In the diagram, $\overrightarrow{OA} = a$ and $\overrightarrow{OB} = b$ . $2\overrightarrow{OB} = 3\overrightarrow{BC},\; \overrightarrow{OD} = 3\overrightarrow{OA},\; \text{and } 2\overrightarrow{AX} = \overrightarrow{XB}.$

(i)
Express, as simply as possible, in terms of a and b,
1. ((a))
  $\overrightarrow{BA}$
2. ((b))
  $\overrightarrow{OX}$
3. ((c))
  $\overrightarrow{CD}$
(ii)
It is given that CM : MD is 10 : 9.
1. ((a))
  Express $\overrightarrow{OM}$ , as simply as possible, in terms of a and b.
2. ((b))
  Show that O, X and M are collinear.
3. ((c))
  Given that MY is parallel to OD, express $\overrightarrow{MY}$ in terms of a and b.
(iii)
Find the ratio of
1. ((a))
  area of $\Delta XMY$ / area of $\Delta XOA'$
2. ((b))
  area of $\Delta OBX$ / area of $\Delta OYA'$

Sum difference Scalar multiplication Position vectors Simplification linear 中文

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Worked solution

3(i)((a))

1. Express $\overrightarrow{BA}$ as the difference of the position vectors of $A$ and $B$.

$\overrightarrow{BA}=\overrightarrow{OA}-\overrightarrow{OB}$
Given $\overrightarrow{OA}=a$ and $\overrightarrow{OB}=b$ ,
$\overrightarrow{BA}=a-b.$

Answer

$\overrightarrow{BA}=a-b$

3(i)((b))

1. Use $2\overrightarrow{AX}=\overrightarrow{XB}$ to determine the fraction of $\overrightarrow{AB}$ represented by $\overrightarrow{AX}$.

Since $2\overrightarrow{AX}=\overrightarrow{XB}$ ,
$\overrightarrow{AB}=\overrightarrow{AX}+\overrightarrow{XB}=\overrightarrow{AX}+2\overrightarrow{AX}=3\overrightarrow{AX}.$
Hence
$\overrightarrow{AX}=\frac{1}{3}\overrightarrow{AB}.$

2. Express $\overrightarrow{OX}$ using position vectors of $O$, $A$ and $X$.

From earlier, $\overrightarrow{BA}=a-b$ , so
$\overrightarrow{AB}=b-a.$
Therefore
$\overrightarrow{OX}=\overrightarrow{OA}+\overrightarrow{AX}$
$=a+\frac{1}{3}(b-a)$
$=\frac{2a+b}{3}.$

Answer

$\overrightarrow{OX}=\dfrac{2a+b}{3}$

3(i)((c))

1. Express $\overrightarrow{OC}$ using the given relation between $\overrightarrow{OB}$ and $\overrightarrow{BC}$.

Given $2\overrightarrow{OB}=3\overrightarrow{BC}$ ,

$\overrightarrow{BC}=\dfrac{2}{3}\overrightarrow{OB}=\dfrac{2}{3}b$ .

Hence

$\overrightarrow{OC}=\overrightarrow{OB}+\overrightarrow{BC}=b+\dfrac{2}{3}b=\dfrac{5}{3}b$ .

2. Find $\overrightarrow{CD}$ as the difference between the position vectors of $D$ and $C$.

Since $\overrightarrow{OD}=3\overrightarrow{OA}=3a$ ,

$\overrightarrow{CD}=\overrightarrow{OD}-\overrightarrow{OC}$

$=3a-\dfrac{5}{3}b$ .

Answer

$\overrightarrow{CD}=3a-\dfrac{5}{3}b$

3(ii)((a))

1. Use the ratio $CM:MD=10:9$ to find $\overrightarrow{CM}$ in terms of $\overrightarrow{CD}$, then form $\overrightarrow{OM}$.

Since $CM:MD=10:9$ ,

$\overrightarrow{CM}=\dfrac{10}{19}\overrightarrow{CD}$ .

From earlier parts,

$\overrightarrow{OC}=\dfrac{5}{3}b \quad \text{and} \quad \overrightarrow{CD}=3a-\dfrac{5}{3}b$ .

Hence

\begin{aligned} \overrightarrow{OM} &=\overrightarrow{OC}+\overrightarrow{CM} \\ &=\dfrac{5}{3}b+\dfrac{10}{19}\left(3a-\dfrac{5}{3}b\right) \\ &=\dfrac{5}{3}b+\dfrac{30}{19}a-\dfrac{50}{57}b \\ &=\dfrac{30}{19}a+\dfrac{15}{19}b \\ &=\dfrac{30a+15b}{19}. \end{aligned}

Answer

$\overrightarrow{OM}=\dfrac{30a+15b}{19}$

3(ii)((b))

1. Write down the position vectors of $X$ and $M$ from earlier parts.

From earlier parts,
$\overrightarrow{OX}=\frac{2a+b}{3}$
and
$\overrightarrow{OM}=\frac{30a+15b}{19}.$

2. Show that $\overrightarrow{OM}$ is a scalar multiple of $\overrightarrow{OX}$.

Factorise $\overrightarrow{OM}$ :
$\overrightarrow{OM}=\frac{15(2a+b)}{19}.$
Since
$\overrightarrow{OX}=\frac{2a+b}{3},$
we have
$\overrightarrow{OM}=\frac{15}{19}(2a+b)=\frac{15}{19}\times 3\overrightarrow{OX}=\frac{45}{19}\overrightarrow{OX}.$

3. Conclude collinearity using the scalar multiple relationship from the common point $O$.

Since $\overrightarrow{OM}$ is a scalar multiple of $\overrightarrow{OX}$ , the vectors have the same direction and both start from $O$ .

Therefore, $O$ , $X$ and $M$ are collinear.

Answer

Since $\overrightarrow{OM}=\dfrac{45}{19}\overrightarrow{OX}$ , $O$ , $X$ and $M$ are collinear.

3(ii)((c))

1. Use the condition $MY \parallel OD$ to eliminate the $b$ component of $\overrightarrow{MY}$.

Since $Y$ lies on $OC$ , let $\overrightarrow{OY}=kb$ .

From part (ii)(a),
$\overrightarrow{OM}=\frac{30a+15b}{19}.$

Hence
$\overrightarrow{MY}=\overrightarrow{OY}-\overrightarrow{OM}$
$=kb-\frac{30a+15b}{19}$
$=-\frac{30}{19}a+\left(k-\frac{15}{19}\right)b.$

Since $MY \parallel OD$ and $\overrightarrow{OD}=3a$ , $\overrightarrow{MY}$ must have no $b$ component.

So
$k-\frac{15}{19}=0,$
$k=\frac{15}{19}.$

Therefore
$\overrightarrow{MY}=-\frac{30}{19}a.$

Answer

$\overrightarrow{MY}=-\dfrac{30}{19}a$

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Worked solution

3(i)((a))

3(i)((b))

3(i)((c))

3(ii)((a))

3(ii)((b))

3(ii)((c))

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