Diketahui balok ABCD.EFGH dengan koordinat titik sudut \( A(3,0,0), \ C(0,\sqrt{7},0), D(0,0,0), F(3,\sqrt{7},4) \) dan \( H(0,0,4) \). Besar sudut antara vektor \( \overrightarrow{DH} \) dan \( \overrightarrow{DF} \) adalah… (UN 2009)
- \( 15^\circ \)
- \( 30^\circ \)
- \( 45^\circ \)
- \( 90^\circ \)
- \( 120^\circ \)
Pembahasan:
Pertama, kita tentukan dulu vektor \( \overrightarrow{DH} \) dan \( \overrightarrow{DF} \) dan panjang masing-masing vektornya, yakni:
\begin{aligned} \overrightarrow{DH} &= H-D = \begin{pmatrix} 0 \\ 0 \\ 4 \end{pmatrix}-\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 4 \end{pmatrix} \\[8pt] |\overrightarrow{DH}| &= \sqrt{0^2+0^2+4^2} = \sqrt{16}=4 \\[8pt] \overrightarrow{DF} &= F-D = \begin{pmatrix} 3 \\ \sqrt{7} \\ 4 \end{pmatrix}-\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 3 \\ \sqrt{7} \\ 4 \end{pmatrix} \\[8pt] |\overrightarrow{DF}| &= \sqrt{3^2+(\sqrt{7})^2+4^2} = \sqrt{9+7+16} \\[8pt] &= \sqrt{32} = 4\sqrt{2} \end{aligned}
Selanjutnya, cari sudut antara vektor \( \overrightarrow{DH} \) dan \( \overrightarrow{DF} \) berdasarkan rumus perkalian titik dua vektor, yakni:
\begin{aligned} \cos \alpha &= \frac{\overrightarrow{DH} \cdot \overrightarrow{DF}}{|\overrightarrow{DH}| |\overrightarrow{DF}|} = \frac{\begin{pmatrix} 0 \\ 0 \\ 4 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ \sqrt{7} \\ 4 \end{pmatrix}}{ (4) (4\sqrt{2}) } \\[8pt] &= \frac{(0)(3)+(0)(\sqrt{7})+(4)(4)}{16\sqrt{2}} \\[8pt] &= \frac{0+0+16}{16\sqrt{2}} = \frac{16}{16\sqrt{2}} \\[8pt] \cos \alpha &= \frac{1}{\sqrt{2}} = \frac{1}{2}\sqrt{2} \\[8pt] \alpha &= 45^\circ \end{aligned}
Jawaban C.