The fundamental identity: $\vec{AB}=\vec{OB}-\vec{OA}$. Think of it as "go from $A$ to the origin, then origin to $B$": $\vec{AB}=\vec{AO}+\vec{OB}=-\vec{OA}+\vec{OB}$. Treat $\mathbf{a}$ and $\mathbf{b}$ exactly like $x$ and $y$ in ordinary algebra — collect like terms the same way: $(3\mathbf{a}+5\mathbf{b})-(2\mathbf{a}-\mathbf{b})=\mathbf{a}+6\mathbf{b}$.

Two vectors are parallel exactly when one is a scalar multiple of the other: $\vec{AB}=k\cdot\vec{CD}$. Find both vector expressions, then factorise to reveal the common bracket.

Worked example: $\vec{AE}=4\mathbf{a}+2\mathbf{b}$, $\vec{BC}=6\mathbf{a}+3\mathbf{b}=1.5(4\mathbf{a}+2\mathbf{b})=1.5\vec{AE}$.

Collinearity needs two conditions, both stated in your conclusion:

1. Two segments between the points are parallel (one is a scalar multiple of the other).
2. They share a common point.

Required concluding statement: "Since $\vec{AC}$ is a scalar multiple of $\vec{AB}$, the vectors are parallel. As they also share the common point $A$, the points $A$, $B$, $C$ are collinear."

Parallelogram ($ABCD$): show one pair of opposite sides is equal and parallel — i.e. $\vec{AB}=\vec{DC}$ exactly (not just a scalar multiple). Equality of the vectors proves both same length and same direction simultaneously.

Trapezium ($OABC$): show one pair of opposite sides is parallel but not equal, i.e. $\vec{AB}=k\cdot\vec{OC}$ where $k\ne1$ — the scalar multiple proves parallel, and $k\ne1$ proves they're different lengths.

If $X$ lies on $AB$ with $AX:XB=m:n$, then $\vec{AX}=\frac{m}{m+n}\vec{AB}$. A midpoint is just the special case $m:n=1:1$, so $\vec{AM}=\frac12\vec{AB}$.

Worked example: $OABC$ is a trapezium with $\vec{OA}=2\mathbf{a}$, $\vec{OC}=\mathbf{c}$, $\vec{AB}=3\mathbf{c}$. $P$ lies on $AC$ with $AP:PC=3:1$. Prove $O,P,B$ are collinear.

Find $\vec{OB}=\vec{OA}+\vec{AB}=2\mathbf{a}+3\mathbf{c}$.

Find $\vec{AC}=\vec{OC}-\vec{OA}=\mathbf{c}-2\mathbf{a}$, so $\vec{AP}=\frac34\vec{AC}=\frac34\mathbf{c}-\frac32\mathbf{a}$.

Find $\vec{OP}=\vec{OA}+\vec{AP} = 2\mathbf{a}+\frac34\mathbf{c}-\frac32\mathbf{a} = \frac12\mathbf{a}+\frac34\mathbf{c}$.

Factorise: $\vec{OP}=\frac14(2\mathbf{a}+3\mathbf{c})=\frac14\vec{OB}$.

Conclusion: Since $\vec{OP}$ is a scalar multiple of $\vec{OB}$, they're parallel; as they share the common point $O$, $O,P,B$ are collinear.

The hardest Grade 9 vector problems: a point $X$ lies at the intersection of two lines. Express $\vec{OX}$ two different ways (one along each line, each using its own unknown scalar), then equate the coefficients of $\mathbf{a}$ and $\mathbf{b}$ separately — since $\mathbf{a}$ and $\mathbf{b}$ aren't parallel, this is the only way two expressions for the same vector can be equal.

Worked example: $\vec{OA}=\mathbf{a}$, $\vec{OB}=\mathbf{b}$. $M$ is the midpoint of $OB$. $P$ on $AB$ with $AP:PB=3:1$. Lines $AM$ and $OP$ meet at $X$. Find $OX:XP$.

Let $\vec{OX}=k\vec{OP}$ and $\vec{AX}=m\vec{AM}$.

$\vec{OP}=\vec{OA}+\frac34\vec{AB}=\mathbf{a}+\frac34(\mathbf{b}-\mathbf{a})=\frac14\mathbf{a}+\frac34\mathbf{b}$, so $\vec{OX}=\frac{k}{4}\mathbf{a}+\frac{3k}{4}\mathbf{b}$.

$\vec{AM}=\frac12\mathbf{b}-\mathbf{a}$, so $\vec{OX}=\vec{OA}+m\vec{AM}=(1-m)\mathbf{a}+\frac{m}{2}\mathbf{b}$.

Equate coefficients: $\frac{k}{4}=1-m$ and $\frac{3k}{4}=\frac{m}{2}$. From the second, $m=\frac{3k}{2}$; substituting into the first: $\frac{k}{4}=1-\frac{3k}{2} \implies k=4-6k \implies 7k=4 \implies k=\frac47$.

Since $\vec{OX}=k\vec{OP}$, the ratio $OX:XP = k:(1-k) = \frac47:\frac37 = \boxed{4:3}$.

Prove that $\vec{AB}=9\mathbf{a}+6\mathbf{b}$ and $\vec{CD}=12\mathbf{a}+8\mathbf{b}$ are parallel.

Factorise each: $\vec{AB}=3(3\mathbf{a}+2\mathbf{b})$, $\vec{CD}=4(3\mathbf{a}+2\mathbf{b})$.

So $\vec{CD}=\frac43\vec{AB}$.

Since $\vec{CD}$ is a scalar multiple of $\vec{AB}$, the two vectors are parallel.