ข้อสอบ PAT 1 - เมษายน 2557

$$\begin{gathered} \mathrm{จากโจทย์} \overline{\mathrm{u}}=\mathrm{a}\overline{\mathrm{i}}+2\overline{\mathrm{j}}+\mathrm{b}\overline{\mathrm{k}} \\ \overline{\mathrm{u}}\times \overline{\mathrm{j}}=\left|\begin{array}{ccc}\overline{\mathrm{i}}& \overline{\mathrm{j}}& \overline{\mathrm{k}}\\ \mathrm{a}& 2& \mathrm{b}\\ 0& 1& 0\end{array}\right| \left(\mathrm{Det} \mathrm{มิติ} 3\times 3\right) \\ =\left|\begin{array}{cc}2& \mathrm{b}\\ 1& 0\end{array}\right|\overline{\mathrm{i}}-\left|\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ 0& 0\end{array}\right|\overline{\mathrm{j}}+\left|\begin{array}{cc}\mathrm{a}& 2\\ 0& 1\end{array}\right|\overline{\mathrm{k}} \\ =-\mathrm{b}\overline{\mathrm{i}}+\mathrm{a}\overline{\mathrm{k}} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} \left|\overline{\mathrm{u}}\times \overline{\mathrm{j}}\right|=2 \\ \mathrm{จะได้} \sqrt{{\left(-\mathrm{b}\right)}^2+{\mathrm{a}}^2}=2 \\ {\mathrm{b}}^2+{\mathrm{a}}^2=4 \end{gathered}$$​

$$\begin{gathered} \mathrm{จากโจทย์} \overline{\mathrm{u}}=\mathrm{a}\overline{\mathrm{i}}+2\overline{\mathrm{j}}+\mathrm{b}\overline{\mathrm{k}} \\ \mathrm{จะได้} \left|\overline{\mathrm{u}}\right|=\sqrt{{\mathrm{a}}^2+2^2+{\mathrm{b}}^2} \\ \mathrm{ดังนั้น} {\left|\overline{\mathrm{u}}\right|}^2={\mathrm{a}}^2+{\mathrm{b}}^2+4 \\ =4+4 \\ =8 \end{gathered}$$