Jika \( \vec{a} = (2,k) \) dan \( \vec{b} = (3,5) \) dan \( \angle \left( \vec{a}, \vec{b} \right) = \frac{\pi}{4} \), maka konstanta positif \(k\) adalah… (UMPTN 2001)
- \( \frac{1}{4} \)
- \( \frac{1}{2} \)
- \( 2 \)
- \( 4 \)
- \( 8 \)
Pembahasan:
Dengan menggunakan rumus aturan perkalian titik dua vektor, diperoleh berikut:
\begin{aligned} \vec{a} \cdot \vec{b} &= |\vec{a}| \cdot |\vec{b}| \cdot \cos \theta \\[8pt] (2 \cdot 3) + (k \cdot 5) &= \sqrt{2^2+k^2} \cdot \sqrt{3^2+5^2} \cdot \cos \left( \frac{\pi}{4} \right) \\[8pt] 6+5k &= \sqrt{4+k^2} \cdot \sqrt{9+25} \cdot \frac{1}{2}\sqrt{2} \\[8pt] (6+5k)^2 &= \left( \sqrt{4+k^2} \cdot \sqrt{34} \cdot \frac{1}{2}\sqrt{2} \right)^2 \\[8pt] 36+60k+25k^2 &= (4+k^2) \cdot 34 \cdot \frac{1}{2} \\[8pt] 36+60k+25k^2 &= 68+17k^2 \\[8pt] 25k^2-17k^2+60k+36-68 &= 0 \\[8pt] 8k^2+60k-32 &= 0 \\[8pt] 2k^2+15k-8 &= 0 \\[8pt] (2k-1)(k+8) &= 0 \\[8pt] k = \frac{1}{2} \ \text{atau} \ k &= -8 \end{aligned}
Jawaban B.