Calculate, correct to the nearest degree, the angle between the vectors \(\begin{pmatrix} 13 \\ 1 \end{pmatrix}\) and \(\begin{pmatrix} 1 \\ 4 \end{pmatrix}\)

A.

58°

B.

72°

C.

74°

D.

87°

Correct answer is B

\(a . b = |a||b| \cos \theta\)

\(\begin{pmatrix} 13 \\ 1 \end{pmatrix}. \begin{pmatrix} 1 \\ 4 \end{pmatrix} = 13 \times 1 + 1 \times 4 = 13 + 4 = 17\)

\(17 = (\sqrt{13^{2} + 1^{2}})(\sqrt{1^{2} + 4^{2}}) \cos \theta\)

\(17 = (\sqrt{170})(\sqrt{17}) \cos \theta\)

\(\cos \theta = \frac{17}{17\sqrt{10}} = \frac{\sqrt{10}}{10} = 0.3162\)

\(\theta = \cos^{-1} 0.3162 = 72°\)