Triangle law with position vectors (4 marks)

$O$ is the origin. $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. $M$ is the midpoint of $AB$. Express $\overrightarrow{OM}$ in terms of $\mathbf{a}$ and $\mathbf{b}$, fully simplified.

Mark scheme (4 marks):

• M1: $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ (finish minus start)
• M1: $\overrightarrow{AM} = \tfrac{1}{2}(\mathbf{b} - \mathbf{a})$
• M1: $\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM} = \mathbf{a} + \tfrac{1}{2}(\mathbf{b} - \mathbf{a})$
• A1: $\overrightarrow{OM} = \tfrac{1}{2}(\mathbf{a} + \mathbf{b})$ or $\tfrac{1}{2}\mathbf{a} + \tfrac{1}{2}\mathbf{b}$

Alternative route via $\overrightarrow{OM} = \overrightarrow{OB} + \tfrac{1}{2}\overrightarrow{BA}$ also full marks.