Position Vector of a Point Dividing a Line

To find a point $P$ dividing $AB$ in a given ratio, travel the right fraction of the way from $A$ to $B$: $\vec{OP} = \vec{OA} + \left(\tfrac{\text{first part}}{\text{total}}\right)\vec{AB}, \qquad \vec{AB} = \vec{OB} - \vec{OA}.$

Points $A$ and $B$ have position vectors $\vec{OA} = \mathbf{i} + 2\mathbf{j}$ and $\vec{OB} = 7\mathbf{i} + 11\mathbf{j}$. The point $P$ lies on $AB$ with $AP:PB = 2:1$. Find the position vector of $P$. [4]

The working

Step 1, find the displacement $\vec{AB}$: $\vec{AB} = \vec{OB} - \vec{OA} = (7\mathbf{i} + 11\mathbf{j}) - (\mathbf{i} + 2\mathbf{j}) = 6\mathbf{i} + 9\mathbf{j} \quad \text{(M1)}$

Step 2, find the fraction. $AP:PB = 2:1$ means $P$ is $\frac{2}{2+1} = \frac{2}{3}$ of the way from $A$ to $B$: $\vec{AP} = \tfrac{2}{3}\vec{AB} = \tfrac{2}{3}(6\mathbf{i} + 9\mathbf{j}) = 4\mathbf{i} + 6\mathbf{j} \quad \text{(M1, A1)}$

Step 3, add to $\vec{OA}$: $\vec{OP} = \vec{OA} + \vec{AP} = (\mathbf{i} + 2\mathbf{j}) + (4\mathbf{i} + 6\mathbf{j}) = 5\mathbf{i} + 8\mathbf{j} \quad \text{(A1)}$

Where the marks are won and lost

• The fraction is $\frac{2}{3}$, not $2$: the "$2$" in $2:1$ is over the total $3$ parts.
• Compute $\vec{AB} = \vec{OB} - \vec{OA}$ (end minus start), then scale it. Scaling $\vec{OB}$ by mistake gives the wrong point.
• Finish by adding $\vec{OA}$; $\vec{AP}$ alone is a displacement, not a position vector.

Common mistakes

• Using $\frac{2}{1}$ or $2$ as the fraction of $\vec{AB}$.
• Computing $\vec{OA} - \vec{OB}$ (wrong direction).
• Giving $\vec{AP} = 4\mathbf{i}+6\mathbf{j}$ as the final answer without adding $\vec{OA}$.

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