ข้อสอบ A-Level คณิตประยุกต์ 1 ปี 2559 (วิชาสามัญเก่า)

$$\begin{gathered} \mathrm{จากสูตร} \overline{\mathrm{u}}·\overline{\mathrm{v}}=\left|\overline{\mathrm{u}}\right|\left|\overline{\mathrm{v}}\right|\mathrm{cos\theta } \\ \left|\overline{\mathrm{u}}\times \overline{\mathrm{v}}\right|=\left|\overline{\mathrm{u}}\right|\left|\overline{\mathrm{v}}\right|\mathrm{sin\theta } \\ \mathrm{จากโจทย์} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\left|\overline{\mathrm{u}}·\overline{\mathrm{v}}\right|}^2+{\left|\overline{\mathrm{u}}\times \overline{\mathrm{v}}\right|}^2 \\ ={\left|\left|\overline{\mathrm{u}}\right|\left|\overline{\mathrm{v}}\right|\mathrm{cos\theta }\right|}^2+{\left|\left|\overline{\mathrm{u}}\right|\left|\overline{\mathrm{v}}\right|\mathrm{sin\theta }\right|}^2 \\ ={\left|\sqrt{5}·\sqrt{3}\mathrm{cos\theta }\right|}^2+{\left|\sqrt{5}·\sqrt{3}\mathrm{sin\theta }\right|}^2 \\ =15{\cos }^2\mathrm{\theta }+15{\sin }^2\mathrm{\theta } \\ =15\left({\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta }\right) \\ \mathbf{โดย} {\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }=1 \\ \mathrm{ดังนั้น} {\left|\overline{\mathrm{u}}·\overline{\mathrm{v}}\right|}^2+{\left|\overline{\mathrm{u}}\times \overline{\mathrm{v}}\right|}^2=15 \end{gathered}$$