| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2011 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Area of triangle using cross product |
| Difficulty | Challenging +1.2 This FP3 question involves 3D vectors and geometric interpretation. Part (i) requires visualization and simple area formulas (½base×height for right triangles). Part (ii) is a direct application of the vector product definition. Part (iii) requires algebraic manipulation of vector products and expanding terms, but follows a clear path once the setup is established. While it involves multiple steps and vector algebra, it's a fairly standard FP3 question without requiring deep geometric insight or novel problem-solving approaches. |
| Spec | 1.10g Problem solving with vectors: in geometry4.04g Vector product: a x b perpendicular vector |
| Answer | Marks | Guidance |
|---|---|---|
| [Sketch of tetrahedron labelled with vertices] | B1 | For sketch of tetrahedron labelled in some way |
| At least one right angle at \(O\) must be indicated or clearly implied | ||
| \(\triangle OPQ = \frac{1}{2}pq, \triangle OQR = \frac{1}{2}qr, \triangle ORP = \frac{1}{2}rp\) | M1 A1 3 | For using \(\triangle = \frac{1}{2}\text{base} \times \text{height}\); For all areas correct CAO |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2} | \overrightarrow{RP} \times \overrightarrow{RQ} | = \frac{1}{2} |
| Answer | Marks | Guidance |
|---|---|---|
| LHS \(= \left(\frac{1}{2}pq\right)^2 + \left(\frac{1}{2}qr\right)^2 + \left(\frac{1}{2}rp\right)^2\) | B1 | For correct expression |
| \(\triangle PQR = \frac{1}{2} | (\vec{pi}-\vec{d}) \times (\vec{pi}-\vec{rk}) | \) |
| OR \(\frac{1}{2} | (\vec{pi}-\vec{k}) \times (\vec{qj}-\vec{rk}) | \) |
| OR \(\frac{1}{2} | (\vec{pi}-\vec{qj}) \times (\vec{qj}-\vec{rk}) | \) |
| \(\triangle PQR = \frac{1}{2} | qr\vec{i}+\vec{pr\vec{j}}+pq\vec{k} | \) |
| RHS \(= \frac{1}{4}\left((pq)^2+(qr)^2+(rp)^2\right)\) | A1 A1 6 | For correct expression; For using \( |
## (i)
[Sketch of tetrahedron labelled with vertices] | B1 | For sketch of tetrahedron labelled in some way
At least one right angle at $O$ must be indicated or clearly implied | |
$\triangle OPQ = \frac{1}{2}pq, \triangle OQR = \frac{1}{2}qr, \triangle ORP = \frac{1}{2}rp$ | M1 A1 3 | For using $\triangle = \frac{1}{2}\text{base} \times \text{height}$; For all areas correct CAO
## (ii)
$\frac{1}{2}|\overrightarrow{RP} \times \overrightarrow{RQ}| = \frac{1}{2}|\overrightarrow{RP}||\overrightarrow{RQ}|\sin R = \triangle PQR$ | B1 1 | For correct justification
## (iii)
LHS $= \left(\frac{1}{2}pq\right)^2 + \left(\frac{1}{2}qr\right)^2 + \left(\frac{1}{2}rp\right)^2$ | B1 | For correct expression
$\triangle PQR = \frac{1}{2}|(\vec{pi}-\vec{d}) \times (\vec{pi}-\vec{rk})|$ | B1 | For $\triangle PQR$ in vector form
OR $\frac{1}{2}|(\vec{pi}-\vec{k}) \times (\vec{qj}-\vec{rk})|$ | |
OR $\frac{1}{2}|(\vec{pi}-\vec{qj}) \times (\vec{qj}-\vec{rk})|$ | |
$\triangle PQR = \frac{1}{2}|qr\vec{i}+\vec{pr\vec{j}}+pq\vec{k}|$ | M1 | For finding vector product of their attempt at $\triangle PQR$
RHS $= \frac{1}{4}\left((pq)^2+(qr)^2+(rp)^2\right)$ | A1 A1 6 | For correct expression; For using $|\vec{a}+\vec{b}+\vec{c}k| = \sqrt{a^2+b^2+c^2}$; For completing proof of AG WWW
---(In this question, the notation $\Delta ABC$ denotes the area of the triangle $ABC$.)
The points $P$, $Q$ and $R$ have position vectors $p\mathbf{i}$, $q\mathbf{j}$ and $r\mathbf{k}$ respectively, relative to the origin $O$, where $p$, $q$ and $r$ are positive. The points $O$, $P$, $Q$ and $R$ are joined to form a tetrahedron.
\begin{enumerate}[label=(\roman*)]
\item Draw a sketch of the tetrahedron and write down the values of $\Delta OPQ$, $\Delta OQR$ and $\Delta ORP$. [3]
\item Use the definition of the vector product to show that $\frac{1}{2}|\overrightarrow{RP} \times \overrightarrow{RQ}| = \Delta PQR$. [1]
\item Show that $(\Delta OPQ)^2 + (\Delta OQR)^2 + (\Delta ORP)^2 = (\Delta PQR)^2$. [6]
\end{enumerate}
\hfill \mbox{\textit{OCR FP3 2011 Q7 [10]}}