(a) Find the angle between the vectors \(a = \begin{pmatrix} -3 \\ 4 \end{pmatrix}\) and \(b = \begin{pmatrix} -8 \\ -15 \end{pmatrix}\).
(b) Given that \(a = (4N, 060°)\) and \(b = (3N, 120°)\), find, in component form, the unit vector along \(a - b\).
(a) \(a = (-3i + 4j) ; b = (-8i - 15j)\)
Let the angle between them be \(\theta\).
\(a \cdot b = |a||b| \cos \theta\)
\(24 - 60 = (\sqrt{(-3)^{2} + 4^{2}})(\sqrt{(-8)^{2} + (-15)^{2}}) \cos \theta\)
\(-36 = 5 \times 17 \times \cos \theta\)
\(\cos \theta = \frac{-36}{85} = -0.4235 \implies \theta = \cos^{-1} (-0.4235)\)
\(\theta = 115.06°\)
(b) \(a = 4 \cos 60° \\ 4 \sin 60° \end{pmatrix} = \begin{pmatrix} 2 \\ 3.464 \end{pmatrix}\)
\(b = \begin{pmatrix} 3 \cos 120° \\ 3 \sin 120° \end{pmatrix} = \begin{pmatrix} -1.5 \\ 2.598 \end{pmatrix}\)
\(a - b = \begin{pmatrix} 2 - (-1.5) \\ 3.464 - 2.598 \end{pmatrix} = \begin{pmatrix} 3.5 \\ 0.866 \end{pmatrix}\)
\(|a - b| = \sqrt{(3.5)^{2} + (0.866)^{2}} = \sqrt{12.25 + 0.75}\)
= \(\sqrt{13}\)
Unit vector along a - b = \(\frac{3.5i + 0.866j}{\sqrt{13}}\).
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