The position vectors of points P, Q and R with respect to the origin are...
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, find:
(a) the position vector of M ;
(b) \(|\overrightarrow{PM}|\) and \(|\overrightarrow{PQ}|\) ;
(c) the acute angle between \(\overrightarrow{PM}\) and \(\overrightarrow{PQ}\), correct to 1 decimal place ;
(d) the area of PQRM.
Explanation
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Since PQRM is a parallelogram,
\(\overrightarrow{PQ} = \overrightarrow{MR}\)
\((4 - 1)i + (-5 - 3)j = (x + 5)i + (y - 2)j\)
Equating components, we have
\(3 = x + 5 \implies x = 3 - 5 = -2\)
\(-8 = y - 2 \implies y = -8 + 2 = -6\)
\(\therefore M = (-2i - 6j)\)
(b) \(|\overrightarrow{PM}| = \sqrt{(-2 - 4)^{2} + (- 6 + 5)^{2}} = \sqrt{37}\)
\(|\overrightarrow{PQ}| = \sqrt{(1 - 4)^{2} + (3 + 5)^{2}} = \sqrt{73}\)
(c) Let the angle be \(\theta\).
\(\overrightarrow{PM} . \overrightarrow{PQ} = |PM| |PQ| \cos \theta\)
\((-6i - j) . (-3i + 8j) = (\sqrt{37})(\sqrt{73}) \cos \theta\)
\(18 - 8 = \sqrt{2701} \cos \theta\)
\(\cos \theta = \frac{10}{\sqrt{2701}} = 0.1924\)
\(\theta = 78.907° \approxeq 78.9°\) (to 1 d.p)
(d) Area of PQRM = \((\sqrt{37})(\sqrt{73}) \sin 78.907°\)
= \(\sqrt{2701} \sin 78.907°\)
= \(51 \text{sq. units}\)