The position vectors of points P, Q and R with respect to the origin are \((4i - 5j), (i + 3j)\) and \((-5i + 2j)\) respectively. If PQRM is a parallelogram...

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Position vectors: \(P=(4,-5),\ Q=(1,3),\ R=(-5,2).\) In parallelogram \(PQRM\) the diagonals \(PR\) and \(QM\) bisect each other, so \(P+R=Q+M.\)

(a) Position vector of M.

\[M=P+R-Q=(4-5-1,\ -5+2-3)=(-2,-6),\quad\text{i.e. }M=-2\mathbf{i}-6\mathbf{j}.\]

(b) \(|\overrightarrow{PM}|\) and \(|\overrightarrow{PQ}|.\)

\(\overrightarrow{PM}=M-P=(-6,-1),\ \ |\overrightarrow{PM}|=\sqrt{36+1}=\sqrt{37}\approx6.08.\)

\(\overrightarrow{PQ}=Q-P=(-3,8),\ \ |\overrightarrow{PQ}|=\sqrt{9+64}=\sqrt{73}\approx8.54.\)

(c) Acute angle between \(\overrightarrow{PM}\) and \(\overrightarrow{PQ}.\)

\[\cos\theta=\frac{\overrightarrow{PM}\cdot\overrightarrow{PQ}}{|\overrightarrow{PM}|\,|\overrightarrow{PQ}|}=\frac{(-6)(-3)+(-1)(8)}{\sqrt{37}\,\sqrt{73}}=\frac{18-8}{\sqrt{2701}}=\frac{10}{51.97}\approx0.1924.\]

\[\theta=\cos^{-1}(0.1924)\approx78.9^\circ.\]

(d) Area of PQRM. Area \(=|\overrightarrow{PM}\times\overrightarrow{PQ}|\) (magnitude of the 2D cross product of adjacent sides):

\[\text{Area}=|(-6)(8)-(-1)(-3)|=|-48-3|=51\text{ square units.}\]