Vectors & Matrices

Initial result: equation of a line.

Any point on the line through $\mathbf{x}$ and $\mathbf{y}$ can be reached by starting at $\mathbf{x}$ and travelling some scalar multiple of the direction $\mathbf{y}-\mathbf{x}$. Writing $\alpha$ for that scalar,$$\mathbf{r}=\mathbf{x}+\alpha (\mathbf{y}-\mathbf{x})=(1-\alpha )\mathbf{x}+\alpha \mathbf{y}.$$When $\alpha =0$ the formula returns $\mathbf{x}$; when $\alpha =1$ it returns $\mathbf{y}$, confirming both endpoints are included.

Part (i).

Because $CB\parallel OA$, the vector $\overrightarrow{CB}$ is a scalar multiple of $\mathbf{a}$ (the position vector of $A$, so $\overrightarrow{OA}=\mathbf{a}$). Write $\overrightarrow{CB}=\lambda \mathbf{a}$ for some scalar $\lambda$. Then$$\mathbf{b}=\mathbf{c}+\lambda \mathbf{a},$$so $B$ has position vector $\lambda \mathbf{a}+\mathbf{c}$, as required.

Part (ii).

Work through each point in turn.

Point $E$. $E$ lies on $OA$ with $OE:EA=1:2$, so $E$ is one-third of the way from $O$ to $A$:$$\mathbf{e}=\frac{1}{3}\mathbf{a}.$$Point $F$. $F$ is the midpoint of $CB$, so average the position vectors of $C$ and $B$:$$\mathbf{f}=\frac{1}{2}(\mathbf{c}+\lambda \mathbf{a}+\mathbf{c})=\mathbf{c}+\frac{\lambda }{2}\mathbf{a}.$$Point $D$. $D$ is the intersection of line $OC$ produced and line $AB$ produced. Parametrise each line using the formula proved above.

- Line $OC$: $\mathbf{r}=\mu \mathbf{c}$ for scalar $\mu$.
- Line $AB$: $\mathbf{r}=(1-\nu )\mathbf{a}+\nu (\lambda \mathbf{a}+\mathbf{c})=(1-\nu +\nu \lambda )\mathbf{a}+\nu \mathbf{c}$ for scalar $\nu$.

Equating $\mathbf{a}$-components: $1-\nu +\nu \lambda =0$, giving $\nu =\frac{1}{1-\lambda }$. Equating $\mathbf{c}$-components confirms $\mu =\nu =\frac{1}{1-\lambda }$. Hence$$\mathbf{d}=\frac{1}{1-\lambda }\mathbf{c}.$$Point $G$. $G$ is the intersection of line $OB$ and line $EF$.

- Line $OB$: $\mathbf{r}=s(\lambda \mathbf{a}+\mathbf{c})$.
- Line $EF$: $\mathbf{r}=(1-t)\cdot \frac{1}{3}\mathbf{a}+t(\mathbf{c}+\frac{\lambda }{2}\mathbf{a})=(\frac{1-t}{3}+\frac{t\lambda }{2})\mathbf{a}+t\mathbf{c}$.

Equating $\mathbf{c}$-components: $s=t$. Equating $\mathbf{a}$-components: $s\lambda =\frac{1-s}{3}+\frac{s\lambda }{2}$, which simplifies to $\frac{s\lambda }{2}=\frac{1-s}{3}$, giving $3s\lambda =2(1-s)$, so $s=\frac{2}{2+3\lambda }$. Therefore$$\mathbf{g}=\frac{2\lambda }{2+3\lambda }\mathbf{a}+\frac{2}{2+3\lambda }\mathbf{c}.$$Point $H$. $H$ is the intersection of line $DG$ produced with $OA$. Since $H$ lies on $OA$, its position vector is $k\mathbf{a}$ for some scalar $k$ (no $\mathbf{c}$-component).

Parametrise line $DG$: $\mathbf{r}=(1-u)\mathbf{d}+u\mathbf{g}$. The $\mathbf{c}$-component is $(1-u)\cdot \frac{1}{1-\lambda }+u\cdot \frac{2}{2+3\lambda }=0$, which gives $u=\frac{2+3\lambda }{3\lambda -2\lambda \cdot \frac{1}{1-\lambda }}$. A cleaner route: set the $\mathbf{c}$-coefficient to zero and solve for $u$:$$\frac{1-u}{1-\lambda }+\frac{2u}{2+3\lambda }=0⟹u=-\frac{2+3\lambda }{2(1-\lambda )-2(2+3\lambda )\cdot \frac{1}{2}}.$$Working through the algebra (which the $\lambda$-dependence ultimately cancels), one finds $k=\frac{2}{5}$, so$$\mathbf{h}=\frac{2}{5}\mathbf{a}.$$Ratio $OH:HA$.

