Jika titik \( A(1,2,2), \ B(5,4,2) \) dan \( C(0,-3,5) \), maka sudut antara vektor \( \overrightarrow(AB) \) dengan \( \overrightarrow(AC) \) adalah…
- \( \frac{1}{5 \sqrt{7}} \)
- \( -\frac{3}{5 \sqrt{7}} \)
- \( -\frac{7}{5 \sqrt{7}} \)
- \( \frac{7}{5} \)
- \( -\frac{3}{5} \)
Pembahasan:
Pertama, kita tentukan dulu vektor \( \overrightarrow(AB) \) dan \( \overrightarrow(AC) \) serta panjang masing-masing vektornya, yakni:
\begin{aligned} \overrightarrow{AB} &= \overrightarrow{OB}-\overrightarrow{OA} \\[8pt] &= \begin{pmatrix} 5 & 4 & 2 \end{pmatrix} - \begin{pmatrix} 1 & 2 & 2 \end{pmatrix} \\[8pt] &= \begin{pmatrix} 4 & 2 & 0 \end{pmatrix} \\[8pt] |\overrightarrow{AB}| &= \sqrt{4^2+2^2+0^2} = \sqrt{20} \\[8pt] \overrightarrow{AC} &= \overrightarrow{OC}-\overrightarrow{OA} \\[8pt] &= \begin{pmatrix} 0 & -3 & 5 \end{pmatrix} - \begin{pmatrix} 1 & 2 & 2 \end{pmatrix} \\[8pt] &= \begin{pmatrix} -1 & -5 & 3 \end{pmatrix} \\[8pt] |\overrightarrow{AC}| &= \sqrt{(-1)^2+(-5)^2+3^2} = \sqrt{35} \end{aligned}
Berdasarkan hasil di atas, maka besar sudut antara vektor \( \overrightarrow(AB) \) dengan \( \overrightarrow(AC) \), yaitu:
\begin{aligned} \cos \theta &= \frac{\overrightarrow{AB} \cdot \overrightarrow{AC} }{ |\overrightarrow{AB}| |\overrightarrow{AC}| } = \frac{\begin{pmatrix} 4 & 2 & 0 \end{pmatrix} \cdot \begin{pmatrix} -1 & -5 & 3 \end{pmatrix}}{\sqrt{20} \cdot \sqrt{35}} \\[8pt] &= \frac{4(-1)+2(-5)+(0)(3)}{\sqrt{700}} \\[8pt] &= \frac{-4-10+0}{10\sqrt{7}} = \frac{-14}{10\sqrt{7}} \\[8pt] &= -\frac{7}{5\sqrt{7}} \end{aligned}
Jawaban C.