2013 H2 Mathematics Paper 1 Question 6

Vectors I: Basics, Dot and Cross Products

Answers

{\overrightarrow{ON} = \frac{4}{7}\mathbf{a} + \frac{3}{7}\mathbf{c}}

{\lambda = \frac{8}{7}\mu}

An equation for the plane passing through

{O,A}

and

{B}

is:

\mathbf{r} = \mathbf{0} + k_1\mathbf{a} + k_2\mathbf{b}, \textrm{ for } k_1,k_2 \in \mathbb{R}

Since

{O, A, B}

and

{C}

lies in the same plane,

{\mathbf{c}}

satisfies the equation of the plane and hence be expressed as

{\mathbf{c} = \lambda \mathbf{a} + \mu \mathbf{b} \; \blacksquare}

By Ratio Theorem

\begin{align*} \overrightarrow{ON} &= \frac{3\mathbf{c}+4\mathbf{a}}{7}\\ & = \frac{4}{7}\mathbf{a} + \frac{3}{7}\mathbf{c} \; \blacksquare \end{align*}

\begin{align*} & \textrm{Area of } \triangle ONC \\ &= \frac{1}{2} \left| \overrightarrow{ON} \times \overrightarrow{OC} \right| \\ &= \frac{1}{2} \left| \left( \frac{4}{7}\mathbf{a} + \frac{3}{7}\mathbf{c}\right) \times \mathbf{c} \right| \\ &= \frac{1}{2} \left| \frac{4}{7} \mathbf{a} \times \mathbf{c} \right| \\ &= \frac{2}{7} \Big | \mathbf{a} \times (\lambda \mathbf{a} + \mu \mathbf{b}) \Big | \\ &= \frac{2}{7} \left| \mu \mathbf{a} \times \mathbf{b} \right| \\ &= \frac{2}{7} | \mu| \left| \mathbf{a} \times \mathbf{b} \right| \\ \end{align*}

{\overrightarrow{OM} = \frac{1}{2}\mathbf{b}}

\begin{align*} & \textrm{Area of } \triangle OMC \\ &= \frac{1}{2} \left| \overrightarrow{OM} \times \overrightarrow{OC} \right| \\ &= \frac{1}{2} \left| \frac{1}{2}\mathbf{b} \times \mathbf{c} \right| \\ &= \frac{1}{4} \Big| \mathbf{b} \times (\lambda \mathbf{a} + \mu \mathbf{b}) \Big| \\ &= \frac{1}{4} \left| \lambda \mathbf{b} \times \mathbf{a} \right| \\ &= \frac{1}{4} |\lambda| \left| \mathbf{a} \times \mathbf{b} \right| \\ \end{align*}

Since the areas of the triangles are equal and

{\lambda}

and

{mu}

are positive,

\begin{align*} \frac{2}{7}\mu &= \frac{1}{4}\lambda \\ \lambda &= \frac{8}{7}\mu \; \blacksquare \end{align*}