| Exam Board | Edexcel |
|---|---|
| Module | FP1 AS (Further Pure 1 AS) |
| Year | 2022 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | 3D geometry applications |
| Difficulty | Standard +0.8 This is a substantial 3D geometry problem requiring multiple techniques: finding intersection points using parametric equations, calculating triangle area using cross product, and computing volume of an irregular solid (likely by subtracting tetrahedron AMNP from ABCD). While the individual steps are standard Further Maths techniques, the multi-part nature, careful coordinate work, and volume calculation for a non-standard solid make this moderately challenging but still within typical FP1 scope. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04g Vector product: a x b perpendicular vector4.04h Shortest distances: between parallel lines and between skew lines |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Correct method to find one coordinate of \(M\), \(N\) or \(P\). E.g. \(\overrightarrow{AB} = \begin{pmatrix}6\\-6\\6\end{pmatrix}\) so \(\overrightarrow{OM} = \begin{pmatrix}3\\2\\-4\end{pmatrix} + \frac{4}{6}\begin{pmatrix}6\\-6\\6\end{pmatrix} = \ldots\) | M1 | Correct method for finding at least one of the three points; allow one slip in coordinates but should have correct fraction to make \(z = 0\) |
| One of \((M=)(7,-2,0)\), \((N=)(0,-2,0)\) or \((P=)(1,0,0)\) | A1 | Any one of the three points correct, ignoring labelling |
| All of \((M=)(7,-2,0)\), \((N=)(0,-2,0)\) and \((P=)(1,0,0)\) | A1 | All three points correct, ignoring labelling |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Correct method, e.g. realises \(MN\) is parallel to \(x\)-axis, so base is 7 and height 2, hence area \(= \frac{1}{2} \times 7 \times 2 = \ldots\) or using \(\frac{1}{2}\ | a \times b\ | \) with \(\overrightarrow{PM} = \pm\begin{pmatrix}6\\-2\\0\end{pmatrix}\), \(\overrightarrow{PN} = \pm\begin{pmatrix}-1\\-2\\0\end{pmatrix}\), e.g. \(\frac{1}{2}\left\ |
| \(= 7\) | A1 | Correct area of 7 from correct vectors |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Vol \(NMPA = \frac{1}{3}A_b h = \frac{1}{3} \times 7 \times 4 = \frac{28}{3}\) or using triple scalar product: \(NMPA = \frac{1}{6}\left | \overrightarrow{AM} \cdot \left(\overrightarrow{AN} \times \overrightarrow{AP}\right)\right | = \frac{28}{3}\) |
| Vol \(ABCD = \frac{1}{6}\left | \overrightarrow{AB} \cdot \left(\overrightarrow{AC} \times \overrightarrow{AD}\right)\right | = \frac{1}{6}\begin{vmatrix}6\\-6\\6\end{vmatrix} \cdot \begin{pmatrix}-84\\42\\-21\end{pmatrix} = 147\) |
| Volume required \(= 147 - \frac{28}{3} = \frac{413}{3}\) | M1 A1 | M1: Finds difference of two volumes using correct method for both; A1: \(\frac{413}{3}\) |
| (6) | ||
| (11 marks total) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{AB} = \begin{pmatrix} 6 \\ -6 \\ 6 \end{pmatrix}\), \(\overrightarrow{AC} = \begin{pmatrix} -9 \\ -12 \\ 12 \end{pmatrix}\), \(\overrightarrow{AD} = \begin{pmatrix} -7 \\ -7 \\ 14 \end{pmatrix}\) | B1 | Correct vectors stated |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{6}\left | \overrightarrow{AB} \cdot \left(\overrightarrow{AC} \times \overrightarrow{AD}\right)\right)\) | M1 |
| \(\overrightarrow{AC} \times \overrightarrow{AD} = \begin{pmatrix} -84 \\ 42 \\ -21 \end{pmatrix}\) | M1 | Attempt at cross product |
| \(\overrightarrow{AB} \cdot \begin{pmatrix} -84 \\ 42 \\ -21 \end{pmatrix} = 6(-84) + (-6)(42) + 6(-21) = -882\) | M1 A1 | Correct dot product |
| \(V = \frac{1}{6} \times 882 = 147\) | A1 | Correct volume = 147 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{6}\begin{vmatrix} 6 & -6 & 6 \\ -9 & -12 & 12 \\ -7 & -7 & 14 \end{vmatrix}\) | M1 | Correct determinant setup |
| \(= \frac{1}{6}\left | 6(-168+84) + 6(-126+84) + 6(63-84)\right | = \frac{1}{6} |
# Question 5:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Correct method to find one coordinate of $M$, $N$ or $P$. E.g. $\overrightarrow{AB} = \begin{pmatrix}6\\-6\\6\end{pmatrix}$ so $\overrightarrow{OM} = \begin{pmatrix}3\\2\\-4\end{pmatrix} + \frac{4}{6}\begin{pmatrix}6\\-6\\6\end{pmatrix} = \ldots$ | **M1** | Correct method for finding at least one of the three points; allow one slip in coordinates but should have correct fraction to make $z = 0$ |
| One of $(M=)(7,-2,0)$, $(N=)(0,-2,0)$ or $(P=)(1,0,0)$ | **A1** | Any one of the three points correct, ignoring labelling |
