ข้อสอบ A-Level คณิตประยุกต์ 1 ปี 2564 (หลักสูตรใหม่)

$$\begin{gathered} \mathrm{จากสูตร} \mathrm{พื้นที่สามเหลี่ยม} = \frac{1}{2}\left|\stackrel{\rightharpoonup }{AB}\times \stackrel{\rightharpoonup }{AC}\right| \\ \mathrm{จากโจทย์} \mathrm{จุดยอดที่} A\left(-2, -4, -4\right) , B\left(0, -2, 0\right) \mathrm{และ} C\left(0, 0, 2\right) \\ \mathrm{จะได้} \stackrel{\rightharpoonup }{AB} = 2\stackrel{\rightharpoonup }{i}+2\stackrel{\rightharpoonup }{j}+4\stackrel{\rightharpoonup }{k} \\ \stackrel{\rightharpoonup }{AC} = 2\stackrel{\rightharpoonup }{i}+4\stackrel{\rightharpoonup }{j}+6\stackrel{\rightharpoonup }{k} \end{gathered}$$

$$\begin{gathered} \mathrm{แสดงว่า} \stackrel{\rightharpoonup }{AB}\times \stackrel{\rightharpoonup }{AC} = \left|\begin{array}{ccc}i& j& k\\ 2& 2& 4\\ 2& 4& 6\end{array}\right| \\ = \left(12\stackrel{\rightharpoonup }{i}+8\stackrel{\rightharpoonup }{j}+8\stackrel{\rightharpoonup }{k}\right)-\left(4\stackrel{\rightharpoonup }{k}+16\stackrel{\rightharpoonup }{i}+12\stackrel{\rightharpoonup }{j}\right) \\ = -4\stackrel{\rightharpoonup }{i}-4\stackrel{\rightharpoonup }{j}+4\stackrel{\rightharpoonup }{k} \end{gathered}$$

$$\begin{gathered} |\stackrel{\rightharpoonup }{AB}\times \stackrel{\rightharpoonup }{AC} | = \sqrt{{(-4)}^2+{(-4)}^2+4^2} \\ = \sqrt{16+16+16} = \sqrt{48} \\ = 4\sqrt{3} \\ \mathrm{ดังนั้น} \mathrm{รูปสามเหลี่ยม} ABC \mathrm{มีพื้นที่} = \frac{1}{2} |\stackrel{\rightharpoonup }{AB}\times \stackrel{\rightharpoonup }{AC} | \\ = \frac{1}{2}\left(4\sqrt{3}\right) = 2\sqrt{3} \mathrm{ตารางหน่วย} \end{gathered}$$