2026-08-31T11:34:15.442705 image/svg+xml Matplotlib v3.10.9, https://matplotlib.org/ © EAGLE BEACON GLOBAL Figure 2 shows the parallelogram \(OACB\), in whi...

Twelve marks of vector geometry in a parallelogram. Figure 2 shows \(OACB\) with \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\), \(M\) the midpoint of \(BC\), and \(X\) the point where \(OM\) crosses the diagonal \(AB\). The whole question turns on one powerful idea: if \(\mathbf{a}\) and \(\mathbf{b}\) are not parallel, then any point of the plane has exactly one expression as \(\lambda\mathbf{a} + \mu\mathbf{b}\). Writing the same point two ways therefore forces the coefficients to match, and that gives simultaneous equations.

(a) \(\overrightarrow{OC}\) [1 mark]

In the parallelogram \(OACB\), the side \(AC\) is equal and parallel to \(OB\), so travelling \(O\) to \(A\) to \(C\) is \(\mathbf{a}\) followed by \(\mathbf{b}\): \[\overrightarrow{OC} = \mathbf{a} + \mathbf{b} \quad \textbf{B1}\]

(b) Show that \(\overrightarrow{OM} = \tfrac{1}{2}\mathbf{a} + \mathbf{b}\) [2 marks]

Since the result is given, the marks are for the route. First find the side \(BC\): \[\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = (\mathbf{a} + \mathbf{b}) - \mathbf{b} = \mathbf{a}, \ \text{so} \ \overrightarrow{BM} = \tfrac{1}{2}\mathbf{a} \quad \textbf{M1}\] Then travel \(O\) to \(B\) to \(M\): \[\overrightarrow{OM} = \overrightarrow{OB} + \overrightarrow{BM} = \mathbf{b} + \tfrac{1}{2}\mathbf{a} = \tfrac{1}{2}\mathbf{a} + \mathbf{b} \quad \textbf{A1}\]

(c) Two expressions for \(\overrightarrow{OX}\) [3 marks]

Along \(OM\), \(X\) is a fraction \(t\) of the way: \[\overrightarrow{OX} = t\left(\tfrac{1}{2}\mathbf{a} + \mathbf{b}\right) = \tfrac{1}{2}t\,\mathbf{a} + t\,\mathbf{b} \quad \textbf{B1}\] Along \(AB\), start at \(A\) and move a fraction \(s\) of the way towards \(B\). The direction is \[\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \mathbf{b} - \mathbf{a} \quad \textbf{B1}\] so \[\overrightarrow{OX} = \overrightarrow{OA} + s\,\overrightarrow{AB} = \mathbf{a} + s(\mathbf{b} - \mathbf{a}) = (1 - s)\mathbf{a} + s\,\mathbf{b} \quad \textbf{B1}\] Three independent B marks. Both expressions must be written purely in terms of \(\mathbf{a}\), \(\mathbf{b}\) and the single parameter, with the brackets expanded ready for comparison.

(d) The values of \(s\) and \(t\) [3 marks]

The two expressions describe the same point, and \(\mathbf{a}\) and \(\mathbf{b}\) are non-parallel, so the coefficients may be equated: \[\text{Coefficients may be compared because } \mathbf{a} \ \text{and} \ \mathbf{b} \ \text{are not parallel.} \quad \textbf{M1}\] \[\mathbf{a}: \ \tfrac{1}{2}t = 1 - s; \qquad \mathbf{b}: \ t = s \quad \textbf{A1}\] Substituting \(t = s\) into the first equation: \[\tfrac{1}{2}s = 1 - s \ \Rightarrow\ \tfrac{3}{2}s = 1 \ \Rightarrow\ s = \tfrac{2}{3}, \qquad t = \tfrac{2}{3} \quad \textbf{A1}\] The M1 rewards the justification as well as the act of comparing; it is the mathematical content of the part. The final A1 needs both values.

