Jika diketahui vektor \( \vec{a} = p \hat{i}+2\hat{j}-\hat{k} \) dan \( \vec{b} = \hat{i}+3\hat{k} \) serta \( | \vec{a}+\vec{b} | = 2 \sqrt{3} \) maka nilai \(p = \cdots\)
- \( -3 \)
- \( -1 \)
- \( 2 \)
- \( 3 \)
- \( 5 \)
Pembahasan:
Dari \( | \vec{a}+\vec{b} | = 2 \sqrt{3} \), kita peroleh berikut:
\begin{aligned} \vec{a}+\vec{b} &= (p, 2, -1)+(1, 0, 3) \\[8pt] &= (p+1, \ 2, \ 2) \\[8pt] |\vec{a}+\vec{b}|^2 &= (p+1)^2+2^2+2^2 \\[8pt] (2\sqrt{3})^2 &= p^2+2p+1+4+4 \\[8pt] 12 &= p^2+2p+9 \\[8pt] 0 &= p^2+2p-3 \\[8pt] 0 &= (p+3)(p-1) \\[8pt] p &= -3 \ \text{atau} \ p = 1 \end{aligned}
Jawaban A.