2015 H2 Maths Paper 1 Q7 — Worked Solution

Part (i)

2 marks

Since $C$ lies on $OA$ with $OC:CA = 3:2$ , $C$ divides $OA$ so that $OC = \frac{3}{5}OA$ :

\overrightarrow{OC} = \frac{3}{5}\mathbf{a}.

Since $D$ lies on $OB$ with $OD:DB = 5:6$ , $D$ divides $OB$ so that $OD = \frac{5}{11}OB$ :

\overrightarrow{OD} = \frac{5}{11}\mathbf{b}.

Part (ii)

3 marks

Approach

A line through two points with known position vectors is written $\mathbf{r} = \mathbf{p} + \lambda\left( \mathbf{q} - \mathbf{p} \right)$ , taking one point as the base and the displacement to the other as the direction. We take $B$ as the base for line $BC$ , so that $\lambda = 0$ gives $B$ and $\lambda = 1$ gives $C$ , matching the form given in the question.

The line $BC$ passes through $B$ (position vector $\mathbf{b}$ ) in the direction of $C - B$ :

\mathbf{r} = \mathbf{b} + \lambda\left( \frac{3}{5}\mathbf{a} - \mathbf{b} \right)

= \frac{3}{5}\lambda\mathbf{a} + (1 - \lambda)\mathbf{b},

as required $\ \$ (shown).

By the same method, the line $AD$ passes through $A$ (position vector $\mathbf{a}$ ) in the direction of $D - A$ :

\mathbf{r} = \mathbf{a} + \mu\left( \frac{5}{11}\mathbf{b} - \mathbf{a} \right)

= (1 - \mu)\mathbf{a} + \frac{5}{11}\mu\mathbf{b}.

Part (iii)

5 marks

Approach

At the point $E$ where the lines meet, the two expressions for $\mathbf{r}$ are equal for some values of $\lambda$ and $\mu$ . Since $\mathbf{a}$ and $\mathbf{b}$ are non-parallel (non-zero, non-collinear position vectors from $O$ ), we can compare coefficients of $\mathbf{a}$ and of $\mathbf{b}$ separately.

At the intersection:

\frac{3}{5}\lambda\mathbf{a} + (1 - \lambda)\mathbf{b} = (1 - \mu)\mathbf{a} + \frac{5}{11}\mu\mathbf{b}.

Comparing coefficients of $\mathbf{a}$ :

\frac{3}{5}\lambda = 1 - \mu.\quad(1)

Comparing coefficients of $\mathbf{b}$ :

1 - \lambda = \frac{5}{11}\mu.\quad(2)

From (2):

\mu = \frac{11}{5}(1 - \lambda).

Substituting into (1):

\frac{3}{5}\lambda = 1 - \frac{11}{5}(1 - \lambda)

= - \frac{6}{5} + \frac{11}{5}\lambda.

Solving:

\frac{3}{5}\lambda - \frac{11}{5}\lambda = - \frac{6}{5}

- \frac{8}{5}\lambda = - \frac{6}{5}

\lambda = \frac{3}{4}.

Then:

\mu = \frac{11}{5}\left( 1 - \frac{3}{4} \right) = \frac{11}{20}.

Substituting $\lambda = \frac{3}{4}$ into the equation of line $BC$ :

\overrightarrow{OE} = \frac{3}{5}\left( \frac{3}{4} \right)\mathbf{a} + \left( 1 - \frac{3}{4} \right)\mathbf{b}

= \frac{9}{20}\mathbf{a} + \frac{1}{4}\mathbf{b}.

To check your answer: substituting

\mu = \frac{11}{20}

into the equation of line

AD

gives

\left( 1 - \frac{11}{20} \right)\mathbf{a} + \frac{5}{11}\left( \frac{11}{20} \right)\mathbf{b} = \frac{9}{20}\mathbf{a} + \frac{1}{4}\mathbf{b}

, the same position vector. ✓

For the ratio $AE:ED$ , let $\mathbf{d} = \overrightarrow{OD} = \frac{5}{11}\mathbf{b}$ , so that line $AD$ is $\mathbf{r} = \mathbf{a} + \mu\left( \mathbf{d} - \mathbf{a} \right)$ , where $\mu = 0$ gives $A$ and $\mu = 1$ gives $D$ . Since $E$ corresponds to $\mu = \frac{11}{20}$ ,

\overrightarrow{AE} = \frac{11}{20}\left( \mathbf{d} - \mathbf{a} \right),\quad\quad\overrightarrow{ED} = \left( 1 - \frac{11}{20} \right)\left( \mathbf{d} - \mathbf{a} \right) = \frac{9}{20}\left( \mathbf{d} - \mathbf{a} \right),

so $\overrightarrow{AE}$ and $\overrightarrow{ED}$ are parallel and in the ratio of these scalars:

AE:ED = \frac{11}{20}:\frac{9}{20} = 11:9.

$\therefore$ $\overrightarrow{OE} = \frac{9}{20}\mathbf{a} + \frac{1}{4}\mathbf{b}$ and $AE:ED = 11:9$ .

2015 H2 Maths Paper 1, Question 7

Part (i)

Part (ii)

Part (iii)

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