Diketahui vektor \( \vec{a} = \hat{i}+2\hat{j}-x\hat{k}, \ \vec{b} = 3\hat{i}-2\hat{j}+\hat{k} \) dan \( \vec{c} = 2\hat{i}+\hat{j}+2\hat{k} \). Jika vektor \( \vec{a} \) tegak lurus vektor \( \vec{c} \), maka \( (\vec{a}+\vec{b})(\vec{a}-\vec{c}) \) adalah... (UN 2012)
- -4
- -2
- 0
- 2
- 4
Pembahasan:
Karena \( \vec{a} \) tegak lurus \( \vec{c} \), maka diperoleh:
\begin{aligned} \vec{a} \cdot \vec{c} = 0 \Leftrightarrow \begin{pmatrix} 1 \\ 2 \\ -x \end{pmatrix} \cdot \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix} &= 0 \\[8pt] 2+2-2x &= 0 \\[8pt] 2x &= 4 \Leftrightarrow x = 2 \\[8pt] \vec{a} = \begin{pmatrix} 1 \\ 2 \\ -x \end{pmatrix} &= \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix} \\[8pt] (\vec{a}+\vec{b})(\vec{a}-\vec{c}) &= \left( \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix} + \begin{pmatrix} 3 \\ -2 \\ 1 \end{pmatrix} \right) \cdot \left( \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix} - \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix} \right) \\[8pt] &= \begin{pmatrix} 4 \\ 0 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 1 \\ -4 \end{pmatrix} \\[8pt] &= 4(-1)+(0)(1)+(-1)(-4) \\[8pt] &= -4+0+4 = 0 \end{aligned}
Jawaban C.