Vectors: Ratios; foot of perp; area — TJC 2025 H2 Math Prelim Paper 1
What this question tests
Question
The points \(A\) and \(B\) lie on a circle with centre \(O\) and radius \(\lambda\) unit. With reference to the origin \(O\), points \(A\) and \(B\) have position vectors \(\mathbf{a}\) and \(\mathbf{b}\) respectively. The point \(X\) on the line segment \(AB\) is such that \(AX : XB = 1 : 3\) and the point \(Y\) is the foot of perpendicular of \(X\) on \(OB\).
- Find the position vector of \(X\).
It is given that the acute angle \(AOB\) is \(\dfrac{\pi}{6}\).
- Find \(\mathbf{a} \cdot \mathbf{b}\) in terms of \(\lambda\).
- Show that the position vector of \(Y\) is \(\dfrac{2 + 3\sqrt{3}}{8}\,\mathbf{b}\).
- Hence find the exact area of \(\triangle OXY\) in terms of \(\lambda\).
Show full worked solution▾
(a) Using the ratio theorem with \(X\) dividing \(AB\) in ratio \(3:1\) (i.e. \(OX = \tfrac{1\cdot OA + 3\cdot OB}{4}\)):
(b) Since \(|\mathbf{a}| = |\mathbf{b}| = \lambda\) (radii of same circle) and the angle between them is \(\dfrac{\pi}{6}\): \[ \mathbf{a}\cdot\mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\frac{\pi}{6} = \lambda^2\cdot\frac{\sqrt{3}}{2} \]
(c) \(Y\) is the foot of perpendicular from \(O\) to line \(OB\)... actually \(Y\) is the projection of \(X\) onto \(OB\). \[ \overrightarrow{OY} = \left(\overrightarrow{OX}\cdot\hat{\mathbf{b}}\right)\hat{\mathbf{b}} = \left(\frac{3\mathbf{a}+\mathbf{b}}{4}\cdot\frac{\mathbf{b}}{|\mathbf{b}|}\right)\frac{\mathbf{b}}{|\mathbf{b}|} = \frac{1}{4}\cdot\frac{\mathbf{b}}{|\mathbf{b}|^2}\cdot(3\mathbf{a}\cdot\mathbf{b}+\mathbf{b}\cdot\mathbf{b}) \] \[ = \frac{1}{4}\cdot\frac{\mathbf{b}}{\lambda^2} \cdot\left(3\lambda^2\cdot\frac{\sqrt{3}}{2} + \lambda^2\right) = \frac{1}{4}\cdot\frac{\mathbf{b}}{\lambda^2} \cdot\lambda^2\!\left(\frac{3\sqrt{3}}{2}+1\right) = \frac{2+3\sqrt{3}}{8}\,\mathbf{b} \quad\text{(Shown)} \]
(d) \[ \text{Area of }\triangle OXY = \frac{1}{2}\,|\overrightarrow{OX}\times\overrightarrow{OY}| = \frac{1}{2}\left|\frac{3\mathbf{a}+\mathbf{b}}{4}\times\frac{(2+3\sqrt{3})}{8}\,\mathbf{b}\right| \] \[ = \frac{2+3\sqrt{3}}{64}\,|{(3\mathbf{a}+\mathbf{b})\times\mathbf{b}}| = \frac{2+3\sqrt{3}}{64}\,|3\mathbf{a}\times\mathbf{b}| \quad(\text{since }\mathbf{b}\times\mathbf{b}=\mathbf{0}) \] \[ = \frac{3(2+3\sqrt{3})}{64}\,|\mathbf{a}||\mathbf{b}|\sin\frac{\pi}{6} = \frac{3(2+3\sqrt{3})}{64}\cdot\lambda^2\cdot\frac{1}{2} = \frac{3(2+3\sqrt{3})}{128}\,\lambda^2 \]