Vector geometry — position vectors, parallel/collinear, ratio proofs

Writing a vector equal to a plain number, or "dividing" one vector by another, loses the communication marks on a top-grade question. A vector answer must stay a vector: keep $\overrightarrow{AB}$ or $\mathbf{a}$-and-$\mathbf{b}$ form throughout, and never set $\overrightarrow{AB}=5$ or add a bare scalar to a vector.

Dropping a bracket when expanding something like $\tfrac12(\mathbf{b}-\mathbf{a})$ flips a sign and ruins the simplification. Keep every bracket until you have multiplied through: $\tfrac12(\mathbf{b}-\mathbf{a})=\tfrac12\mathbf{b}-\tfrac12\mathbf{a}$, not $\tfrac12\mathbf{b}-\mathbf{a}$. Sign and bracket slips are the most common route errors.

Guessing a relationship such as $\overrightarrow{OQ}=2\,\overrightarrow{OP}$ from the look of the diagram, instead of deriving it. Find two genuinely independent expressions for the same vector — usually one route with a parameter and one geometric constraint — and equate their coefficients of $\mathbf{a}$ and $\mathbf{b}$.

With $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OB}=\mathbf{b}$, the displacement is $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}$ (go $A\to O\to B$, i.e. $-\mathbf{a}+\mathbf{b}$), NOT $\mathbf{a}-\mathbf{b}$. Reversing this is the single most common route error; always start the journey at the FIRST letter and end at the second.

For $M$ on $AB$ with $AM:MB=m:n$, the fraction of the way along is $\dfrac{m}{m+n}$, NOT $\dfrac{m}{n}$. So $AP:PB=2:3$ gives $\dfrac{2}{5}$, not $\dfrac{2}{3}$: the denominator is the SUM of the ratio parts. Then $\overrightarrow{AM}=\dfrac{m}{m+n}\overrightarrow{AB}$.

Showing $\overrightarrow{OQ}=k\,\overrightarrow{OP}$ proves the two vectors are parallel, but that alone does NOT prove $O$, $Q$, $P$ are collinear. You must also state that they share a common point (here $O$). Parallel plus a shared point ⇒ one straight line; parallel with no shared point is just two parallel lines.