, 2\right) \\ \mathrm{หา} \overline{AD} \mathrm{จาก} \overline{AD} = \left(14, 5, 1\right)-\left(0, 3, 2\right) \\ = 14\stackrel{\rightharpoonup }{i}+2\stackrel{\rightharpoonup }{j}-1\stackrel{\rightharpoonup }{k} \\ \mathrm{ขนาด} \overline{AD} = \sqrt{{\left(14\right)}^2+{\left(2\right)}^2+{\left(-1\right)}^2} \\ = \sqrt{201} \\ \mathrm{ดังนั้น} \mathrm{ขนาด} \overline{AD} \mathrm{เท่ากับ} \sqrt{201} \end{gathered}$$