Diketahui titik R terletak pada ruas garis PQ sehingga \( \overrightarrow{PR}: \overrightarrow{PQ} = 1:2 \). Jika vektor \( \vec{p} = 3 \hat{i}+\hat{j}+\hat{k} \) dan \( \vec{q} = 9 \hat{i}+5\hat{j}+7\hat{k} \) maka \( |\vec{r}| = \cdots \)
- \( \sqrt{21} \)
- \( \sqrt{61} \)
- \( \sqrt{38} \)
- \( 2 \sqrt{21} \)
- \( 2 \sqrt{15} \)
Pembahasan:
Dari perbandingan \( \overrightarrow{PR}: \overrightarrow{PQ} = 1:2 \), dapat kita peroleh:
\begin{aligned} 2 \overrightarrow{PR} = \overrightarrow{PQ} \Leftrightarrow 2 (\vec{r}-\vec{p}) &= \vec{q}-\vec{p} \\[8pt] 2 \vec{r} - 2 \vec{p} &= \vec{q} - \vec{p} \\[8pt] 2 \vec{r} &= \vec{q}-\vec{p}+2\vec{p} \\[8pt] 2 \vec{r} &= \vec{p}+\vec{q} \\[8pt] 2 \vec{r} &= (3, 1, 1)+(9,5,7) \\[8pt] 2 \vec{r} &= (12, \ 6, \ 8) \\[8pt] \vec{r} &= (6, \ 3, \ 4) \\[8pt] |\vec{r}| &= \sqrt{6^2+3^2+4^2} \\[8pt] &= \sqrt{36+9+16} \\[8pt] &= \sqrt{61} \end{aligned}
Jawaban B.