ข้อสอบ PAT 1 - มีนาคม 2559

$$\begin{gathered} \mathrm{จากโจทย์} \overline{a} = \overline{i}+\overline{j}+\overline{k} \\ \mathrm{จะได้} \left|\overline{a}\right| = \sqrt{1^2+1^2+1^2} = \sqrt{3} \\ \mathrm{จากโจทย์} \left|\overline{b}\right| = {\left|\overline{a}\right|}^2 \\ \mathrm{จะได้} \left|\overline{b}\right| = {\left(\sqrt{3}\right)}^2 = 3 \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} \overline{a}·\overline{b}+\overline{b}·\overline{c}+\overline{c}·\overline{a} = 9 \\ \mathrm{จะได้} \overline{b}·\overline{c}+\overline{c}·\overline{a} = 9-\overline{a}·\overline{b} \\ \left(\overline{a}+\overline{b}\right)·\overline{c} = 9-\overline{a}·\overline{b} \\ \mathrm{โดย} \overline{a}+\overline{b} =t\overline{c} ; \left(t\overline{c} \right)·\overline{c} = 9-\overline{a}·\overline{b} \\ t{\left|\overline{c}\right|}^2 = 9-\overline{a}·\overline{b} \\ \mathrm{โดย} \left|\overline{c}\right|=\sqrt{2} ; t{\left(\sqrt{2}\right)}^2 = 9-\overline{a}·\overline{b} \\ 2t = 9-\overline{a}·\overline{b} \\ \overline{a}·\overline{b} = 9-2t \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} \overline{a}+\overline{b} = t\overline{c} \\ \mathrm{จะได้} \left|\overline{a}+\overline{b}\right| = \left|t\overline{c}\right| \\ {\left|\overline{a}+\overline{b}\right|}^2 = {\left|t\overline{c}\right|}^2 \\ {\left|\overline{a}\right|}^2+2\overline{a}·\overline{b}+{\left|\overline{b}\right|}^2 = t^2{\left|\overline{c}\right|}^2 \\ \mathbf{โดย} \left|\overline{a}\right|=\sqrt{3} , \left|\overline{b}\right|=3 , \left|\overline{c}\right|=\sqrt{2}, \overline{a}·\overline{b} = 9-2t \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} 3+18-4t+9 = 2t^2 \\ 2t^2+4t-30 = 0 \\ t^2+2t-15 = 0 \\ \left(t+5\right)\left(t-3\right) = 0 \\ t = -5, 3 \\ \mathrm{จากโจทย์} t \mathrm{เป็นจำนวนจริงบวก} \\ \mathbf{ดังนั้น}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathit{t}\mathbf{=}\mathbf{3} \end{gathered}$$