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 \(\overrightarrow{OM} = \dfrac{8}{5}\mathbf{i}+ \dfrac{16}{5}\mathbf{j}\).

Since \(l\) is parallel to \(\mathbf{i}\) (or the \(x\)-axis), the vector \(\overrightarrow{OL}\) is the \(\mathbf{i}\)-component of \(\overrightarrow{OA}\). So the position vector of \(L\) is \(2\mathbf{i}\).

The point \(M\) is on \(m\), so its position vector can be written as \(\mu(\mathbf{i}+2\mathbf{j})\) for some number \(\mu\). Since it is the projection, it follows that \(m\) and \(\overrightarrow{AM}\) are perpendicular. In other words, their dot product is zero. \[\begin{align*} (\mathbf{i}+2\mathbf{j}) . ((\mu-2)\mathbf{i}+(2\mu-3)\mathbf{j}-5\mathbf{k}) &= 0 \\ \implies\quad (\mu-2)+2(2\mu-3) &= 0 \\ \implies\quad \mu &= \frac{8}{5} \end{align*}\]So \(\overrightarrow{OM} = \dfrac{8}{5}\mathbf{i}+ \dfrac{16}{5}\mathbf{j}\) as required.

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

We require the angle between the vectors \(\overrightarrow{AM}\) and \(\overrightarrow{AL}\) and we can find this using the dot product.

\[\begin{align*} \overrightarrow{AM} &= \overrightarrow{OM} -\overrightarrow{OA} = \dfrac{8}{5}\mathbf{i}+ \dfrac{16}{5}\mathbf{j} - (2\mathbf{i}+3\mathbf{j}+5\mathbf{k}) = -\frac{2}{5}\mathbf{i} + \frac{1}{5}\mathbf{j} - 5\mathbf{k} \\ \overrightarrow{AL} &= \overrightarrow{OL} -\overrightarrow{OA} = 2\mathbf{i} - (2\mathbf{i}+3\mathbf{j}+5\mathbf{k}) = -3\mathbf{j}-5\mathbf{k} \end{align*}\] So now, \[\begin{align*} \overrightarrow{AM} . \overrightarrow{AL} &= 0-\frac{3}{5}+25 = \frac{122}{5} \\ &= \big\vert AM\big\vert\times\big\vert AL\big\vert\cos\theta \\ &= \sqrt{\left(\tfrac{2}{5}\right)^2 + \left(\tfrac{1}{5}\right)^2 +5^2}\sqrt{3^2+5^2}\cos\theta \\ &= \sqrt{\tfrac{126}{5}}\sqrt{34}\cos\theta \\ \implies\quad \cos\theta &=\frac{122/5}{\sqrt{126/5}\sqrt{34}}\approx0.8336 \\ \implies\quad \theta &\approx33.53^\circ \end{align*}\]The angle \(MAL\) is \(34^\circ\).

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 \(\dfrac{OX}{ON} = \dfrac{16}{19}\).

We can now work in two dimensions. We have that \(\overrightarrow{OL} = 2\mathbf{i}\), \(\overrightarrow{OM} = \frac{8}{5}\mathbf{i}+\frac{16}{5}\mathbf{j}\) and \(\overrightarrow{ON} = 2\mathbf{i}+3\mathbf{j}\). Let \(\overrightarrow{OX} = \alpha (2\mathbf{i}+3\mathbf{j})\) so that \(\alpha\) is the required ratio.

The vector \(\overrightarrow{LM} = \frac{8}{5}\mathbf{i}+\frac{16}{5}\mathbf{j} - 2\mathbf{i} = -\frac{2}{5}\mathbf{i}+\frac{16}{5}\mathbf{j}\).

The line \(LM\) is \(\mathbf{r} = 2\mathbf{i} + \beta(-\frac{2}{5}\mathbf{i}+\frac{16}{5}\mathbf{j})\). So at \(X\), \[ \alpha (2\mathbf{i}+3\mathbf{j}) = 2\mathbf{i} + \beta\left(-\tfrac{2}{5}\mathbf{i}+\tfrac{16}{5}\mathbf{j}\right) \] Comparing coefficients, we have \[\begin{align*} 2\alpha &= 2-\tfrac{2}{5}\beta \text{ and } 3\alpha = \tfrac{16}{5}\beta \\ \implies\quad \alpha &= \frac{16}{19} \end{align*}\]as required.