Vectors: Line Intersection, Triangle Area and Projection — MI 2025 H2 Math Prelim P1

(i) Position vectors of \(C\) and \(D\)

Use the given ratios to write each point as a fraction along its line.

\(OC:CA = 1:2\) means \(C\) divides \(OA\) in the ratio \(1:2\), so

\[ \overrightarrow{OC} = \tfrac{1}{3}\overrightarrow{OA} = \tfrac{1}{3}\mathbf{a}. \]

\(OB:DB = 3:1\) means \(D\) divides \(OB\) such that \(OD:DB = 2:1\), so

\[ \overrightarrow{OD} = \tfrac{2}{3}\overrightarrow{OB} = \tfrac{2}{3}\mathbf{b}. \]

(ii) Intersection point \(E\)

Write \(\overrightarrow{OE}\) two ways along lines \(AD\) and \(BC\), then compare coefficients of \(\mathbf{a}\) and \(\mathbf{b}\).

\[ \begin{aligned} \overrightarrow{OE} &= \overrightarrow{OA}+\lambda\overrightarrow{AD} \quad\text{--- (1)}\\ \overrightarrow{OE} &= \overrightarrow{OB}+\mu\overrightarrow{BC} \quad\text{--- (2)} \end{aligned} \]

Setting (1) = (2):

\[ \begin{aligned} \overrightarrow{OA}+\lambda(\overrightarrow{OD}-\overrightarrow{OA}) &= \overrightarrow{OB}+\mu(\overrightarrow{OC}-\overrightarrow{OB})\\ \mathbf{a}+\lambda\!\left(\tfrac{2}{3}\mathbf{b}-\mathbf{a}\right) &= \mathbf{b}+\mu\!\left(\tfrac{1}{3}\mathbf{a}-\mathbf{b}\right)\\ (1-\lambda)\mathbf{a}+\tfrac{2}{3}\lambda\mathbf{b} &= \tfrac{1}{3}\mu\mathbf{a}+(1-\mu)\mathbf{b} \end{aligned} \]

Comparing coefficients of \(\mathbf{a}\): \(\;1-\lambda = \tfrac{1}{3}\mu \Rightarrow \lambda + \tfrac{1}{3}\mu = 1\) --- (3)

Comparing coefficients of \(\mathbf{b}\): \(\;\tfrac{2}{3}\lambda = 1-\mu \Rightarrow \tfrac{2}{3}\lambda+\mu = 1\) --- (4)

Solving (3) and (4): \(\lambda = \dfrac{6}{7},\ \mu = \dfrac{3}{7}\). Substituting \(\lambda = \dfrac{6}{7}\) into (1):

\[ \begin{aligned} \overrightarrow{OE} &= \mathbf{a}+\tfrac{6}{7}\!\left(\tfrac{2}{3}\mathbf{b}-\mathbf{a}\right)\\ &= \mathbf{a}+\tfrac{4}{7}\mathbf{b}-\tfrac{6}{7}\mathbf{a}\\ &= \tfrac{1}{7}\mathbf{a}+\tfrac{4}{7}\mathbf{b} \quad\text{(shown)} \end{aligned} \]

(iii) Area of triangle \(OAE\)

Half the magnitude of the cross product of two sides from \(O\); the \(\mathbf{a}\times\mathbf{a}\) term vanishes.

\[ \begin{aligned} \text{Area} &= \tfrac{1}{2}\left|\overrightarrow{OA}\times\overrightarrow{OE}\right| = \tfrac{1}{2}\left|\mathbf{a}\times\left(\tfrac{1}{7}\mathbf{a}+\tfrac{4}{7}\mathbf{b}\right)\right|\\ &= \tfrac{1}{2}\left|\tfrac{1}{7}(\mathbf{a}\times\mathbf{a})+\tfrac{4}{7}(\mathbf{a}\times\mathbf{b})\right|\\ &= \tfrac{1}{2}\left|\tfrac{4}{7}(\mathbf{a}\times\mathbf{b})\right| \qquad (\because\ \mathbf{a}\times\mathbf{a}=\mathbf{0})\\ &= \tfrac{2}{7}|\mathbf{a}\times\mathbf{b}| \end{aligned} \]

so \(k = \dfrac{2}{7}\) (shown).

(iv) Geometrical meaning of \(|\mathbf{a}\cdot\mathbf{b}|\)

With \(\mathbf{b}\) a unit vector, the scalar product gives a length of projection.

\(|\mathbf{a}\cdot\mathbf{b}|\) is the length of projection of \(\overrightarrow{OA}\) onto the line \(OB\).

(v) Perpendicular distance from \(A\) to \(OB\)

Use Pythagoras: the perpendicular distance is found from \(|\mathbf{a}|\) and the projection length.

\[ \begin{aligned} \text{Distance} &= \sqrt{|\mathbf{a}|^2 - |\mathbf{a}\cdot\hat{\mathbf{b}}|^2}\\ &= \sqrt{|\mathbf{a}|^2 - \left(\tfrac{1}{2}|\mathbf{a}|\right)^2} \qquad \left(\because\ |\mathbf{a}\cdot\hat{\mathbf{b}}| = |\mathbf{a}\cdot\mathbf{b}| = \tfrac{1}{2}|\mathbf{a}|\right)\\ &= \sqrt{|\mathbf{a}|^2 - \tfrac{1}{4}|\mathbf{a}|^2} = \sqrt{\tfrac{3}{4}|\mathbf{a}|^2} = \tfrac{\sqrt{3}}{2}|\mathbf{a}| \end{aligned} \]