Diketahui vektor \( \overrightarrow{OR}=(2,3) \) dan vektor \( \overrightarrow{OS}=(1,2) \) dengan titik \(T\) terletak pada \(RS\) sehingga \( RT:TS = 2:3 \), maka panjang vektor \( \overrightarrow{OT} \) adalah…
- \( \frac{3}{2} \sqrt{26} \)
- \( \frac{3}{5} \sqrt{212} \)
- \( \frac{1}{2} \sqrt{21} \)
- \( \frac{1}{5} \sqrt{233} \)
- \( \frac{3}{5} \sqrt{28} \)
Pembahasan:
Diketahui \( RT:TS = 2:3 \) sehingga kita peroleh:
\begin{aligned} \vec{t} &= \frac{2\vec{s}+3\vec{r}}{2+3} = \frac{2\vec{s}+3\vec{r}}{5} \\[8pt] &= \frac{1}{5} \left( 2 (1, 2) + 3(2, 3) \right) \\[8pt] &= \frac{1}{5} \left( (2,4)+(6,9) \right) \\[8pt] &= \frac{1}{5} \left( 8, 13 \right) = \left( \frac{8}{5}, \frac{13}{5} \right) \\[8pt] |\overrightarrow{OT}| &= |\vec{t}| = \sqrt{ \left( \frac{8}{5} \right)^2 + \left( \frac{13}{5} \right)^2 } \\[8pt] &= \sqrt{ \frac{64}{25} + \frac{169}{25} } = \sqrt{\frac{233}{25}} \\[8pt] &= \frac{1}{5} \sqrt{233} \end{aligned}
Jawaban D.