Diketahui titik \( A(2,7,8), \ B(-1,1,-1) \) dan \( C(0,3,2) \). Jika \( \overrightarrow{AB} \) wakil \( \vec{u} \) dan \( \overrightarrow{BC} \) wakil \( \vec{v} \) maka proyeksi orthogonal vektor \( \vec{u} \) pada \( \vec{v} \) adalah… (UN 2009)
- \( -3\hat{i}-6\hat{j}-9\hat{k} \)
- \( \hat{i}+2\hat{j}+3\hat{k} \)
- \( \frac{1}{3} \hat{i}+ \frac{2}{3} \hat{j}+\hat{k} \)
- \( -9\hat{i}-18\hat{j}-27\hat{k} \)
- \( 3\hat{i}+6\hat{j}+9\hat{k} \)
Pembahasan:
Misalkan \( \vec{p} \) adalah proyeksi orthogonal vektor \( \vec{u} \) pada \( \vec{v} \), maka kita peroleh berikut:
\begin{aligned} \vec{u} &= \overrightarrow{AB}=B-A \\[8pt] &= (-1,1,-1)-(2,7,8) = (-3,-6,-9) \\[8pt] \vec{v} &= \overrightarrow{BC}=C-B \\[8pt] &= (0,3,2)-(-1,1,-1) = (1,2,3) \\[8pt] \vec{p} &= \left( \frac{ \vec{u} \cdot \vec{v} }{|\vec{v}|^2} \right) \cdot \vec{v} = \left( \frac{(-3,-6,-9) \cdot (1,2,3) }{ \left( \sqrt{1^2+2^2+3^2} \right)^2 } \right) \cdot (1,2,3) \\[8pt] &= \left( \frac{(-3)(1)+(-6)(2)+(-9)(3)}{1+4+9} \right)\cdot (1,2,3) \\[8pt] &= \left( \frac{-3-12-27}{14} \right) \cdot (1,2,3) = \frac{-42}{14} \cdot (1,2,3) \\[8pt] &= -3(1,2,3) = (-3,-6,-9) \\[8pt] &= -3\hat{i}-6\hat{j}-9\hat{k} \end{aligned}
Jawaban A.