The straight lines \(l\) and \(m\) intersect at the origin \(O\), and their directions are parallel to \(\mathbf{i}\) and \(\mathbf{i}+2\mathbf{j}\) respectively. The point \(A\) has position vector \(\overrightarrow{OA}\) given by \(2\mathbf{i}+3\mathbf{j}+5\mathbf{k}\). The point \(L\) on \(l\) is such that \(\overrightarrow{OL}\) is the projection of \(\overrightarrow{OA}\) on \(l\), and the point \(M\) on \(m\) is such that \(\overrightarrow{OM}\) is the projection of \(\overrightarrow{OA}\) on \(m\). State the position vector of \(L\), and show that \(\overrightarrow{OM} = \dfrac{8}{5}\mathbf{i}+ \dfrac{16}{5}\mathbf{j}\).

Drawing a diagram should help to clarify what is being described here.

If \(\overrightarrow{OL}\) is the projection of \(\overrightarrow{OA}\) on \(l\) it means that \(\overrightarrow{OL}\) is the component of \(\overrightarrow{OA}\) in the direction of \(l\). So we can find \(L\) by dropping a perpendicular…

Calculate the angle \(MAL\), giving your answer correct to the nearest degree.

To find the angle between vectors we can often use the dot product.

The point \(N\), where \(\overrightarrow{ON} = 2\mathbf{i} + 3\mathbf{j}\), is the foot of the perpendicular from \(A\) on to the plane \(OLM\). The lines \(ON\) and \(LM\) intersect at \(X\). Show that \(\dfrac{OX}{ON} = \dfrac{16}{19}\).

Can we find the position vector of \(N\)? From there, can we find the position vector of \(X\)?