ข้อสอบคณิต 9 สามัญ - ปี 2564

$$\begin{gathered} \mathrm{จากสูตร} \stackrel{\rightharpoonup }{u}·\stackrel{\rightharpoonup }{v} = \left|\stackrel{\rightharpoonup }{u}\right|\left|\stackrel{\rightharpoonup }{v}\right|\cos \theta \\ \cos \theta = \frac{\stackrel{\rightharpoonup }{u}·\stackrel{\rightharpoonup }{v}}{\left|\stackrel{\rightharpoonup }{u}\right|\left|\stackrel{\rightharpoonup }{v}\right|} \end{gathered}$$

$$\begin{gathered} ก) \mathrm{จากโจทย์} \mathrm{มุมระหว่าง} 3\stackrel{\rightharpoonup }{u} \mathrm{และ} 3\stackrel{\rightharpoonup }{v} \mathrm{มีขนาดเป็น} 3 \mathrm{เท่าของมุมระหว่าง} \stackrel{\rightharpoonup }{u} \mathrm{และ} \stackrel{\rightharpoonup }{v} \\ - \mathrm{หามุมระหว่าง} \stackrel{\rightharpoonup }{u} \mathrm{และ} \stackrel{\rightharpoonup }{v} \\ \mathrm{จาก} \overline{)1} \mathrm{จะได้} \cos \theta = \frac{\stackrel{\rightharpoonup }{u}·\stackrel{\rightharpoonup }{v}}{\left|\stackrel{\rightharpoonup }{u}\right|\left|\stackrel{\rightharpoonup }{v}\right|} \end{gathered}$$

$$\begin{gathered} -\mathrm{หามุมระหว่าง} 3\stackrel{\rightharpoonup }{u} \mathrm{และ} 3\stackrel{\rightharpoonup }{v} \\ \mathrm{จาก} \overline{)1} \mathrm{จะได้} \cos \theta = \frac{\left(3\stackrel{\rightharpoonup }{u}\right)·\left(3\stackrel{\rightharpoonup }{v}\right)}{\left|3\stackrel{\rightharpoonup }{u}\right|\left|3\stackrel{\rightharpoonup }{v}\right|} \\ \cos \theta = \frac{\cancel{9}\stackrel{\rightharpoonup }{u}·\stackrel{\rightharpoonup }{v}}{\cancel{9}\left|\stackrel{\rightharpoonup }{u}\right|\left|\stackrel{\rightharpoonup }{v}\right|} = \frac{\stackrel{\rightharpoonup }{u}·\stackrel{\rightharpoonup }{v}}{\left|\stackrel{\rightharpoonup }{u}\right|\left|\stackrel{\rightharpoonup }{v}\right|} \\ \mathrm{ดังนั้น} \mathrm{มุมระหว่าง} \stackrel{\rightharpoonup }{u} \mathrm{และ} \stackrel{\rightharpoonup }{v} \mathrm{มีขนาดเท่ากับมุมระหว่าง} 3\stackrel{\rightharpoonup }{u} \mathrm{และ} 3\stackrel{\rightharpoonup }{v} \\ \mathrm{ข้อ} ก) \mathrm{ผิด} \end{gathered}$$

$$\begin{gathered} ข) \mathrm{จากโจทย์} \mathrm{มุมระหว่าง} \stackrel{\rightharpoonup }{u}-\stackrel{\rightharpoonup }{v} \mathrm{และ} \stackrel{\rightharpoonup }{v}-\stackrel{\rightharpoonup }{u} \mathrm{มีขนาด} 180° \\ \mathrm{กำหนดให้} \stackrel{\rightharpoonup }{a} = \stackrel{\rightharpoonup }{u}-\stackrel{\rightharpoonup }{v} \mathrm{และ} -\stackrel{\rightharpoonup }{a} = -\left(\stackrel{\rightharpoonup }{u}-\stackrel{\rightharpoonup }{v}\right) \\ -\stackrel{\rightharpoonup }{a} = \stackrel{\rightharpoonup }{v}-\stackrel{\rightharpoonup }{u} \\ \mathrm{จะได้} \mathrm{มุมระหว่าง} \stackrel{\rightharpoonup }{u}-\stackrel{\rightharpoonup }{v} \mathrm{และ} \stackrel{\rightharpoonup }{v}-\stackrel{\rightharpoonup }{u} \mathrm{คือ} \\ \mathrm{มุมระหว่าง} \stackrel{\rightharpoonup }{a} \mathrm{และ} -\stackrel{\rightharpoonup }{a} \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \overline{)1} \mathrm{จะได้} \cos \theta = \frac{\stackrel{\rightharpoonup }{a}·\left(-\stackrel{\rightharpoonup }{a} \right)}{\left|\stackrel{\rightharpoonup }{a}\right|\left|-\stackrel{\rightharpoonup }{a}\right|} \\ \cos \theta = \frac{-\stackrel{\rightharpoonup }{a}·\stackrel{\rightharpoonup }{a}}{\left|\stackrel{\rightharpoonup }{a}\right|\left|\stackrel{\rightharpoonup }{a}\right|} = \frac{-\cancel{{\left|\stackrel{\rightharpoonup }{a}\right|}^2}}{\cancel{{\left|\stackrel{\rightharpoonup }{a}\right|}^2}} \\ \cos \theta = -1 \\ \mathrm{แสดงว่า} \theta = 180° \left(\cos 180°=-1\right) \end{gathered}$$