(e) The ratio \(AX : XB\) [1 mark]

Since \(\overrightarrow{AX} = s\,\overrightarrow{AB} = \tfrac{2}{3}\overrightarrow{AB}\), the point \(X\) is two thirds of the way from \(A\) to \(B\), leaving one third: \[AX : XB = 2 : 1 \quad \textbf{B1}\] Read from \(s\) directly as \(s : (1 - s)\). Quoting \(1 : 2\) reverses the segments.

(f) \(\left|\overrightarrow{OX}\right|\) with \(\mathbf{a} = 4\mathbf{i} + \mathbf{j}\), \(\mathbf{b} = -2\mathbf{i} + 5\mathbf{j}\) [2 marks]

Use either expression with \(s = t = \tfrac{2}{3}\). Taking the second gives \(\overrightarrow{OX} = \tfrac{1}{3}\mathbf{a} + \tfrac{2}{3}\mathbf{b}\): \[\overrightarrow{OX} = \tfrac{1}{3}(4\mathbf{i} + \mathbf{j}) + \tfrac{2}{3}(-2\mathbf{i} + 5\mathbf{j}) = \left(\tfrac{4}{3} - \tfrac{4}{3}\right)\mathbf{i} + \left(\tfrac{1}{3} + \tfrac{10}{3}\right)\mathbf{j} = \tfrac{11}{3}\mathbf{j} \quad \textbf{M1}\] The \(\mathbf{i}\) components cancel exactly, so \(\overrightarrow{OX}\) points straight up the \(\mathbf{j}\) direction and its modulus is simply the size of that one component: \[\left|\overrightarrow{OX}\right| = \tfrac{11}{3} \quad \textbf{A1}\] Using the first expression is an equally good check: \(\tfrac{2}{3}\left(\tfrac{1}{2}\mathbf{a} + \mathbf{b}\right) = \tfrac{1}{3}\mathbf{a} + \tfrac{2}{3}\mathbf{b}\), the same vector.

Common wrong turns on this question

• Taking \(M\) as the midpoint of \(OC\) or of \(AC\). Read the figure: \(M\) is the midpoint of \(BC\), which is why \(\overrightarrow{OM}\) carries a full \(\mathbf{b}\) and only half of \(\mathbf{a}\).
• Writing \(\overrightarrow{AB} = \mathbf{a} - \mathbf{b}\). "To minus from" gives \(\mathbf{b} - \mathbf{a}\); the reversed version leads to \(s = \tfrac{1}{3}\) and a wrong ratio.
• Equating coefficients without stating that \(\mathbf{a}\) and \(\mathbf{b}\) are non-parallel. That justification is what the M1 in part (d) is for; the step is invalid for parallel vectors.
• Assuming \(X\) is the midpoint of \(AB\) because \(OM\) looks central in the figure. The diagonal is cut in the ratio \(2 : 1\), not \(1 : 1\).
• Being surprised by the cancelling \(\mathbf{i}\) components in part (f) and assuming an error. The numbers are chosen so that \(\overrightarrow{OX}\) is vertical.

How to check

With the given components, \(\overrightarrow{OM} = \tfrac{1}{2}(4\mathbf{i} + \mathbf{j}) + (-2\mathbf{i} + 5\mathbf{j}) = 0\mathbf{i} + \tfrac{11}{2}\mathbf{j}\), so \(OM\) itself lies along the \(\mathbf{j}\)-axis and \(\tfrac{2}{3}\) of it is \(\tfrac{11}{3}\mathbf{j}\), confirming part (f) independently. The ratio in part (e) is also verifiable numerically: \(A\) is at \((4, 1)\), \(B\) at \((-2, 5)\), and the point two thirds of the way along is \(\left(4 - \tfrac{2}{3}(6),\ 1 + \tfrac{2}{3}(4)\right) = \left(0,\ \tfrac{11}{3}\right)\), exactly the \(X\) found.