(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...
Question 1 Report
(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.
Answer Details
(a) Angle between \(p=\binom{3}{-1}\) and \(q=\binom{3}{4}\).
\[\cos\theta=\frac{p\cdot q}{|p|\,|q|}=\frac{(3)(3)+(-1)(4)}{\sqrt{3^{2}+(-1)^{2}}\,\sqrt{3^{2}+4^{2}}}=\frac{9-4}{\sqrt{10}\cdot 5}=\frac{5}{5\sqrt{10}}=\frac{1}{\sqrt{10}}\]\[\theta=\cos^{-1}(0.3162)=71.6^{\circ}\ (\text{1 d.p.})\](b) Forces on square A(0,0), B(2,0), C(2,2), D(0,2).
Write each force as its magnitude times the unit vector of its line of action:
- \(10\,\text{N}\) along \(\overrightarrow{BA}=(-1,0):\ (-10,0)\)
- \(15\,\text{N}\) along \(\overrightarrow{BC}=(0,1):\ (0,15)\)
- \(20\,\text{N}\) along \(\overrightarrow{DC}=(1,0):\ (20,0)\)
- \(5\,\text{N}\) along \(\overrightarrow{AD}=(0,1):\ (0,5)\)
Resultant components:
\[R_x=-10+20=10,\qquad R_y=15+5=20\](i) Magnitude:
\[|R|=\sqrt{10^{2}+20^{2}}=\sqrt{500}=10\sqrt{5}\approx 22.4\,\text{N}\](ii) Direction (measured from the positive x-axis, i.e. along AB):
\[\tan\alpha=\frac{20}{10}=2\;\Rightarrow\;\alpha=63.4^{\circ}\]The resultant is about \(22.4\,\text{N}\) at \(63.4^{\circ}\) to AB.
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