Position Vectors

A position vector locates a point from the origin: $\overrightarrow{OA} = \mathbf{a}$ puts $A$ at the tip of $\mathbf{a}$. From two position vectors comes the rule the whole subtopic stands on:

$\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$, destination minus start

Getting this backwards ($\mathbf{a} - \mathbf{b}$) flips every subsequent sign; say “destination minus start” as you write it, every time.

The derived facts

• Midpoint of $AB$: position vector $\frac{1}{2}(\mathbf{a} + \mathbf{b})$
• $P$ divides $AB$ in ratio $\lambda:\mu$ (from $A$): $\overrightarrow{OP} = \mathbf{a} + \dfrac{\lambda}{\lambda+\mu}(\mathbf{b} - \mathbf{a}) = \dfrac{\mu\mathbf{a} + \lambda\mathbf{b}}{\lambda + \mu}$

$P$ divides $AB$ in ratio $2:1$, $\mathbf{a} = \mathbf{i} + \mathbf{j}$, $\mathbf{b} = 4\mathbf{i} + 7\mathbf{j}$: $\overrightarrow{OP} = \dfrac{1\cdot\mathbf{a} + 2\cdot\mathbf{b}}{3} = \dfrac{9\mathbf{i} + 15\mathbf{j}}{3} =$ $3\mathbf{i} + 5\mathbf{j}$

Build the ratio point from the definition (start + fraction of the journey) rather than the memorised formula, the construction line is method, and it can’t be mis-recalled.

Collinearity: scalar multiples plus a shared point

“Show $P$, $Q$, $R$ are collinear”: compute $\overrightarrow{PQ}$ and $\overrightarrow{QR}$; show $\overrightarrow{PQ} = k\cdot\overrightarrow{QR}$ (state $k$); note they share $Q$; conclude in words. ”$\overrightarrow{PQ}$ is parallel to $\overrightarrow{QR}$ and both pass through $Q$, so $P$, $Q$, $R$ are collinear.” The arithmetic earns half; the stated conclusion earns the rest. The ratio $PQ:QR = k:1$ often follows as a free part (ii).

Equating coefficients: the heavy machinery

The hardest standard question expresses one point two ways and equates:

$\overrightarrow{OX}$ lies on both $OC$ (so $\overrightarrow{OX} = \lambda\cdot\overrightarrow{OC}$) and on $AB$ (so $\overrightarrow{OX} = \mathbf{a} + \mu(\mathbf{b} - \mathbf{a})$). With $\overrightarrow{OC}$, $\mathbf{a}$, $\mathbf{b}$ all in terms of non-parallel $\mathbf{a}$ and $\mathbf{b}$: Equate coefficients of $\mathbf{a}$ and of $\mathbf{b}$ separately, valid precisely because $\mathbf{a}$ and $\mathbf{b}$ are not parallel, giving two equations in $\lambda$, $\mu$. Solve, substitute back.

Write the justification once (“equating coefficients, since $\mathbf{a}$ and $\mathbf{b}$ are non-parallel”), it’s a stated-reason mark, and it’s the line that separates method from luck.

Common mistakes

• $\overrightarrow{AB} = \mathbf{a} - \mathbf{b}$ (backwards)
• Ratio fractions inverted ($2:1$ giving $\frac{1}{3}$ of the journey instead of $\frac{2}{3}$)
• Collinearity shown numerically but never concluded in words
• Coefficients equated without the non-parallel justification
• Position vectors and displacement vectors blurred mid-solution

Full topic context: Vectors notes.