Diketahui \( A(4,0,0), \ B(0,-4,0) \) dan \( C(0,0,8) \). Panjang vektor proyeksi \( \overrightarrow{AC} \) ke vektor \( \overrightarrow{AB} \) adalah… (SBMPTN 2013)
- \( 2 \sqrt{2} \)
- \( 3 \frac{\sqrt{2}}{2} \)
- \( \frac{\sqrt{2}}{3} \)
- \( \sqrt{2} \)
- \( \frac{\sqrt{3}}{2} \)
Pembahasan:
Misalkan \( \vec{p} \) adalah hasil vektor proyeksi \( \overrightarrow{AC} \) ke vektor \( \overrightarrow{AB} \), maka panjang proyeksi vektornya dapat dicari sebagai berikut:
\begin{aligned} \overrightarrow{AC} &= C-A = (0,0,8)-(4,0,0) = (-4,0,8) \\[8pt] \overrightarrow{AB} &= B-A = (0,-4,0)-(4,0,0) = (-4,-4,0) \\[8pt] |\overrightarrow{AB}| &= \sqrt{(-4)^2+(-4)^2+0^2} \\[8pt] &= \sqrt{16+16+0} = \sqrt{32} \\[8pt] &= 4\sqrt{2} \\[8pt] |\vec{p}| &= \frac{\overrightarrow{AC} \cdot \overrightarrow{AB}}{|\overrightarrow{AB}|} = \frac{(-4,0,8) \cdot (-4,-4,0) }{ 4\sqrt{2} } \\[8pt] &= \frac{(-4)(-4)+(0)(-4)+(8)(0)}{4\sqrt{2}} \\[8pt] &= \frac{16+0+0}{4\sqrt{2}} = \frac{16}{4\sqrt{2}} = \frac{4}{\sqrt{2}} \\[8pt] &= \frac{4}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{4\sqrt{2}}{2} \\[8pt] &= 2\sqrt{2} \end{aligned}
Jawaban A.