| Exam Board | OCR |
|---|---|
| Module | FP1 AS (Further Pure 1 AS) |
| Year | 2021 |
| Session | June |
| Marks | 6 |
| Topic | Vectors: Cross Product & Distances |
| Type | Find parameter value for geometric condition |
| Difficulty | Standard +0.3 This is a straightforward FP1 vectors question requiring cross product calculation, dot product verification, and solving for a parameter. Part (a) is routine cross product, part (b) is algebraic verification using the intersection condition, and part (c) involves simple equation solving. While it requires multiple techniques, each step follows standard procedures with no novel insight needed, making it slightly easier than average. |
| Spec | 4.04c Scalar product: calculate and use for angles4.04g Vector product: a x b perpendicular vector |
3 The equations of two intersecting lines are\\
$\mathbf { r } = \left( \begin{array} { c } - 12 \\ a \\ - 1 \end{array} \right) + \lambda \left( \begin{array} { l } 2 \\ 2 \\ 1 \end{array} \right) \quad \mathbf { r } = \left( \begin{array} { l } 2 \\ 0 \\ 5 \end{array} \right) + \mu \left( \begin{array} { c } - 3 \\ 1 \\ - 1 \end{array} \right)$\\
where $a$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Find a vector, $\mathbf { b }$, which is perpendicular to both lines.
\item Show that b. $\left( \begin{array} { c } - 12 \\ a \\ - 1 \end{array} \right) =$ b. $\left( \begin{array} { l } 2 \\ 0 \\ 5 \end{array} \right)$.
\item Hence, or otherwise, find the value of $a$.
\end{enumerate}
\hfill \mbox{\textit{OCR FP1 AS 2021 Q3 [6]}}