The position vectors of 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 coord...

Answer Details

Write the position vectors as coordinates: \(P(4,-5)\), \(Q(1,3)\), \(R(-5,2)\).

(a) Coordinates of M

In the parallelogram \(PQRM\) the vertices are taken in order \(P\to Q\to R\to M\), so the diagonals \(PR\) and \(QM\) bisect each other. Equating their midpoints:

\[\text{mid}(PR)=\left(\tfrac{4+(-5)}{2},\tfrac{-5+2}{2}\right)=\left(-\tfrac{1}{2},-\tfrac{3}{2}\right).\]

\[\text{mid}(QM)=\left(\tfrac{1+m_1}{2},\tfrac{3+m_2}{2}\right).\]

So \(1+m_1=-1\Rightarrow m_1=-2\) and \(3+m_2=-3\Rightarrow m_2=-6\).

\[\boxed{M(-2,-6)}.\]

(b) Acute angle between \(\overline{PM}\) and \(\overline{PQ}\)

\[\overline{PM}=M-P=(-2-4,\,-6-(-5))=(-6,-1),\]

\[\overline{PQ}=Q-P=(1-4,\,3-(-5))=(-3,8).\]

Using the scalar (dot) product,

\[\overline{PM}\cdot\overline{PQ}=(-6)(-3)+(-1)(8)=18-8=10,\]

\[|\overline{PM}|=\sqrt{(-6)^2+(-1)^2}=\sqrt{37},\qquad |\overline{PQ}|=\sqrt{(-3)^2+8^2}=\sqrt{73}.\]

\[\cos\theta=\frac{10}{\sqrt{37}\,\sqrt{73}}=\frac{10}{\sqrt{2701}}\approx 0.1924.\]

\[\theta=\cos^{-1}(0.1924)\approx 78.9^{\circ}\approx 79^{\circ}.\]

Since this value is already acute, the acute angle is \(79^{\circ}\).