(a) Find, correct to one decimal place, the angle between \(p = \begin{pmatrix} 3 \\ -1 \end{pmatrix}\) and \(q = \begin{pmatrix} 3 \\ 4 \end{pmatrix}\).
(b) ABCD is a square with vertices at A(0, 0), B(2, 0), C(2, 2) and D(0, 2). Forces of magnitude 10 N, 15 N, 20 N and 5 N act along \(\overrightarrow{BA}, \overrightarrow{BC}, \overrightarrow{DC}\) and \(\overrightarrow{AD}\) respectively. Find the (i) magnitude (ii) direction; of the resultant.
(a) \(p = \begin{pmatrix} 3 \\ -1 \end{pmatrix} = 3i - j ; q = \begin{pmatrix} 3 \\ 4 \end{pmatrix} = 3i + 4j\)
Let the angle be \(\theta\),
\(p . q = |p||q| \cos \theta\)
\((3i - j).(3i + 4j) = 9 - 4 = 5\)
\(|3i - j| = \sqrt{3^{2} + (-1)^{2}} = \sqrt{10}\)
\(|3i + 4j| = \sqrt{3^{2} + 4^{2}} = \sqrt{25} = 5\)
\(5 = (\sqrt{10})(5) \cos \theta \implies \cos \theta = \frac{\sqrt{10}}{10} = 0.3162\)
\(\theta = \cos^{-1} 0.3162 = 71.567° \approxeq 71.6°\)
(b) 
(i) Resolving the forces along the x and y axes, we obtain
Resultant (R) = (10i + 20j) N.
Magnitude = \(\sqrt{10^{2} + 20^{2}} = \sqrt{500}\)
= \(10\sqrt{5} \approxeq 22.4 N\)
(ii) Direction : \(\tan \theta = \frac{20}{10} = 2\)
\(\theta = \tan^{-1} 2 = 63.4°\).
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