Vectors: Ratio Theorem, Foot of Perpendicular and Triangle Area — TJC 2025 H2 Math Prelim P1
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 \(OB\).
(a) Find the position vector of \(X\).
It is given that the acute angle \(AOB\) is \(\dfrac{\pi}{6}\).
(b) Find \(\mathbf{a} \cdot \mathbf{b}\) in terms of \(\lambda\).
(c) Show that the position vector of \(Y\) is \(\dfrac{2 + 3\sqrt{3}}{8}\,\mathbf{b}\).
(d) Hence find the exact area of \(\triangle OXY\) in terms of \(\lambda\).
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(a) Position vector of \(X\)
Apply the ratio theorem: \(X\) divides \(AB\) in the ratio \(AX:XB = 1:3\).
Since \(AX:XB = 1:3\), the ratio theorem weights the endpoints \(A\) and \(B\) by the opposite segments, giving
\[ \overrightarrow{OX} = \frac{3\cdot OA + 1\cdot OB}{4} = \frac{3\mathbf{a}+\mathbf{b}}{4} \](b) Value of \(\mathbf{a}\cdot\mathbf{b}\)
Both \(A\) and \(B\) lie on the same circle, so \(|\mathbf{a}| = |\mathbf{b}| = \lambda\), 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) Position vector of \(Y\)
\(Y\) is the projection of \(X\) onto the line \(OB\), so project \(\overrightarrow{OX}\) onto the unit vector \(\hat{\mathbf{b}}\).
\[ \overrightarrow{OY} = \left(\overrightarrow{OX}\cdot\hat{\mathbf{b}}\right)\hat{\mathbf{b}} = \left(\frac{3\mathbf{a}+\mathbf{b}}{4}\cdot\frac{\mathbf{b}}{|\mathbf{b}|}\right)\frac{\mathbf{b}}{|\mathbf{b}|} = \frac{1}{4}\cdot\frac{\mathbf{b}}{|\mathbf{b}|^2}\cdot(3\mathbf{a}\cdot\mathbf{b}+\mathbf{b}\cdot\mathbf{b}) \] \[ = \frac{1}{4}\cdot\frac{\mathbf{b}}{\lambda^2} \cdot\left(3\lambda^2\cdot\frac{\sqrt{3}}{2} + \lambda^2\right) = \frac{1}{4}\cdot\frac{\mathbf{b}}{\lambda^2} \cdot\lambda^2\!\left(\frac{3\sqrt{3}}{2}+1\right) = \frac{2+3\sqrt{3}}{8}\,\mathbf{b} \quad\text{(Shown)} \](d) Exact area of \(\triangle OXY\)
Use the cross product: the area is half the magnitude of \(\overrightarrow{OX}\times\overrightarrow{OY}\).
\[ \text{Area of }\triangle OXY = \frac{1}{2}\,\left|\overrightarrow{OX}\times\overrightarrow{OY}\right| = \frac{1}{2}\left|\frac{3\mathbf{a}+\mathbf{b}}{4}\times\frac{2+3\sqrt{3}}{8}\,\mathbf{b}\right| \] \[ = \frac{2+3\sqrt{3}}{64}\,\left|(3\mathbf{a}+\mathbf{b})\times\mathbf{b}\right| = \frac{2+3\sqrt{3}}{64}\,|3\mathbf{a}\times\mathbf{b}| \quad(\text{since }\mathbf{b}\times\mathbf{b}=\mathbf{0}) \] \[ = \frac{3(2+3\sqrt{3})}{64}\,|\mathbf{a}||\mathbf{b}|\sin\frac{\pi}{6} = \frac{3(2+3\sqrt{3})}{64}\cdot\lambda^2\cdot\frac{1}{2} = \frac{3(2+3\sqrt{3})}{128}\,\lambda^2 \]