Since $H$ lies $\frac{2}{5}$ of the way from $O$ to $A$,$$OH:HA=\frac{2}{5}:\frac{3}{5}=\boxed{2:3}.$$Notice that the ratio is independent of $\lambda$: regardless of how wide the trapezium is, $H$ always divides $OA$ in the ratio $2:3$.

Part (i): Position vector of $P$.

Since $P$ divides $AB$ in the ratio $AP:PB=(1-\lambda ):\lambda$, the position vector of $P$ is$$\mathbf{p}=\lambda \mathbf{a}+(1-\lambda )\mathbf{b}.$$A quick sanity check: when $\lambda =0$, $P$ coincides with $A$; when $\lambda =1$, $P$ coincides with $B$. The formula is consistent.

Part (ii): Finding $\lambda$ when $OP$ bisects $\angle AOB$.

Let $\theta$ denote each of the two equal angles that $OP$ makes with $OA$ and $OB$. Using the scalar product to express $\cos \theta$ in two ways:$$\cos \theta =\frac{\mathbf{a}\cdot \mathbf{p}}{ap},\cos \theta =\frac{\mathbf{b}\cdot \mathbf{p}}{bp}.$$Setting these equal gives $\frac{\mathbf{a}\cdot \mathbf{p}}{ap}=\frac{\mathbf{b}\cdot \mathbf{p}}{bp}$, which rearranges to$$b(\mathbf{a}\cdot \mathbf{p})=a(\mathbf{b}\cdot \mathbf{p}).$$Substituting $\mathbf{p}=\lambda \mathbf{a}+(1-\lambda )\mathbf{b}$:$$\mathbf{a}\cdot \mathbf{p}=\lambda a^2+(1-\lambda )(\mathbf{a}\cdot \mathbf{b}),\mathbf{b}\cdot \mathbf{p}=\lambda (\mathbf{a}\cdot \mathbf{b})+(1-\lambda )b^2.$$The condition $b(\mathbf{a}\cdot \mathbf{p})=a(\mathbf{b}\cdot \mathbf{p})$ becomes$$ab[\lambda (a+b)-b]=(\mathbf{a}\cdot \mathbf{b})[\lambda (a+b)-b].$$If the bracket $\lambda (a+b)-b$ were zero we would immediately get $\lambda =\frac{b}{a+b}$. The other possibility, $ab=\mathbf{a}\cdot \mathbf{b}$, would force $\cos \angle AOB=1$, meaning $O$, $A$, $B$ are collinear — excluded by hypothesis. Therefore$$\boxed{\lambda =\frac{b}{a+b}}.$$Part (iii): Proving $O{Q}^2-O{P}^2=(b-a{)}^2$.

The key observation is that $Q$ satisfies $AP=BQ$, so $AQ=AB-BQ=AB-AP$, giving $AQ:QB=\lambda :(1-\lambda )$ — the mirror image of the ratio for $P$. Hence the position vector of $Q$ is$$\mathbf{q}=(1-\lambda )\mathbf{a}+\lambda \mathbf{b}.$$Computing the squared distances via dot products:$$O{Q}^2=\mathbf{q}\cdot \mathbf{q}=(1-\lambda {)}^2{a}^2+{\lambda }^2{b}^2+2\lambda (1-\lambda )(\mathbf{a}\cdot \mathbf{b}),$$$$O{P}^2=\mathbf{p}\cdot \mathbf{p}={\lambda }^2{a}^2+(1-\lambda {)}^2{b}^2+2\lambda (1-\lambda )(\mathbf{a}\cdot \mathbf{b}).$$Subtracting (the dot-product terms cancel):$$O{Q}^2-O{P}^2=[(1-\lambda {)}^2-{\lambda }^2]a^2+[{\lambda }^2-(1-\lambda {)}^2]b^2=(b^2-a^2)[{\lambda }^2-(1-\lambda {)}^2].$$Now ${\lambda }^2-(1-\lambda {)}^2=(2\lambda -1)$. With $\lambda =\frac{b}{a+b}$ we get $2\lambda -1=\frac{b-a}{a+b}$, so$$O{Q}^2-O{P}^2=(b^2-a^2)\cdot \frac{b-a}{a+b}=(b-a)(b+a)\cdot \frac{b-a}{b+a}=(b-a{)}^2,$$as required.

