Diketahui vektor \( \vec{a} = \hat{i}-2\hat{j}+\hat{k} \) dan \( \vec{b} = 3\hat{i}+\hat{j}-2\hat{k} \). Vektor \( \vec{c} \) mewakili vektor hasil proyeksi orthogonal vektor \( \vec{b} \) pada vektor \( \vec{a} \) maka vektor \( \vec{c} = \cdots \) (UN 2013)
- \( -\frac{1}{6} ( \hat{i}-2\hat{j}+\hat{k} ) \)
- \( -\frac{1}{6} ( 3\hat{i}-2\hat{j}+2\hat{k} ) \)
- \( -\frac{1}{14} ( \hat{i}-2\hat{j}+\hat{k} ) \)
- \( -\frac{1}{14} ( 3\hat{i}-\hat{j}+2\hat{k} ) \)
- \( \frac{1}{6} ( \hat{i}-2\hat{j}+\hat{k} ) \)
Pembahasan:
Vektor \( \vec{c} \) atau hasil proyeksi vektor \( \vec{b} \) pada vektor \( \vec{a} \) dapat dicari sebagai berikut:
\begin{aligned} \vec{c} &= \left( \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} \right) \cdot \vec{a} = \left( \frac{(3,1,-2) \cdot (1,-2,1)}{ \left( \sqrt{1^2+(-2)^2+1^2} \right)^2 } \right) \cdot (1, -2, 1) \\[8pt] &= \left( \frac{(3)(1)+(1)(-2)+(-2)(1)}{ 1+4+1 } \right) \cdot (1,-2,1) \\[8pt] &= \left( \frac{3-2-2}{6} \right) \cdot (1,-2,1) = -\frac{1}{6} \ (1,-2,1) \\[8pt] &= -\frac{1}{6} (\hat{i}-2\hat{j}+\hat{k}) \end{aligned}
Jawaban A.