N1994 P1 Q14

In the diagram, \(O\) is centre of the square base \(ABCD\) of a right pyramid, vertex \(V\). Perpendicular unit vectors \(\mathbf{i}\), \(\mathbf{j}\), \(\mathbf{k}\) are parallel to \(AB\), \(AD\), \(OV\) respectively. The length of \(AB\) is 4 units and the length of \(OV\) is \(2h\) units. \(P\), \(Q\), \(M\) and \(N\) are mid-points of \(AB\), \(BC\), \(CV\) and \(VA\) respectively. The point \(O\) is taken as the origin for position vectors.

1. Show that the equation of the line \(PM\) may be expressed as \(\mathbf{r}=\left( \begin{matrix} 0 \\ -2 \\ 0 \\ \end{matrix} \right)+t\left( \begin{matrix} 1 \\ 3 \\ h \\ \end{matrix} \right)\), where \(t\) is a parameter.[2]
2. Find the equation for the line \(QN\).[2]
3. Show that the lines \(PM\) and \(QN\) intersect, and that the position vector \(\overrightarrow{OX}\) of their point of intersection is \(\left( \begin{matrix} \frac{1}{2} \\ -\frac{1}{2} \\ \frac{1}{2}h \\ \end{matrix} \right)\).[3]
4. Given that \(OX\) is perpendicular to \(VB\), find the value of \(h\) and calculate the acute angle between \(PM\) and \(QN\), giving your answer correct to the nearest \(0.1{}^\circ \).[4]