$$\begin{gathered} \mathrm{ดังนั้น} \mathrm{มุมระหว่าง} \stackrel{\rightharpoonup }{u}-\stackrel{\rightharpoonup }{v} \mathrm{และ} \stackrel{\rightharpoonup }{v}-\stackrel{\rightharpoonup }{u} \mathrm{มีขนาด} 180° \\ \mathrm{ข้อ} ข) \mathrm{ถูก} \end{gathered}$$

$$\begin{gathered} ค) \mathrm{จากโจทย์} \mathrm{มุมระหว่าง} \stackrel{\rightharpoonup }{u}+\stackrel{\rightharpoonup }{v} \mathrm{และ} \stackrel{\rightharpoonup }{u}-\stackrel{\rightharpoonup }{v} \mathrm{มีขนาด} 90° \\ \mathrm{จาก} \overline{)1} \mathrm{จะได้} \cos \theta = \frac{\left(\stackrel{\rightharpoonup }{u}+\stackrel{\rightharpoonup }{v}\right)·\left(\stackrel{\rightharpoonup }{u}-\stackrel{\rightharpoonup }{v}\right)}{\left|\stackrel{\rightharpoonup }{u}+\stackrel{\rightharpoonup }{v}\right|\left|\stackrel{\rightharpoonup }{u}-\stackrel{\rightharpoonup }{v}\right|} \\ \cos \theta = \frac{\stackrel{\rightharpoonup }{u}·\stackrel{\rightharpoonup }{u}-\stackrel{\rightharpoonup }{u}·\stackrel{\rightharpoonup }{v}+\stackrel{\rightharpoonup }{v}·\stackrel{\rightharpoonup }{u}-\stackrel{\rightharpoonup }{v}·\stackrel{\rightharpoonup }{v}}{\left|\stackrel{\rightharpoonup }{u}+\stackrel{\rightharpoonup }{v}\right|\left|\stackrel{\rightharpoonup }{u}-\stackrel{\rightharpoonup }{v}\right|} \\ -\mathrm{โดย} \stackrel{\rightharpoonup }{u}·\stackrel{\rightharpoonup }{v} =\stackrel{\rightharpoonup }{v}·\stackrel{\rightharpoonup }{u} , \stackrel{\rightharpoonup }{u}·\stackrel{\rightharpoonup }{u} = {\left|\stackrel{\rightharpoonup }{u}\right|}^2 \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} \cos \theta = \frac{{\left|\stackrel{\rightharpoonup }{u}\right|}^2-{\left|\stackrel{\rightharpoonup }{v}\right|}^2}{\left|\stackrel{\rightharpoonup }{u}+\stackrel{\rightharpoonup }{v}\right|\left|\stackrel{\rightharpoonup }{u}-\stackrel{\rightharpoonup }{v}\right|} \\ \mathrm{จากโจทย์} \left|\stackrel{\rightharpoonup }{u}\right| \ne \left|\stackrel{\rightharpoonup }{v}\right| \\ \mathrm{แสดงว่า} \cos \theta \ne 0 \left(\cos 90° = 0\right) \\ \mathrm{ดังนั้น} \mathrm{มุมระหว่าง} \stackrel{\rightharpoonup }{u}+\stackrel{\rightharpoonup }{v} \mathrm{และ} \stackrel{\rightharpoonup }{u}-\stackrel{\rightharpoonup }{v} \mathrm{ไม่ตั้งฉากกัน} \\ \mathrm{ข้อ} ค) \mathrm{ผิด} \end{gathered}$$