2026-08-31T11:35:14.531460 image/svg+xml Matplotlib v3.10.9, https://matplotlib.org/ © EAGLE BEACON GLOBAL Figure 1 shows the quadrilateral \(OABC\), in whi...

This is the standard method for finding where two lines meet in vector form: express the same point by two different routes and equate. The statement that \(\mathbf{a}\) and \(\mathbf{c}\) are not parallel is not incidental. It is what makes them a basis for the plane, so that a vector has only one expression in terms of them and the coefficients can be compared. Without that condition the whole method collapses.

(a) The two diagonals [2 marks]

Build each vector by walking along routes that are already known.

(i) From \(O\) to \(B\) via \(A\):

\[\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{AB} = \mathbf{a} + 3\mathbf{c} \qquad \textbf{[1]}\]

(ii) From \(A\) to \(C\) via \(O\), remembering that reversing a vector reverses its sign:

\[\overrightarrow{AC} = \overrightarrow{AO} + \overrightarrow{OC} = -\mathbf{a} + \mathbf{c} = \mathbf{c} - \mathbf{a} \qquad \textbf{[1]}\]

Each is one independent mark. The order matters: \(\mathbf{a} - \mathbf{c}\) is \(\overrightarrow{CA}\) and does not score. Note that \(\overrightarrow{AB} = 3\mathbf{c}\) is given, so \(AB\) is parallel to \(OC\) and three times as long, which is what makes \(OABC\) a trapezium and explains the answer to part (c) in advance.

(b) Showing \(\lambda = \dfrac{1}{4}\) and finding \(\mu\) [4 marks]

The point \(X\) lies on both diagonals, so reach it twice. Along \(OB\):

\[\overrightarrow{OX} = \lambda\,\overrightarrow{OB} = \lambda\left(\mathbf{a} + 3\mathbf{c}\right) = \lambda\mathbf{a} + 3\lambda\mathbf{c} \qquad \textbf{M1}\]

Along \(AC\), starting from \(A\) and therefore adding the position vector of \(A\) first:

\[\overrightarrow{OX} = \overrightarrow{OA} + \mu\,\overrightarrow{AC} = \mathbf{a} + \mu\left(\mathbf{c} - \mathbf{a}\right) = \left(1 - \mu\right)\mathbf{a} + \mu\mathbf{c} \qquad \textbf{M1}\]

These two M marks are for the two routes and survive the algebra that follows. The commonest error is writing the second route as \(\mu\left(\mathbf{c} - \mathbf{a}\right)\) alone, forgetting that \(\overrightarrow{AX}\) is measured from \(A\) and so must be added to \(\overrightarrow{OA}\) to give a position vector from \(O\).

Because \(\mathbf{a}\) and \(\mathbf{c}\) are not parallel, the two expressions can only agree if the coefficients of \(\mathbf{a}\) match and the coefficients of \(\mathbf{c}\) match separately:

\[\lambda = 1 - \mu, \qquad 3\lambda = \mu \qquad \textbf{M1}\]

This is the step to state explicitly, together with its justification, since the third M mark is for equating components. Substituting the second equation into the first:

\[\lambda = 1 - 3\lambda \quad \Longrightarrow \quad 4\lambda = 1 \quad \Longrightarrow \quad \lambda = \frac{1}{4}, \qquad \mu = 3\lambda = \frac{3}{4} \qquad \textbf{A1}\]

The value of \(\lambda\) is printed in the question, so it functions as a check: a candidate who obtains anything else knows at once that a slip has occurred, and one who writes \(\lambda = \tfrac{1}{4}\) down without the two routes earns nothing. The single A1 covers both constants, so \(\mu\) must be right as well. Substituting into either expression gives the same point, which is the final check:

\[\tfrac{1}{4}\mathbf{a} + \tfrac{3}{4}\mathbf{c} \quad \text{and} \quad \left(1 - \tfrac{3}{4}\right)\mathbf{a} + \tfrac{3}{4}\mathbf{c} = \tfrac{1}{4}\mathbf{a} + \tfrac{3}{4}\mathbf{c}.\]

(c) The ratio \(AX : XC\) and the position vector of \(X\) [2 marks]

Since \(\overrightarrow{AX} = \dfrac{3}{4}\overrightarrow{AC}\), the point \(X\) lies three quarters of the way from \(A\) to \(C\). The remaining quarter is \(XC\), so

\[AX : XC = \frac{3}{4} : \frac{1}{4} = 3 : 1 \qquad \textbf{[1]}\]

The trap is quoting \(3 : 4\), which is the ratio \(AX : AC\) rather than \(AX : XC\). The part after the colon is the remaining piece, not the whole. From part (b), using \(\lambda = \dfrac{1}{4}\):

\[\overrightarrow{OX} = \tfrac{1}{4}\left(\mathbf{a} + 3\mathbf{c}\right) = \tfrac{1}{4}\mathbf{a} + \tfrac{3}{4}\mathbf{c} \qquad \textbf{[1]}\]

The two answers are consistent with the geometry. Because \(AB\) is three times \(OC\) and parallel to it, triangles \(OXC\) and \(BXA\) are similar with a scale factor of \(3\), so the diagonals cut each other in the ratio \(3 : 1\) with the longer part next to the longer parallel side. That similar-triangles view is a quick way to predict the answer before doing any algebra, and a useful way to confirm it afterwards. It also explains \(\lambda = \dfrac{1}{4}\): \(X\) is one quarter of the way along \(OB\) from \(O\), the same quarter that \(OC\) represents of \(OC\) plus \(AB\) taken together.

Examination takeaway. To find an intersection in vector form, write the same position vector by two routes and equate the coefficients of the two non-parallel base vectors, always adding the position vector of the starting point when a displacement begins away from the origin. When converting a fraction of a segment into a ratio, remember that the second number is what is left over.