Free practice / JC2 H2 Maths Prelim
Vectors: Line intersection; area; proj. — MI 2025 H2 Math Prelim Paper 1
Attempt the question on paper, then open the worked solution to check your method.
What this question tests
Question
Referred to the origin \(O\), the points \(A\) and \(B\) have position vectors \(\mathbf{a}\) and \(\mathbf{b}\) respectively, where \(\mathbf{a}\) and \(\mathbf{b}\) are non-zero and non-parallel vectors. The point \(C\) lies on \(OA\) such that \(OC:CA = 1:2\). The point \(D\) lies on \(OB\) such that \(OB:DB = 3:1\).
- Find the position vectors \(\overrightarrow{OC}\) and \(\overrightarrow{OD}\), giving your answers in terms of \(\mathbf{a}\) and \(\mathbf{b}\).
- Show that the point \(E\) where the lines \(AD\) and \(BC\) meet has position vector \(\dfrac{1}{7}\mathbf{a} + \dfrac{4}{7}\mathbf{b}\).
- Show that the area of triangle \(OAE\) can be written as \(k|\mathbf{a}\times\mathbf{b}|\), where \(k\) is a constant to be found.
It is further given that \(\mathbf{b}\) is a unit vector.
- Give the geometrical meaning of \(|\mathbf{a}\cdot\mathbf{b}|\).
- If \(|\mathbf{a}\cdot\mathbf{b}| = \dfrac{1}{2}|\mathbf{a}|\), find the perpendicular distance of point \(A\) to \(OB\), leaving your answer in terms of \(|\mathbf{a}|\).
Show full worked solution▾
(i) \(OC:CA = 1:2\) means \(C\) divides \(OA\) in the ratio \(1:2\).
\(\overrightarrow{OC} = \dfrac{1}{3}\overrightarrow{OA} = \dfrac{1}{3}\mathbf{a}\).
\(OB:DB = 3:1\) means \(D\) divides \(OB\) such that \(OD:DB = 2:1\).
\(\overrightarrow{OD} = \dfrac{2}{3}\overrightarrow{OB} = \dfrac{2}{3}\mathbf{b}\).
\(\overrightarrow{OC} = \dfrac{1}{3}\mathbf{a}\), \(\overrightarrow{OD} = \dfrac{2}{3}\mathbf{b}\)
(ii) \[\begin{aligned} \overrightarrow{OE} &= \overrightarrow{OA}+\lambda\overrightarrow{AD} \quad\text{--- (1)}\\ \overrightarrow{OE} &= \overrightarrow{OB}+\mu\overrightarrow{BC} \quad\text{--- (2)} \end{aligned}\]
Solve (1) = (2): \[\begin{aligned} \overrightarrow{OA}+\lambda\overrightarrow{AD} &= \overrightarrow{OB}+\mu\overrightarrow{BC}\\ \overrightarrow{OA}+\lambda(\overrightarrow{OD}-\overrightarrow{OA}) &= \overrightarrow{OB}+\mu(\overrightarrow{OC}-\overrightarrow{OB})\\ \mathbf{a}+\lambda\!\left(\frac{2}{3}\mathbf{b}-\mathbf{a}\right) &= \mathbf{b}+\mu\!\left(\frac{1}{3}\mathbf{a}-\mathbf{b}\right)\\ (1-\lambda)\mathbf{a}+\frac{2}{3}\lambda\mathbf{b} &= \frac{1}{3}\mu\mathbf{a}+(1-\mu)\mathbf{b} \end{aligned}\]
Comparing coefficients of \(\mathbf{a}\): \(1-\lambda = \dfrac{1}{3}\mu \;\Rightarrow\; \lambda + \dfrac{1}{3}\mu = 1\) — (3)
Comparing coefficients of \(\mathbf{b}\): \(\dfrac{2}{3}\lambda = 1-\mu \;\Rightarrow\; \dfrac{2}{3}\lambda+\mu = 1\) — (4)
Solving (3) & (4) by GC: \(\lambda = \dfrac{6}{7}\), \(\mu = \dfrac{3}{7}\).
Substituting \(\lambda = \dfrac{6}{7}\) into (1): \[\begin{aligned} \overrightarrow{OE} &= \mathbf{a}+\frac{6}{7}\!\left(\frac{2}{3}\mathbf{b}-\mathbf{a}\right)\\ &= \mathbf{a}+\frac{4}{7}\mathbf{b}-\frac{6}{7}\mathbf{a}\\ &= \frac{1}{7}\mathbf{a}+\frac{4}{7}\mathbf{b} \quad\text{(shown)} \end{aligned}\]
(iii) Area of \(\triangle OAE = \dfrac{1}{2}|\overrightarrow{OA}\times\overrightarrow{OE}|\).
\[\begin{aligned} &= \frac{1}{2}\left|\mathbf{a}\times\left(\frac{1}{7}\mathbf{a}+\frac{4}{7}\mathbf{b}\right)\right|\\ &= \frac{1}{2}\left|\frac{1}{7}(\mathbf{a}\times\mathbf{a})+\frac{4}{7}(\mathbf{a}\times\mathbf{b})\right|\\ &= \frac{1}{2}\left|0+\frac{4}{7}(\mathbf{a}\times\mathbf{b})\right| \quad (\because\;\mathbf{a}\times\mathbf{a}=\mathbf{0})\\ &= \frac{2}{7}|\mathbf{a}\times\mathbf{b}| \end{aligned}\]where \(k = \dfrac{2}{7}\) (shown).
\(k = \dfrac{2}{7}\) (shown)
(iv) \(|\mathbf{a}\cdot\mathbf{b}|\) is the length of projection of \(\overrightarrow{OA}\) onto the line \(OB\).
The length of projection of \(\overrightarrow{OA}\) onto \(\overrightarrow{OB}\).
(v) Perpendicular distance from \(A\) to \(OB\):
