ข้อสอบ PAT 1 - มีนาคม 2556

$$\begin{gathered} \mathrm{จากโจทย์} \left|\overline{\mathrm{b}}-\overline{\mathrm{c}}\right|=5 \\ \mathrm{จะได้} \left|\left(6\overline{\mathrm{i}}+\mathrm{y}\overline{\mathrm{j}}\right)-\left(2\overline{\mathrm{i}}+\overline{\mathrm{j}}\right)\right|=5 \\ \left|4\overline{\mathrm{i}}+\left(\mathrm{y}-1\right)\overline{\mathrm{j}}\right|=5 \\ \sqrt{4^2+{\left(\mathrm{y}-1\right)}^2}=5 \end{gathered}$$

$$\begin{gathered} 16+{(\mathrm{y}-1)}^2=25 \\ {(\mathrm{y}-1)}^2=9 \\ \mathrm{y}-1=\pm 3 \\ \mathrm{y}=4,-2 \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} \mathrm{เวกเตอร์} \overline{ \mathrm{a} } \mathrm{ตั้งฉากกับเวกเตอร์} \overline{\mathrm{b}} \\ \mathrm{จะได้} \overline{\mathrm{a}}·\overline{\mathrm{b}}=0 \\ \left(\mathrm{x}\overline{\mathrm{i}}+\frac{12}{5}\overline{\mathrm{j}}\right)\left(6\overline{\mathrm{i}}+\mathrm{y}\overline{\mathrm{j}}\right)=0 \\ \mathrm{x}\left(6\right)+\frac{12}{5}\left(\mathrm{y}\right)=0 \end{gathered}$$

$$\begin{gathered} \mathrm{y}=4 \mathrm{จะได้} 6\mathrm{x}+\left(\frac{12}{5}\right)\left(4\right)=0 \\ \mathrm{x}=-\frac{8}{5} \\ \mathrm{y}=-2 \mathrm{จะได้} 6\mathrm{x}+\left(\frac{12}{5}\right)\left(-2\right)=0 \\ \mathrm{x}=\frac{4}{5} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} \overline{\mathrm{a}}·\overline{\mathrm{c}}>0 \\ \left(\mathrm{x}\overline{\mathrm{i}}+\frac{12}{5}\overline{\mathrm{j}}\right)\left(2\overline{\mathrm{i}}+\overline{\mathrm{j}}\right)>0 \\ 2\mathrm{x}+\frac{12}{5}\left(1\right)>0 \\ \mathrm{x}=-\frac{8}{5} \mathrm{จะได้} \overline{\mathrm{a}}·\overline{\mathrm{c}}=\left(-\frac{8}{5}\right)\left(2\right)+\left(\frac{12}{5}\right)\left(1\right) \\ \overline{\mathrm{a}}·\overline{\mathrm{c}}=\frac{-4}{5}<0 \end{gathered}$$

$$\begin{gathered} \mathrm{แสดงว่า} \mathrm{ใช้ไม่ได้} \\ \mathrm{x}=\frac{4}{5} \mathrm{จะได้} \overline{\mathrm{a}}·\overline{\mathrm{c}}=\left(\frac{4}{5}\right)\left(2\right)+\left(\frac{12}{5}\right)\left(1\right) \\ \overline{\mathrm{a}}·\overline{\mathrm{c}}=4>0 \\ \mathrm{แสดงว่า} \mathrm{ใช้ได้} \end{gathered}$$

$$\begin{gathered} \mathrm{ดังนั้น} \overline{\mathrm{a}}=\frac{4}{5}\overline{\mathrm{i}}+\frac{12}{5} \overline{\mathrm{j}} \\ \overline{\mathrm{b}}=6\overline{\mathrm{i}}-2\overline{\mathrm{j}} \\ \mathrm{ดังนั้น} {\left|5\overline{\mathrm{a}}+\overline{\mathrm{b}}\right|}^2={\left|5\left(\frac{4}{5}\overline{\mathrm{i}}+\frac{12}{5}\overline{\mathrm{j}}\right)+\left(6\overline{\mathrm{i}}-2\overline{\mathrm{j}}\right)\right|}^2 \\ ={\left|\left(4\overline{\mathrm{i}}+12\overline{\mathrm{j}}\right)+\left(6\overline{\mathrm{i}}-2\overline{\mathrm{j}}\right)\right|}^2 \end{gathered}$$

$$\begin{gathered} ={\left|10\overline{\mathrm{i}}+10\overline{\mathrm{j}}\right|}^2 \\ ={\left(\sqrt{{10}^2+{10}^2}\right)}^2 \\ ={\left(\sqrt{200}\right)}^2 \\ =200 \end{gathered}$$