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 coordinates of M;
(b) the acute angle between \(\overline{PM}\) and \(\overline{PQ}\), correct to the nearest degree.
Explanation
\(\overline{PQ}\) = \(\begin{pmatrix} 1 \\ 3 \end{pmatrix}\) - \(\begin{pmatrix} 4 \\ -5 \end{pmatrix}\) = \(\begin{pmatrix} -3 \\ 8 \end{pmatrix}\)
\(\overline{MR}\) = \(\begin{pmatrix} -5 \\ 2 \end{pmatrix}\) - \(\begin{pmatrix} x \\ y \end{pmatrix}\) = \(\begin{pmatrix} -5 & -x \\ 2 & - y \end{pmatrix}\)
\(\begin{pmatrix} -3 \\ 8 \end{pmatrix}\) = \(\begin{pmatrix} -5 & -x \\ 2 & - y \end{pmatrix}\)
⇒ -5 - x = -3; x = -2
⇒ 2 - y = 8; y = -6
The coordinate of M (-2,-6)
(b) \(\overline{PM}\) = \(\overline{OM}\) - \(\overline{OP}\)
= \(\begin{pmatrix} -2 \\ -6 \end{pmatrix}\) - \(\begin{pmatrix} 4 \\ -5 \end{pmatrix}\) = \(\begin{pmatrix} -6 \\ -1 \end{pmatrix}\)
\(\overline{PQ}\) = \(\overline{OQ}\) - \(\overline{OP}\)
= \(\begin{pmatrix} 1 \\ 3 \end{pmatrix}\) - \(\begin{pmatrix} 4 \\ -5 \end{pmatrix}\) = \(\begin{pmatrix} -3 \\ 8 \end{pmatrix}\)
|\(\overline{PM}\)| = √(-6\(^2\) + [-1\(^2\)]) = √37
|\(\overline{PQ}\)| = √(-3\(^2\) + 8\(^2\)) = √73
cosØ = \(\frac{PM.PQ}{|\overline{PM}||\overline{PQ}|}\)
cosØ = \(\frac{[-6i - j].[-3i+8j]}{√37 . √73}\) = \(\frac{10}{√2710}\)
⇒ Ø = cos\(^{-1}\) \(\frac{10}{√2710}\)
: Ø = 78.91° ≈ 79°