Diketahui vektor-vektor \( \vec{a}=(1,3,3), \ \vec{b}=(3,2,1) \) dan \( \vec{c} = (1,-5,0) \). Sudut antara vektor \( \left( \vec{a}-\vec{b} \right) \) dan \( \left( \vec{a}+\vec{c} \right) \) adalah… (SPMB 2005)
- \( 30^\circ \)
- \( 45^\circ \)
- \( 60^\circ \)
- \( 90^\circ \)
- \( 120^\circ \)
Pembahasan:
Kita cari dulu vektor berikut:
\begin{aligned} \vec{a}-\vec{b} &= (1,3,3)-(3,2,1) \\[8pt] &= (1-3,3-2,3-1) \\[8pt] &= (-2,1,2) \\[8pt] \vec{a}+\vec{c} &= (1,3,3)+(1,-5,0) \\[8pt] &= (1+1,3-5,3+0) \\[8pt] &= (2,-2,3) \end{aligned}
Kemudian dengan menggunakan rumus perkalian titik dua vektor, diperoleh berikut:
\begin{aligned} (\vec{a}-\vec{b}) \cdot (\vec{a}+\vec{c}) &= | (\vec{a}-\vec{b}) | \cdot |\vec{a}+\vec{c}| \cdot \cos \beta \\[8pt] \cos \beta &= \frac{(\vec{a}-\vec{b}) \cdot (\vec{a}+\vec{c})}{| (\vec{a}-\vec{b}) | \cdot |\vec{a}+\vec{c}|} \\[8pt] &= \frac{(-2)(2)+(1)(-2)+(2)(3)}{\sqrt{(-2)^2+1^2+2^2} \cdot \sqrt{2^2+(-2)^2+3^2} } \\[8pt] &= \frac{-4-2+6}{\sqrt{4+1+4} \cdot \sqrt{4+4+9}} \\[8pt] &= \frac{0}{\sqrt{9} \cdot \sqrt{17}} = 0 \\[8pt] \cos \beta &= 0 \Leftrightarrow \beta = \frac{\pi}{2}=90^\circ \end{aligned}
Jawaban D.