| Exam Board | OCR |
|---|---|
| Module | H240/02 (Pure Mathematics and Statistics) |
| Year | 2018 |
| Session | December |
| Marks | 8 |
| Topic | Vectors Introduction & 2D |
| Type | Ratio division of line segment |
| Difficulty | Moderate -0.3 Part (a) is a standard proof of the section formula that appears in most textbooks. Parts (b) and (c) require setting up equations using the condition that C lies on the y-axis (x and z components equal zero), then solving for p—straightforward application of the formula with minimal problem-solving. The question is slightly easier than average due to its routine nature and clear structure, though it does require multiple steps and careful algebraic manipulation. |
| Spec | 1.10e Position vectors: and displacement1.10g Problem solving with vectors: in geometry |
Points $A$ and $B$ have position vectors $\mathbf{a}$ and $\mathbf{b}$. Point $C$ lies on $AB$ such that $AC : CB = p : 1$.
\begin{enumerate}[label=(\alph*)]
\item Show that the position vector of $C$ is $\frac{1}{p+1}(\mathbf{a} + p\mathbf{b})$. [3]
\end{enumerate}
It is now given that $\mathbf{a} = 2\mathbf{i} + 3\mathbf{j} - 4\mathbf{k}$ and $\mathbf{b} = -6\mathbf{i} + 4\mathbf{j} + 12\mathbf{k}$, and that $C$ lies on the $y$-axis.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the value of $p$. [4]
\item Write down the position vector of $C$. [1]
\end{enumerate}
\hfill \mbox{\textit{OCR H240/02 2018 Q5 [8]}}