| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure (Further Additional Pure) |
| Year | 2018 |
| Session | September |
| Marks | 12 |
| Topic | Vectors: Cross Product & Distances |
| Type | Volume of tetrahedron using scalar triple product |
| Difficulty | Challenging +1.2 This is a structured Further Maths question on 3D vectors requiring scalar triple product for volumes and vector methods for distance to plane. While it involves multiple techniques (scalar triple product, normal vectors, distance formula), each part is clearly signposted and follows standard procedures. The geometric setup is straightforward (coordinate axes intercepts), and the 'show that' in part (i)(b) provides a target to verify rather than requiring independent discovery. More routine than typical Further Maths proof questions but above average A-level difficulty due to the topic and multi-step nature. |
| Spec | 4.04g Vector product: a x b perpendicular vector4.04j Shortest distance: between a point and a plane |
| Answer | Marks | Guidance |
|---|---|---|
| (i) (a) Use of volume formula \(V = \frac{1}{6} \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})\) in any valid setting | M1 | Any arrangement of \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) |
| Genuine attempt at a suitable scalar triple product \((\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}))\) for \(OABC\) \(\Rightarrow V_{OABC} = \frac{1}{6}abc\) | M1, A1 | By the determinant method or long-hand |
| Answer | Marks | Guidance |
|---|---|---|
| \((\mathbf{p} - \mathbf{a}) \cdot ((\mathbf{p} - \mathbf{b}) \times (\mathbf{p} - \mathbf{c})) = \begin{vmatrix} 0 & a & a \\ b & 0 & b \\ c & c & 0 \end{vmatrix} = (2abc)\) | M1 | Must have appropriate subtracted components |
| \(\Rightarrow V_{PABC} = \frac{1}{6}abc\) | A1 | |
| (i) (b) Since \(V(OABC) : V(PABC) = 1 : 2\) and the area of the base-plane \(ABC\) is common to both tetrahedra, the "heights" are also in the ratio \(1 : 2\) | B1, B1 | |
| (ii) (a) \(\mathbf{n} = (\mathbf{b} - \mathbf{a}) \times (\mathbf{c} - \mathbf{a})\) (e.g.) \(= \begin{vmatrix} -a & -a & bc \\ b & 0 & ca \\ 0 & c & ab \end{vmatrix}\) | M1, A1 | |
| (ii) (b) Area \(\triangle ABC = \frac{1}{2} | \mathbf{n} | \) where \(\mathbf{n} =\) vector answer to (a) |
| Then \(\frac{1}{6}abc = \frac{1}{4}(\text{area } \triangle ABC)\) (dist. \(O\) to \(\Pi\)) | M1 | |
| \(\Rightarrow \text{dist. } P \text{ to } \Pi = 2 \times \text{dist. } O \text{ to } \Pi = \frac{2abc}{\sqrt{a^2b^2 + b^2c^2 + c^2a^2}}\) | A1 | |
| Alternative: Plane \(ABC\) has eqn. \(\mathbf{r} \cdot \mathbf{a} = \mathbf{a} \cdot \mathbf{n} = abc\) (e.g.) using \(\mathbf{n}\) from (a) | B1, M1 | Use of formula |
| Dist. \(O\) to \(\Pi =\) this RHS \(\div\) [their \(\mathbf{n}\)] \(\Rightarrow \text{dist. } P \text{ to } \Pi = 2 \times \text{dist. } O \text{ to } \Pi = \frac{2abc}{\sqrt{a^2b^2 + b^2c^2 + c^2a^2}}\) | A1 |
**(i) (a)** Use of volume formula $V = \frac{1}{6} \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$ in any valid setting | M1 | Any arrangement of $\mathbf{a}, \mathbf{b}, \mathbf{c}$
Genuine attempt at a suitable scalar triple product $(\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}))$ for $OABC$ $\Rightarrow V_{OABC} = \frac{1}{6}abc$ | M1, A1 | By the determinant method or long-hand
Second scalar triple product attempted for $PABC$:
$(\mathbf{p} - \mathbf{a}) \cdot ((\mathbf{p} - \mathbf{b}) \times (\mathbf{p} - \mathbf{c})) = \begin{vmatrix} 0 & a & a \\ b & 0 & b \\ c & c & 0 \end{vmatrix} = (2abc)$ | M1 | Must have appropriate subtracted components
$\Rightarrow V_{PABC} = \frac{1}{6}abc$ | A1 |
**(i) (b)** Since $V(OABC) : V(PABC) = 1 : 2$ and the area of the base-plane $ABC$ is common to both tetrahedra, the "heights" are also in the ratio $1 : 2$ | B1, B1 |
**(ii) (a)** $\mathbf{n} = (\mathbf{b} - \mathbf{a}) \times (\mathbf{c} - \mathbf{a})$ (e.g.) $= \begin{vmatrix} -a & -a & bc \\ b & 0 & ca \\ 0 & c & ab \end{vmatrix}$ | M1, A1 |
**(ii) (b)** Area $\triangle ABC = \frac{1}{2}|\mathbf{n}|$ where $\mathbf{n} =$ vector answer to (a) | B1 | Not required (yet) in terms of $a, b, c$
Then $\frac{1}{6}abc = \frac{1}{4}(\text{area } \triangle ABC)$ (dist. $O$ to $\Pi$) | M1 |
$\Rightarrow \text{dist. } P \text{ to } \Pi = 2 \times \text{dist. } O \text{ to } \Pi = \frac{2abc}{\sqrt{a^2b^2 + b^2c^2 + c^2a^2}}$ | A1 |
**Alternative:** Plane $ABC$ has eqn. $\mathbf{r} \cdot \mathbf{a} = \mathbf{a} \cdot \mathbf{n} = abc$ (e.g.) using $\mathbf{n}$ from (a) | B1, M1 | Use of formula
Dist. $O$ to $\Pi =$ this RHS $\div$ [their $\mathbf{n}$] $\Rightarrow \text{dist. } P \text{ to } \Pi = 2 \times \text{dist. } O \text{ to } \Pi = \frac{2abc}{\sqrt{a^2b^2 + b^2c^2 + c^2a^2}}$ | A1 |The points $A$, $B$, $C$ and $P$ have coordinates $(a, 0, 0)$, $(0, b, 0)$, $(0, 0, c)$ and $(a, b, c)$ respectively, where $a$, $b$ and $c$ are positive constants.
The plane $\Pi$ contains $A$, $B$ and $C$.
\begin{enumerate}[label=(\roman*)]
\item \begin{enumerate}[label=(\alph*)]
\item Use the scalar triple product to determine
\begin{itemize}
\item the volume of tetrahedron $OABC$,
\item the volume of tetrahedron $PABC$. [5]
\end{itemize}
\item Hence show that the distance from $P$ to $\Pi$ is twice the distance from $O$ to $\Pi$. [2]
\end{enumerate}
\item \begin{enumerate}[label=(\alph*)]
\item Determine a vector which is normal to $\Pi$. [2]
\item Hence determine, in terms of $a$, $b$ and $c$ only, the distance from $P$ to $\Pi$. [3]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure 2018 Q4 [12]}}