An extended segment and coefficient comparison · source reference

Question

Relative to origin O, P and Q have position vectors $\mathbf p,\mathbf q$, nonparallel and nonzero. R lies on PQ produced with $PQ:QR=1:\lambda$. M is the midpoint of OR.

(i) Find the position vector of R in terms of $\lambda,\mathbf p,\mathbf q$. [1]

F lies on OQ and F, P, M are collinear.

(ii) Find $OF:FQ$ in terms of $\lambda$. [4]

[2018/EJC/Prelim/I/5]

Source working

(i) Ratio theorem:

$$\overrightarrow{OQ}=\frac{\overrightarrow{OR}+\lambda\overrightarrow{OP}}{1+\lambda}\Rightarrow\overrightarrow{OR}=(1+\lambda)\mathbf q-\lambda\mathbf p.$$

(ii) F on OQ gives $\overrightarrow{OF}=t\mathbf q$ for some $t\in\mathbb R$.

F on PM: let $PF:FM=s:1-s$, $s\in\mathbb R$. Ratio theorem gives

$$\begin{aligned}\overrightarrow{OF}&=s\overrightarrow{OM}+(1-s)\overrightarrow{OP}\\&=\tfrac s2\overrightarrow{OR}+(1-s)\mathbf p\\&=\tfrac s2((1+\lambda)\mathbf q-\lambda\mathbf p)+(1-s)\mathbf p\\&=(1-s\lambda/2-s)\mathbf p+\tfrac{s(1+\lambda)}2\mathbf q.\end{aligned}$$

Thus $t\mathbf q=(1-s\lambda/2-s)\mathbf p+s(1+\lambda)\mathbf q/2$.

Nonparallel, nonzero p and q imply

$$1-s\lambda/2-s=0\Rightarrow s(\lambda/2+1)=1,$$ $$s=\frac1{\lambda/2+1}=\frac2{\lambda+2}.$$ $$\overrightarrow{OF}=0\mathbf p+\tfrac{s(1+\lambda)}2\mathbf q=\frac2{\lambda+2}\frac{1+\lambda}2\mathbf q=\frac{1+\lambda}{2+\lambda}\mathbf q.$$ $$\overrightarrow{FQ}=\overrightarrow{OQ}-\overrightarrow{OF}=\mathbf q-\frac{1+\lambda}{2+\lambda}\mathbf q=\frac1{2+\lambda}\mathbf q.$$ $$\overrightarrow{OF}=\frac{1+\lambda}{2+\lambda}\mathbf q=(1+\lambda)\frac1{2+\lambda}\mathbf q=(1+\lambda)\overrightarrow{FQ}.$$

Hence $OF:FQ=(1+\lambda):1$.

Teacher notes

R lies outside segment PQ, nearer Q, because it is on PQ produced.

Form two expressions for OF in terms of p and q, using F on OQ and F on PM. Equate these expressions to find s in terms of λ. Alternatively, consider F as the intersection of lines OQ and PM.

Native diagrams

The first source panel shows P–Q–R with ratios $1,\lambda$ and origin O; its ratio-theorem arrows identify the weighted vectors. The second adds midpoint M on OR, F at PM∩OQ, $PF:FM=s:1-s$ and the equal lengths $OM=MR$ marked $1,1$.

Printed answer key

(i) $\overrightarrow{OR}=(1+\lambda)\mathbf q-\lambda\mathbf p$; (ii) $OF:FQ=(1+\lambda):1$.

Provenance

2025 TJC H2 Math (9758) JC2 MSM: VECTORS. Question document pages 1; solution document pages 3. Original page numbers are retained as text metadata.