Calculate, correct to the nearest degree, the angle between the vectors \(\begin{pmatrix} 13 \\ 1 \end{pmatrix}\) and \(\begin{pmatrix} 1 \\ 4 \end{pmatrix}\).
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Correct Answer: Option B
Explanation:
\(a . b = |a||b| \cos \theta\)
\(\begin{pmatrix} 13 \\ 1 \end{pmatrix}. \begin{pmatrix} 1 \\ 4 \end{pmatrix} = 13 \times 1 + 1 \times 4 = 13 + 4 = 17\)
\(17 = (\sqrt{13^{2} + 1^{2}})(\sqrt{1^{2} + 4^{2}}) \cos \theta\)
\(17 = (\sqrt{170})(\sqrt{17}) \cos \theta\)
\(\cos \theta = \frac{17}{17\sqrt{10}} = \frac{\sqrt{10}}{10} = 0.3162\)
\(\theta = \cos^{-1} 0.3162 = 72°\)
\(a . b = |a||b| \cos \theta\)
\(\begin{pmatrix} 13 \\ 1 \end{pmatrix}. \begin{pmatrix} 1 \\ 4 \end{pmatrix} = 13 \times 1 + 1 \times 4 = 13 + 4 = 17\)
\(17 = (\sqrt{13^{2} + 1^{2}})(\sqrt{1^{2} + 4^{2}}) \cos \theta\)
\(17 = (\sqrt{170})(\sqrt{17}) \cos \theta\)
\(\cos \theta = \frac{17}{17\sqrt{10}} = \frac{\sqrt{10}}{10} = 0.3162\)
\(\theta = \cos^{-1} 0.3162 = 72°\)