| All of $(M=)(7,-2,0)$, $(N=)(0,-2,0)$ and $(P=)(1,0,0)$ | **A1** | All three points correct, ignoring labelling |
| | **(3)** | |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Correct method, e.g. realises $MN$ is parallel to $x$-axis, so base is 7 and height 2, hence area $= \frac{1}{2} \times 7 \times 2 = \ldots$ **or** using $\frac{1}{2}\|a \times b\|$ with $\overrightarrow{PM} = \pm\begin{pmatrix}6\\-2\\0\end{pmatrix}$, $\overrightarrow{PN} = \pm\begin{pmatrix}-1\\-2\\0\end{pmatrix}$, e.g. $\frac{1}{2}\left\|\overrightarrow{MP} \times \overrightarrow{PN}\right\| = \frac{1}{2}\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\6&-2&0\\-1&-2&0\end{vmatrix} = \frac{1}{2}\left\|\begin{pmatrix}0\\0\\-14\end{pmatrix}\right\| = \ldots$ | **M1** | Correct method for finding area of triangle; condone sign slips except must use $-\mathbf{j}$ in cross product |
| $= 7$ | **A1** | Correct area of 7 from correct vectors |
| | **(2)** | |
## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Vol $NMPA = \frac{1}{3}A_b h = \frac{1}{3} \times 7 \times 4 = \frac{28}{3}$ or using triple scalar product: $NMPA = \frac{1}{6}\left|\overrightarrow{AM} \cdot \left(\overrightarrow{AN} \times \overrightarrow{AP}\right)\right| = \frac{28}{3}$ | **M1 A1** | M1: Correct method for volume of $NMPA$; A1: $\frac{28}{3}$ |
| Vol $ABCD = \frac{1}{6}\left|\overrightarrow{AB} \cdot \left(\overrightarrow{AC} \times \overrightarrow{AD}\right)\right| = \frac{1}{6}\begin{vmatrix}6\\-6\\6\end{vmatrix} \cdot \begin{pmatrix}-84\\42\\-21\end{pmatrix} = 147$ | **M1 A1** | M1: Complete attempt at volume of $ABCD$ with correct cross product method; A1: 147 |
| Volume required $= 147 - \frac{28}{3} = \frac{413}{3}$ | **M1 A1** | M1: Finds difference of two volumes using correct method for both; A1: $\frac{413}{3}$ |
| | **(6)** | |
| | **(11 marks total)** | |
## Volume of Tetrahedron Mark Scheme
---
### Question (Volume of Tetrahedron):
**Vectors from A:**
$\overrightarrow{AB} = \begin{pmatrix} 6 \\ -6 \\ 6 \end{pmatrix}$, $\overrightarrow{AC} = \begin{pmatrix} -9 \\ -12 \\ 12 \end{pmatrix}$, $\overrightarrow{AD} = \begin{pmatrix} -7 \\ -7 \\ 14 \end{pmatrix}$ | B1 | Correct vectors stated
---
**Method — scalar triple product:**
$\frac{1}{6}\left|\overrightarrow{AB} \cdot \left(\overrightarrow{AC} \times \overrightarrow{AD}\right)\right)$ | M1 | Correct volume formula using scalar triple product
$\overrightarrow{AC} \times \overrightarrow{AD} = \begin{pmatrix} -84 \\ 42 \\ -21 \end{pmatrix}$ | M1 | Attempt at cross product
$\overrightarrow{AB} \cdot \begin{pmatrix} -84 \\ 42 \\ -21 \end{pmatrix} = 6(-84) + (-6)(42) + 6(-21) = -882$ | M1 A1 | Correct dot product
$V = \frac{1}{6} \times 882 = 147$ | A1 | Correct volume = **147**
---
**Alternative — determinant method:**
$\frac{1}{6}\begin{vmatrix} 6 & -6 & 6 \\ -9 & -12 & 12 \\ -7 & -7 & 14 \end{vmatrix}$ | M1 | Correct determinant setup
$= \frac{1}{6}\left|6(-168+84) + 6(-126+84) + 6(63-84)\right| = \frac{1}{6}|-882| = 147$ | A1 | **Volume = 147**
---
**Note:** All four vertex-based calculations give the same result of $\frac{1}{6} \times 882 = \mathbf{147}$, confirming consistency regardless of which vertex is used as the base.5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{1be33445-e669-49af-a97e-a8ae84d63463-12_762_1129_246_468}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The points $A ( 3,2 , - 4 ) , B ( 9 , - 4,2 ) , C ( - 6 , - 10,8 )$ and $D ( - 4 , - 5,10 )$ are the vertices of a tetrahedron.
The plane with equation $z = 0$ cuts the tetrahedron into two pieces, one on each side of the plane.
The edges $A B , A C$ and $A D$ of the tetrahedron intersect the plane at the points $M , N$ and $P$ respectively, as shown in Figure 1.
Determine
\begin{enumerate}[label=(\alph*)]
\item the coordinates of the points $M , N$ and $P$,
\item the area of triangle $M N P$,
\item the exact volume of the solid $B C D P N M$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP1 AS 2022 Q5 [11]}}