In $\triangle OAB$, P and Q are the midpoints of sides $OA$ and $OB$ respectively. $AQ$ intersects $BP$ at $M$

Key takeaway: This question uses midpoint vectors and line intersections in a triangle. The key idea is to write the position vector of $M$ in two different ways and then compare them.

(a) Sketch the diagram

Draw triangle $OAB$ with:

• $P$ as the midpoint of $OA$
• $Q$ as the midpoint of $OB$
• line $AQ$ and line $BP$ intersecting at $M$

(b) Express $\overrightarrow{OM}$ in terms of $a$, $b$ and $\lambda$

Given

• $\overrightarrow{OA}=a$
• $\overrightarrow{OB}=b$
• $P$ is the midpoint of $OA$, so $\overrightarrow{OP}=\tfrac12 a$
• $Q$ is the midpoint of $OB$, so $\overrightarrow{OQ}=\tfrac12 b$

Since $\overrightarrow{AM}=\lambda\,\overrightarrow{AQ}$,

$$\overrightarrow{AQ}=\overrightarrow{OQ}-\overrightarrow{OA}=\tfrac12 b-a$$

So

$$\overrightarrow{OM}=\overrightarrow{OA}+\overrightarrow{AM} =a+\lambda\left(\tfrac12 b-a\right)$$

Hence

$$\overrightarrow{OM}=(1-\lambda)a+\frac{\lambda}{2}b$$

(c) Express $\overrightarrow{OM}$ in terms of $a$, $b$ and $\mu$

Similarly,

$$\overrightarrow{BP}=\overrightarrow{OP}-\overrightarrow{OB}=\tfrac12 a-b$$

Since $\overrightarrow{BM}=\mu\,\overrightarrow{BP}$,

$$\overrightarrow{OM}=\overrightarrow{OB}+\overrightarrow{BM} =b+\mu\left(\tfrac12 a-b\right)$$

So

$$\overrightarrow{OM}=\frac{\mu}{2}a+(1-\mu)b$$

(d) Prove that $\overrightarrow{OA}+\overrightarrow{OB}=3\overrightarrow{OM}$

Because $M$ lies on both $AQ$ and $BP$, the two expressions for $\overrightarrow{OM}$ are equal:

$$(1-\lambda)a+\frac{\lambda}{2}b=\frac{\mu}{2}a+(1-\mu)b$$

Matching coefficients of $a$ and $b$ gives

$$1-\lambda=\frac{\mu}{2},\qquad \frac{\lambda}{2}=1-\mu$$

Solving these simultaneously gives

$$\lambda=\mu=\frac{2}{3}$$

Substitute into either expression for $\overrightarrow{OM}$:

$$\overrightarrow{OM}=\left(1-\frac23\right)a+\frac{1}{3}b=\frac13 a+\frac13 b$$

Therefore,

$$3\overrightarrow{OM}=a+b$$

So

$$\overrightarrow{OA}+\overrightarrow{OB}=3\overrightarrow{OM}$$

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Pitfalls the pros know 👇 A common mistake is to write $\overrightarrow{AQ}=\tfrac12\,\overrightarrow{OB}$ directly. That is not correct because $AQ$ is not the same direction as $OB$. You must first express $\overrightarrow{AQ}$ as $\overrightarrow{OQ}-\overrightarrow{OA}$.

What if the problem changes? If $P$ and $Q$ were not midpoints, the same method still works, but the coefficients change. You would first write $\overrightarrow{OP}=t\,a$ and $\overrightarrow{OQ}=s\,b$ for the given division ratios, then express $\overrightarrow{OM}$ from both lines and equate them.

Tags: position vector, midpoint theorem, line intersection