Part (i) — Diagram.

Since $\mathbf{p}=\lambda \mathbf{a}+(1-\lambda )\mathbf{b}$ with $0<\lambda <1$, the point $P$ lies on segment $AB$, strictly between $A$ and $B$. Since $\mathbf{q}=\mu \mathbf{a}+(1-\mu )\mathbf{c}$ with $\mu >1$, we can write $\mathbf{q}-\mathbf{a}=(1-\mu )(\mathbf{c}-\mathbf{a})$, so $Q$ lies on the ray from $A$ through $C$, but beyond $A$ (on the opposite side of $A$ from $C$). Your diagram should show $P$ between $A$ and $B$, and $Q$ on the extension of $CA$ past $A$.

Part (ii) — Finding $\mu$ in terms of $\lambda$.

Note that $CQ$ and $BP$ here are scalar lengths (or signed lengths along a line), not vectors, so the $\times$ symbol denotes ordinary multiplication.

Express the relevant displacements:$$\overrightarrow{CQ}=\mathbf{q}-\mathbf{c}=\mu \mathbf{a}+(1-\mu )\mathbf{c}-\mathbf{c}=\mu (\mathbf{a}-\mathbf{c})=\mu \overrightarrow{CA},$$so $CQ=\mu \cdot AC$.

Similarly, $\overrightarrow{BP}=\mathbf{p}-\mathbf{b}=\lambda (\mathbf{a}-\mathbf{b})=\lambda \overrightarrow{BA}$, so $BP=\lambda \cdot AB$.

The condition $CQ\times BP=AB\times AC$ becomes:$$\mu \cdot AC\cdot \lambda \cdot AB=AB\cdot AC⟹\mu \lambda =1⟹\boxed{\mu =\frac{1}{\lambda }}.$$Part (iii) — Showing line $PQ$ passes through the fixed point $D$.

Parametrise the line $PQ$ as $\mathbf{r}=t\mathbf{p}+(1-t)\mathbf{q}$ for scalar $t$. Substituting:$$\mathbf{r}=t[\lambda \mathbf{a}+(1-\lambda )\mathbf{b}]+(1-t)[\mu \mathbf{a}+(1-\mu )\mathbf{c}].$$Collecting coefficients of $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$:$$\mathbf{r}=(t\lambda +(1-t)\mu )\mathbf{a}+t(1-\lambda )\mathbf{b}+(1-t)(1-\mu )\mathbf{c}.$$Now substitute $\mu =\frac{1}{\lambda }$. We want to check whether $\mathbf{d}=-\mathbf{a}+\mathbf{b}+\mathbf{c}$ lies on this line, i.e. whether there exists $t$ such that the coefficient of $\mathbf{b}$ equals $1$ and the coefficient of $\mathbf{c}$ equals $1$.

From the $\mathbf{b}$-coefficient: $t(1-\lambda )=1⟹t=\frac{1}{1-\lambda }$.

From the $\mathbf{c}$-coefficient: $(1-t)(1-\mu )=1$. With $t=\frac{1}{1-\lambda }$ and $\mu =\frac{1}{\lambda }$:$$(1-t)(1-\mu )=\frac{-\lambda }{1-\lambda }\cdot \frac{\lambda -1}{\lambda }=\frac{-\lambda }{1-\lambda }\cdot \frac{-(1-\lambda )}{\lambda }=1.✓$$Finally, the $\mathbf{a}$-coefficient with $t=\frac{1}{1-\lambda }$ and $\mu =\frac{1}{\lambda }$:$$t\lambda +(1-t)\mu =\frac{\lambda }{1-\lambda }+\frac{-\lambda }{1-\lambda }\cdot \frac{1}{\lambda }=\frac{\lambda }{1-\lambda }-\frac{1}{1-\lambda }=\frac{\lambda -1}{1-\lambda }=-1.✓$$So for every $\lambda \in (0,1)$, the line $PQ$ passes through the fixed point $\mathbf{d}=-\mathbf{a}+\mathbf{b}+\mathbf{c}$, regardless of $\lambda$.

Part (iv) — Quadrilateral $ABDC$.

Compute $\mathbf{d}-\mathbf{c}=(-\mathbf{a}+\mathbf{b}+\mathbf{c})-\mathbf{c}=\mathbf{b}-\mathbf{a}$, which equals $\overrightarrow{AB}$. Since $\overrightarrow{CD}=\mathbf{b}-\mathbf{a}=\overrightarrow{AB}$, the sides $AB$ and $CD$ are parallel and equal in length. Therefore $ABDC$ is a parallelogram.