The four points \(P\), \(Q\), \(R\) and \(S\) are the vertices of a plane quadrilateral. What is the geometrical shape of \(PQRS\) if \(\vec{PQ} = \vec{SR}\)? What is the geometrical shape of \(PQRS\) if \(\vec{PQ} = \vec{SR}\) and \(|\vec{PQ}| = |\vec{PS}|\)?
A cube with edges of unit length has opposite vertices at \((0,0,0)\) and \((1,1,1)\). The points
$$P(p,0,0), \quad Q(1,q,0), \quad R(r,1,1) \quad \text{and} \quad S(0,s,1)$$
lie on edges of the cube. Given that the four points lie in the same plane, show that
$$rq = (1-s)(1-p).$$
Show that \(\vec{PQ} = \vec{SR}\) if and only if the centroid of the quadrilateral \(PQRS\) is at the centre of the cube.
Note: the centroid of the quadrilateral \(PQRS\) is the point with position vector
$$\frac{1}{4}(\vec{OP} + \vec{OQ} + \vec{OR} + \vec{OS}),$$
where \(O\) is the origin.
Given that \(\vec{PQ} = \vec{SR}\) and \(|\vec{PQ}| = |\vec{PS}|\), express \(q\), \(r\) and \(s\) in terms of \(p\).
Show that
$$\cos PQR = \frac{4p-1}{5-4p+8p^2}.$$
Write down the values of \(p\), \(q\), \(r\) and \(s\) if \(PQRS\) is a square, and show that the length of each side of this square is greater than \(\frac{21}{20}\).
Solution:
If \(\vec{PQ} = \vec{SR}\) we have a parallelogram.
\(\vec{PQ} = \vec{SR}\) and \(|\vec{PQ}| = |\vec{PS}|\) then we have a rhombus.
If the four points lie in a plane then
\((\vec{RS} \times \vec{RP}) \cdot \vec{RQ} =0\), so
\begin{align*}
&& 0 &=\left ( \begin{pmatrix}-r\\ s-1 \\ 0 \end{pmatrix} \times \begin{pmatrix}p-r\\ -1 \\ -1 \end{pmatrix}\right) \cdot \begin{pmatrix}1-r\\ q-1 \\ -1 \end{pmatrix} \\
&& &= \begin{pmatrix}1-s \\ -r \\r+(p-r)(1-s) \end{pmatrix} \cdot \begin{pmatrix}1-r\\ q-1 \\ -1 \end{pmatrix} \\
&&&= (1-s)(1-r)-r(q-1)-r-(p-r)(1-s) \\
&&&=(1-s)(1-r-p+r)-rq \\
\Rightarrow && rq &= (1-s)(1-p)
\end{align*}
\(\,\) \begin{align*}
&& \vec{PQ} &= \vec{SR} \\
\Leftrightarrow && \begin{pmatrix}1-p\\q \\ 0 \end{pmatrix} &= \begin{pmatrix}r\\1-s \\ 0 \end{pmatrix} \\
\Leftrightarrow && 1-p = r & \quad ; \quad q = 1-s\\
\Leftrightarrow && 1= r+p & \quad ; \quad 1 = q+s\\
\end{align*}
The centroid is \(\frac14 (p+1+r, q+s+1, 2)\) which is clearly \(\frac12(1,1,1)\) iff those equations are true.
\(\,\) \begin{align*}
&& |\vec{PQ}| &= |\vec{PS}| \\
\Leftrightarrow && (1-p)^2+q^2+ 0^2 &= p^2+s^2+1)\\
\Leftrightarrow && 1-2p+p^2+q^2 &= p^2 + s^2 + 1 \\
\Leftrightarrow && -2p+q^2 &= s^2
\end{align*}
From the previous equations we have \(r = 1-p\), and
\(-2p+(1-s)^2 = s^2 \Rightarrow -2p + 1 -2s = 0 \Rightarrow s = \frac12 - p\) and \(q = \frac12 + p\)
\begin{align*}
&& \cos PQR &= \frac{\vec{QP}\cdot \vec{QR}}{|\vec{QP}||\vec{QR}|} \\
&&&= \frac{ \begin{pmatrix}p-1\\ -q \\ 0 \end{pmatrix} \cdot \begin{pmatrix}r-1\\ 1-q \\ 1 \end{pmatrix}}{\sqrt{(p-1)^2+q^2}\sqrt{(r-1)^2+(1-q)^2+1^2}} \\
&&&= \frac{ \begin{pmatrix}p-1\\ -\frac12-p \\ 0 \end{pmatrix} \cdot \begin{pmatrix}-p\\ \frac12-p \\ 1 \end{pmatrix}}{\sqrt{(p-1)^2+(-\frac12-p)^2}\sqrt{p^2+(\frac12-p)^2+1^2}} \\
&&&= \frac{ p-p^2-\frac14+p^2}{\sqrt{p^2-2p+1+\frac14+p+p^2}\sqrt{p^2+\frac14-p+p^2+1}} \\
&&&= \frac{4p-1}{\sqrt{8p^2-4p+5}\sqrt{8p^2-4p+5}}\\
&&&= \frac{4p-1}{8p^2-4p+5}\\
\end{align*}
For \(PQRS\) to be a square \(\cos PQR = 0\), ie \(p = \frac14\) and so
\((p,q,r,s) = (\frac14, \frac34, \frac34, \frac14)\) and \(|PQ| = \sqrt{(1-p)^2+q^2} = \sqrt{\left ( \frac34 \right)^2 + \left ( \frac34 \right)^2 } = \frac{3\sqrt{2}}4\), notice that \(\left ( \frac{21}{20} \right)^2 = \frac{441}{400} < \frac{9}{8}\) (\(441 < 450\)) therefore the sides are at least as long as \(\frac{21}{20}\)