ข้อสอบ PAT 1 - พฤศจิกายน 2557

$$\begin{gathered} \mathrm{จากโจทย์} \overline{\mathrm{u}}=\overline{\mathrm{AB}} \left|\overline{\mathrm{u}}\right|=5 \\ \overline{\mathrm{v}}=\overline{\mathrm{BC}} \left|\overline{\mathrm{v}}\right|=12 \\ \mathrm{วาดรูปสามเหลี่ยม} \mathrm{ABC} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} \overline{\mathrm{w}}=\overline{\mathrm{CA}} \\ =\overline{\mathrm{CB}}+\overline{\mathrm{BA}} \\ =-\overline{\mathrm{BC}}+-\overline{\mathrm{AB}}\hspace{0.17em}\hspace{0.17em} ;\hspace{0.17em}\mathrm{แทน} \overline{\mathrm{u}}=\overline{\mathrm{AB}} , \overline{\mathrm{v}}=\overline{\mathrm{BC}} \\ =-\overline{\mathrm{v}}+\left(-\overline{\mathrm{u}}\right) \\ =-\overline{\mathrm{v}}-\overline{\mathrm{u}} \end{gathered}$$

$$\begin{gathered} \mathrm{หาค่าของ} \left(2\overline{\mathrm{u}}-\overline{\mathrm{v}}\right)·\overline{\mathrm{w}} ; \mathrm{แทน} \overline{\mathrm{w}}=-\overline{\mathrm{v}}-\overline{\mathrm{u}} \\ \mathrm{จะได้} \left(2\overline{\mathrm{u}}-\overline{\mathrm{v}}\right)·\overline{\mathrm{w}}=\left(2\overline{\mathrm{u}}-\overline{\mathrm{v}}\right)·\left(-\overline{\mathrm{v}}-\overline{\mathrm{u}}\right) \\ \mathrm{dot} \mathrm{กระจาย} \\ =-2\overline{\mathrm{u}}·\overline{\mathrm{v}}-2\overline{\mathrm{u}}·\overline{\mathrm{u}}+\overline{\mathrm{v}}·\overline{\mathrm{v}}+\overline{\mathrm{v}}·\overline{\mathrm{u}} \\ \hspace{0.17em}\mathbf{โดย} \overline{\mathrm{v}}·\overline{\mathrm{u}}=\overline{\mathrm{u}}·\overline{\mathrm{v}} , \overline{\mathrm{u}}·\overline{\mathrm{u}}={\left|\overline{\mathrm{u}}\right|}^2 , \overline{\mathrm{v}}·\overline{\mathrm{v}}={\left|\stackrel{\rightharpoonup }{\mathrm{v}}\right|}^2 \end{gathered}$$
$$\begin{gathered} =-2\overline{\mathrm{u}}·\overline{\mathrm{v}}-2{\left|\overline{\mathrm{u}}\right|}^2+{\left|\overline{\mathrm{v}}\right|}^2+\overline{\mathrm{u}}·\overline{\mathrm{v}} \\ =-\overline{\mathrm{u}}·\overline{\mathrm{v}}-2{\left|\overline{\mathrm{u}}\right|}^2+{\left|\overline{\mathrm{v}}\right|}^2 \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{สูตร} \overline{\mathrm{u}}·\overline{\mathrm{v}}=\left|\overline{\mathrm{u}}\right|\left|\overline{\mathrm{v}}\right|\mathrm{cos\theta } \\ =-\left|\overline{\mathrm{u}}\right|\left|\overline{\mathrm{v}}\right|\mathrm{cos\theta }-2{\left|\overline{\mathrm{u}}\right|}^2+{\left|\overline{\mathrm{v}}\right|}^2 \\ \mathrm{\theta } \mathrm{เป็นมุมที่} \mathrm{u} \mathrm{ทำกับ} \mathrm{v} \left(\mathrm{หางต่อหาง} \mathrm{ดูรูปภาพประกอบ}\right) \end{gathered}$$


$$\begin{gathered} -\mathrm{แทน} \left|\overline{\mathrm{u}}\right|=5 , \left|\overline{\mathrm{v}}\right|=12 , \mathrm{\theta }=120° \\ \mathrm{ดังนั้น} \left(2\overline{\mathrm{u}}-\overline{\mathrm{v}}\right)·\overline{\mathrm{w}}=-\left(5\right)\left(12\right)(-\frac{1}{2})-2{\left(5\right)}^2+{\left(12\right)}^2 \\ =30-50+144 \\ =124 \end{gathered}$$