ข้อสอบ PAT 1 - มีนาคม 2564

$$\begin{gathered} \mathrm{จากโจทย์} \overline{O{P}_2} = \left[\begin{array}{c}-6\\ 3\\ 6\end{array}\right] \mathrm{และ} \overline{O{P}_3} = \left[\begin{array}{c}7\\ 4\\ -4\end{array}\right] \\ \mathrm{จะได้} \overline{O{P}_2} \times \overline{O{P}_3} = \left(-6\overline{i}+3\overline{j}+6\overline{k}\right)\times \left(7\overline{i}+4\overline{j}-4\overline{k}\right) \\ =\left|\begin{array}{ccc}1& j& k\\ -6& 3& 6\\ 7& 4& -4\end{array}\right| = \left|\begin{array}{cc}3& 6\\ 4& -4\end{array}\right|\overline{i} - \left|\begin{array}{cc}-6& 6\\ 7& -4\end{array}\right|\overline{j} +\left|\begin{array}{cc}-6& 3\\ 7& 4\end{array}\right|\overline{k} \end{gathered}$$

           $$\begin{gathered} = \left(-12-24\right)\overline{i}-\left(24-42\right)\overline{j}+\left(-24-21\right)\overline{k} \\ = -36\overline{i}+18\overline{j}-45\overline{k} \\ = -9\left(4\overline{i}-2\overline{j}+5\overline{k}\right) = -9\left[\begin{array}{c}4\\ -2\\ 5\end{array}\right] \end{gathered}$$

$$\begin{gathered} \mathrm{แสดงว่า} \left|\overline{O{P}_2}\times \overline{O{P}_3}\right| = \left|-9\left(4\overline{i}-2\overline{j}+5\overline{k}\right)\right| \\ = 9\sqrt{4^2+{\left(-2\right)}^2+5^2} = 9\sqrt{45} \\ = 27\sqrt{5} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} \overline{O{P}_1} = \left[\begin{array}{c}6\\ -6\\ 3\end{array}\right] \\ \mathrm{จะได้} \overline{O{P}_1} = 6\overline{i}-6\overline{j}+3\overline{k} \\ = 3\left(2\overline{i}-2\overline{j}+\overline{k}\right) \\ \left|\overline{O{P}_1}\right| = \left| 3\left(2\overline{i}-2\overline{j}+\overline{k}\right)\right| \\ = 3\sqrt{2^2+{\left(-2\right)}^2+1^2} = 3\sqrt{9} \\ = 9 \end{gathered}$$

$$\begin{gathered} \mathrm{จากสูตร} \overline{u}·\overline{v} = \left|\overline{u}\right|\left|\overline{v}\right|\cos \theta \\ \mathrm{จากโจทย์} \theta \mathrm{เป็นมุมระหว่าง} \overline{O{P}_1} \mathrm{และ} \overline{O{P}_2}\times \overline{O{P}_3} \\ \mathrm{จะได้} \left( \overline{O{P}_1} \right)·\left( \overline{O{P}_2}\times \overline{O{P}_3}\right) = \left|\overline{O{P}_1} \right|\left|\overline{O{P}_2}\times \overline{O{P}_3} \right|\cos \theta \\ \left[\begin{array}{c}6\\ -6\\ 3\end{array}\right]·\left(-9\right)\left[\begin{array}{c}4\\ -2\\ 5\end{array}\right] = 9\left(27\sqrt{5}\right)\cos \theta \end{gathered}$$

$$\begin{gathered} -9\left[6\left(4\right)+\left(-6\right)\left(-2\right)+3\left(5\right)\right]= 9\left(27\sqrt{5}\right)\cos \theta \\ -\cancel{9}\left(51\right) = \cancel{9}\left(27\sqrt{5}\right)\cos \theta \\ \cos \theta = \frac{-17}{9\sqrt{5}} \to \cos \theta <0 \end{gathered}$$

$$\begin{gathered} \cos \theta \mathrm{ติดลบ} \to \theta \mathrm{อยู่} Q_2, Q_3 \\ \mathrm{แต่เวกเตอร์} \mathrm{พิจารณาแค่} Q_1, Q_2 \mathrm{เท่านั้น} \\ \mathrm{ดังนั้น} \theta \mathrm{อยู่} Q_2 \to 90°<\theta <180° \end{gathered}$$