ข้อสอบ PAT 1 - เมษายน 2557

$$\begin{gathered} \mathrm{นิยาม} \left|\overline{\mathrm{u}}\right|=\left|-\overline{\mathrm{u}}\right| \\ \overline{\mathrm{u}}·\overline{\mathrm{u}}={\left|\overline{\mathrm{u}}\right|}^2 \\ \mathrm{จากโจทย์} \overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}=\overline{0} \\ \mathrm{จะได้} \overline{\mathrm{a}}+\overline{\mathrm{b}}=-\overline{\mathrm{c}} \\ \left|\overline{\mathrm{a}}+\overline{\mathrm{b}}\right|=\left|-\overline{\mathrm{c}}\right| ; \mathrm{โดย} \left|\overline{\mathrm{a}}+\overline{\mathrm{b}}\right|=5 \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}5=\left|\overline{\mathrm{c}}\right| \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์}\hspace{0.17em} \overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}=\overline{0} \\ \mathrm{จะได้} \overline{\mathrm{b}}+\overline{\mathrm{c}}=-\overline{\mathrm{a}} \\ \left|\overline{\mathrm{b}}+\overline{\mathrm{c}}\right|=\left|-\overline{\mathrm{a}}\right| ; \mathrm{โดย} \left|\overline{\mathrm{b}}+\overline{\mathrm{c}}\right|=3 \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}3=\left|\overline{\mathrm{a}}\right| \end{gathered}$$​

$$\begin{gathered} \mathrm{จากโจทย์} \overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}=\overline{0} \\ \mathrm{จะได้} \overline{\mathrm{a}}+\overline{\mathrm{c}}=-\overline{\mathrm{b}} \\ \left|\overline{\mathrm{a}}+\overline{\mathrm{c}}\right|=\left|-\overline{\mathrm{b}}\right| ; \mathrm{โดย} \left|\overline{\mathrm{b}}\right|=\sqrt{10} \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left|\overline{\mathrm{a}}+\mathrm{c}\right|=\sqrt{10} \\ \underline{\mathrm{พิจารณา} \left(\mathrm{ก}\right)} \\ \mathrm{จากโจทย์} \mathrm{เวกเตอร์} \overline{\mathrm{a}} \mathrm{ทำมุม} \mathrm{\theta } \mathrm{กับเวกเตอร์} \overline{\mathrm{b}} \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{เมื่อ} 0\le \mathrm{\theta }\le \mathrm{\pi } \\ \mathrm{จะได้} \left(\overline{\mathrm{a}}+\overline{\mathrm{b}}\right)·\left(\overline{\mathrm{a}}+\overline{\mathrm{b}}\right)=\overline{\mathrm{a}}·\overline{\mathrm{a}}+2\overline{\mathrm{a}}·\overline{\mathrm{b}}+\overline{\mathrm{b}}·\overline{\mathrm{b}} \end{gathered}$$

$$\begin{gathered} {\left|\overline{\mathrm{a}}+\overline{\mathrm{b}}\right|}^2={\left|\overline{\mathrm{a}}\right|}^2+2\overline{\mathrm{a}}\overline{\mathrm{b}}+{\left|\overline{\mathrm{b}}\right|}^2 \\ 5^2=3^2+2\overline{\mathrm{a}}\overline{\mathrm{b}}+{\left(\sqrt{10}\right)}^2 \\ 25=19+2\overline{\mathrm{a}}\overline{\mathrm{b}} \\ \overline{\mathrm{a}}\overline{\mathrm{b}}=\frac{6}{2} \\ \overline{\mathrm{a}}\overline{\mathrm{b}}=3 \end{gathered}$$

$$\begin{gathered} \mathrm{จากสูตร} \overline{\mathrm{a}}\overline{\mathrm{b}}=\left|\overline{\mathrm{a}}\right|\left|\overline{\mathrm{b}}\right|\cos \mathrm{\theta } \\ \cos \mathrm{\theta }=\frac{\overline{\mathrm{a}}\overline{\mathrm{b}}}{\left|\overline{\mathrm{a}}\right|\left|\overline{\mathrm{b}}\right|} \\ =\frac{3}{\left(3\right)\left(\sqrt{10}\right)}=\frac{1}{\sqrt{10}} \end{gathered}$$​

$$\begin{gathered} \mathrm{จะได้} \tan \mathrm{\theta }=3 \\ \mathrm{ดังนั้น} \mathrm{ข้อ} \left(\mathrm{ก}\right) \mathrm{ถูก} \end{gathered}$$

$$\begin{gathered} \underline{\mathrm{พิจารณา} \left(\mathrm{ข}\right)} \\ \mathrm{จากโจทย์}\hspace{0.17em} \overline{\mathrm{a}}·\overline{\mathrm{c}}=-12 \\ \mathrm{โดย} \left(\overline{\mathrm{a}}+\overline{\mathrm{c}}\right)·\left(\overline{\mathrm{a}}+\overline{\mathrm{c}}\right)=\overline{\mathrm{a}}·\overline{\mathrm{a}}+2\overline{\mathrm{a}}·\overline{\mathrm{c}}+\overline{\mathrm{c}}·\overline{\mathrm{c}} \\ \mathrm{จะได้} {\left|\overline{\mathrm{a}}+\overline{\mathrm{c}}\right|}^2={\left|\overline{\mathrm{a}}\right|}^2+2\overline{\mathrm{a}}·\overline{\mathrm{c}}+{\left|\overline{\mathrm{c}}\right|}^2 \\ {\left(\sqrt{10}\right)}^2=3^2+2\overline{\mathrm{a}}·\overline{\mathrm{c}}+5^2 \\ 10=9+25+2\overline{\mathrm{a}}·\overline{\mathrm{c}} \\ -24=2\overline{\mathrm{a}}·\overline{\mathrm{c}} \\ \overline{\mathrm{a}}·\overline{\mathrm{c}}=-12 \\ \mathrm{ดังนั้น} \mathrm{ข้อ} \left(\mathrm{ข}\right) \mathrm{ถูก} \end{gathered}$$