| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 2 (Further Pure Core 2) |
| Year | 2018 |
| Session | September |
| Marks | 6 |
| Topic | Vectors: Lines & Planes |
| Type | Perpendicular distance point to plane |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question testing standard vector methods for planes. Part (i) requires computing a cross product of the direction vectors, part (ii) uses the normal vector to convert to Cartesian form, and part (iii) applies the standard distance formula. All three parts are routine applications of well-practiced techniques with no problem-solving insight required, making it slightly easier than average even for Further Maths. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04g Vector product: a x b perpendicular vector4.04j Shortest distance: between a point and a plane |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{pmatrix} 1 \\ 1 \\ 3 \end{pmatrix} \times \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix} = \ldots\) | M1 | Attempt to find cross product (can be implied by one correct component or all 3 with wrong sign) |
| \(\begin{pmatrix} -5 \\ -4 \\ 3 \end{pmatrix}\) | A1 | Or any non-zero multiple. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{r} \cdot \begin{pmatrix} -5 \\ -4 \\ 3 \end{pmatrix} = \begin{pmatrix} -5 \\ -4 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} -5 \\ -4 \\ 3 \end{pmatrix} = 2x - 5 + (-3x - 4) + 5 \times 3\) | M1 | Use their \(\mathbf{n}\) from part (i) in \(\mathbf{r.n} = \mathbf{a.n}\) and attempt to find dot product. "Dot product" must be a scalar. |
| \(\mathbf{r} \cdot \begin{pmatrix} -5 \\ -4 \\ 3 \end{pmatrix} = 17\) | A1 | Or any non-zero multiple of entire equation |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{vmatrix} -5 \\ -4 \\ 3 \end{vmatrix} = \sqrt{25 + 16 + 9} = \sqrt{50}\) | M1 | Correct method for finding modulus of their normal vector |
| \(d = \frac{17}{\sqrt{50}}\) cao | A1 | Must be positive |
## (i)
$\begin{pmatrix} 1 \\ 1 \\ 3 \end{pmatrix} \times \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix} = \ldots$ | M1 | Attempt to find cross product (can be implied by one correct component or all 3 with wrong sign)
$\begin{pmatrix} -5 \\ -4 \\ 3 \end{pmatrix}$ | A1 | Or any non-zero multiple.
[2]
## (ii)
$\mathbf{r} \cdot \begin{pmatrix} -5 \\ -4 \\ 3 \end{pmatrix} = \begin{pmatrix} -5 \\ -4 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} -5 \\ -4 \\ 3 \end{pmatrix} = 2x - 5 + (-3x - 4) + 5 \times 3$ | M1 | Use their $\mathbf{n}$ from part (i) in $\mathbf{r.n} = \mathbf{a.n}$ and attempt to find dot product. "Dot product" must be a scalar.
$\mathbf{r} \cdot \begin{pmatrix} -5 \\ -4 \\ 3 \end{pmatrix} = 17$ | A1 | Or any non-zero multiple of entire equation
[2]
## (iii)
$\begin{vmatrix} -5 \\ -4 \\ 3 \end{vmatrix} = \sqrt{25 + 16 + 9} = \sqrt{50}$ | M1 | Correct method for finding modulus of their normal vector
$d = \frac{17}{\sqrt{50}}$ cao | A1 | Must be positive
[2]
---The equation of a plane, $\Pi$, is
$$\Pi: \mathbf{r} = \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ 3 \end{pmatrix} + \mu \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix}.$$
\begin{enumerate}[label=(\roman*)]
\item Find a vector which is perpendicular to $\Pi$. [2]
\item Hence find an equation for $\Pi$ in the form $\mathbf{r} \cdot \mathbf{n} = p$. [2]
\item Find in the form $\sqrt{q}$ the shortest distance between $\Pi$ and the origin, where $q$ is a rational number. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 2 2018 Q3 [6]}}