ข้อสอบคณิต 9 สามัญ - ปี 2555

$\mathrm{จากโจทย์} \mathrm{นำมาวาดรูป}$

$$\begin{gathered} \mathrm{จะได้} \overline{\mathrm{OA}}=\mathrm{i}-4\mathrm{j}-3\mathrm{k} \\ \overline{\mathrm{OB}}=3\mathrm{i}-6\mathrm{j}+2\mathrm{k} \\ \mathrm{จากรูป} \mathrm{cos\theta }=\frac{\left|\overline{\mathrm{OC}} \right| }{\overline{\left|\mathrm{OA}\right|}} \\ \left|\overline{\mathrm{OC}}\right|=\left|\overline{\mathrm{OA}}\right|\mathrm{cos\theta }\to \boxed{1} \\ \mathbf{โดย} \mathrm{cos\theta }=\frac{\overline{\mathrm{OA}}·\overline{\mathrm{OB}}}{\left|\overline{\mathrm{OA}}\right|\left|\overline{\mathrm{OB}}\right|} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}; \mathrm{นำ} \mathrm{cos\theta } \mathrm{ไปแทนใน} \boxed{1} \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \boxed{1} \left|\overline{\mathrm{OC}}\right|=\cancel{\left|\overline{\mathrm{OA}}\right|}·\frac{\overline{\mathrm{OA}}·\overline{\mathrm{OB}}}{\cancel{\left|\overline{\mathrm{OA}}\right|}\left|\overline{\mathrm{OB}}\right|} \\ =\frac{\overline{\mathrm{OA}}·\overline{\mathrm{OB}}}{\left|\overline{\mathrm{OB}}\right|} \\ =\frac{\left(1\right)\left(3\right)+\left(-4\right)\left(-6\right)+\left(-3\right)\left(2\right)}{\sqrt{3^2+{\left(-6\right)}^2+2^2}} \\ =\frac{21}{7}=3 \end{gathered}$$