| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Volume of tetrahedron using scalar triple product |
| Difficulty | Standard +0.8 This is a Further Maths question requiring (a) finding a plane equation via cross product of two vectors in the plane, and (b) using the scalar triple product formula for tetrahedron volume (V = 1/6|a·(b×c)|) to solve for k. While the techniques are standard for FM students, the multi-step nature, coordinate manipulation, and application of the volume formula place it moderately above average difficulty. |
| Spec | 4.04b Plane equations: cartesian and vector forms |
| VIIIV SIHI NI IIIIM I I O N OA | VI4V SIHI NI JIIIM ION OC | VJYV SIHI NI JIIIM ION OO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\overrightarrow{AB} = \begin{pmatrix}-2\\1\\1\end{pmatrix}\), \(\overrightarrow{AC} = \begin{pmatrix}1\\-1\\3\end{pmatrix}\), \(\overrightarrow{BC} = \begin{pmatrix}3\\-2\\2\end{pmatrix}\) | B1 | Two correct vectors in \(\Pi\). Can be negatives of those shown |
| \(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 1 & 1 \\ 1 & -1 & 3\end{vmatrix} = \begin{pmatrix}4\\7\\1\end{pmatrix}\) | M1A1 | M1: Attempt cross product of two vectors lying in \(\Pi\) (at least one number correct). A1: Correct normal vector |
| \(\begin{pmatrix}4\\7\\1\end{pmatrix} \cdot \begin{pmatrix}1\\2\\3\end{pmatrix} = 4+14+3\) | dM1 | Attempt scalar product with their normal and a point in the plane |
| \(4x + 7y + z = 21\) | A1 | Cao (oe) |
| Alternative (substitution method): \(a+2b+3c=d\); \(-a+3b+4c=d\); \(2a+b+6c=d\) | B1 | Correct equations |
| \(a=\frac{4}{21}d,\ b=\frac{1}{3}d,\ c=\frac{1}{21}d\) | M1A1 | M1: Solve for \(a,b,c\) in terms of \(d\). A1: Correct equations |
| \(d=21 \Rightarrow a=\ldots,\ b=\ldots,\ c=\ldots\) | M1 | Obtains values for \(a,b,c\) and \(d\) |
| \(4x+7y+z=21\) | A1 | Cao (oe) |
| Alternative (vector form): \(\mathbf{r} = \begin{pmatrix}1\\2\\3\end{pmatrix} + s\begin{pmatrix}-2\\1\\1\end{pmatrix} + t\begin{pmatrix}1\\-1\\3\end{pmatrix}\) | M1 | Two correct vectors in the plane |
| Deduce \(x=1-2s+t\), \(y=2+s-t\), \(z=3+s+3t\) | A1 | Deduce 3 correct equations |
| \(4x+7y+z=21\) | M1A1 | M1: Eliminate \(s,t\). A1: Cao |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(AD \cdot AB \times AC\) | M1 | Attempt suitable triple product |
| \(= \begin{pmatrix}4\\7\\1\end{pmatrix} \cdot \begin{pmatrix}k-1\\2\\11\end{pmatrix} = 4k-4+14+11\) | ||
| \(\therefore \frac{1}{6}(4k+21) = 6\) | dM1A1 | M1: Set \(\frac{1}{6}\)(their triple product) \(= 6\). A1: Correct equation |
| \(k = \frac{15}{4}\) | A1 | Cao (oe) |
| Alternative: Area \(ABC = \frac{1}{2} | \overrightarrow{AB} | |
| Distance \(D\) to \(\Pi\) is \(\frac{4k+28+14-21}{\sqrt{16+49+1}}\) | ||
| \(\frac{1}{6}\sqrt{6}\sqrt{11}\cdot\frac{4k+28+14-21}{\sqrt{16+49+1}} = 6\) | dM1A1 | M1: Set \(\frac{1}{3}\)(their area \(\times\) their distance) \(= 6\). A1: Correct equation |
| \(k = \frac{15}{4}\) | A1 | Cao (oe) |
| (4) | ||
| Total 9 |
## Question 6:
**Part (a)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\overrightarrow{AB} = \begin{pmatrix}-2\\1\\1\end{pmatrix}$, $\overrightarrow{AC} = \begin{pmatrix}1\\-1\\3\end{pmatrix}$, $\overrightarrow{BC} = \begin{pmatrix}3\\-2\\2\end{pmatrix}$ | B1 | Two correct vectors in $\Pi$. Can be negatives of those shown |
| $\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 1 & 1 \\ 1 & -1 & 3\end{vmatrix} = \begin{pmatrix}4\\7\\1\end{pmatrix}$ | M1A1 | M1: Attempt cross product of two vectors lying in $\Pi$ (at least one number correct). A1: Correct normal vector |
| $\begin{pmatrix}4\\7\\1\end{pmatrix} \cdot \begin{pmatrix}1\\2\\3\end{pmatrix} = 4+14+3$ | dM1 | Attempt scalar product with their normal and a point in the plane |
| $4x + 7y + z = 21$ | A1 | Cao (oe) |
| **Alternative (substitution method):** $a+2b+3c=d$; $-a+3b+4c=d$; $2a+b+6c=d$ | B1 | Correct equations |
| $a=\frac{4}{21}d,\ b=\frac{1}{3}d,\ c=\frac{1}{21}d$ | M1A1 | M1: Solve for $a,b,c$ in terms of $d$. A1: Correct equations |
| $d=21 \Rightarrow a=\ldots,\ b=\ldots,\ c=\ldots$ | M1 | Obtains values for $a,b,c$ and $d$ |
| $4x+7y+z=21$ | A1 | Cao (oe) |
| **Alternative (vector form):** $\mathbf{r} = \begin{pmatrix}1\\2\\3\end{pmatrix} + s\begin{pmatrix}-2\\1\\1\end{pmatrix} + t\begin{pmatrix}1\\-1\\3\end{pmatrix}$ | M1 | Two correct vectors in the plane |
| Deduce $x=1-2s+t$, $y=2+s-t$, $z=3+s+3t$ | A1 | Deduce 3 correct equations |
| $4x+7y+z=21$ | M1A1 | M1: Eliminate $s,t$. A1: Cao |
| | | (5) |
**Part (b)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $AD \cdot AB \times AC$ | M1 | Attempt suitable triple product |
| $= \begin{pmatrix}4\\7\\1\end{pmatrix} \cdot \begin{pmatrix}k-1\\2\\11\end{pmatrix} = 4k-4+14+11$ | | |
| $\therefore \frac{1}{6}(4k+21) = 6$ | dM1A1 | M1: Set $\frac{1}{6}$(their triple product) $= 6$. A1: Correct equation |
| $k = \frac{15}{4}$ | A1 | Cao (oe) |
| **Alternative:** Area $ABC = \frac{1}{2}|\overrightarrow{AB}||\overrightarrow{AC}| = \frac{1}{2}\sqrt{6}\sqrt{11}$ | M1 | Attempt area $ABC$ and distance between $D$ and $\Pi$ |
| Distance $D$ to $\Pi$ is $\frac{4k+28+14-21}{\sqrt{16+49+1}}$ | | |
| $\frac{1}{6}\sqrt{6}\sqrt{11}\cdot\frac{4k+28+14-21}{\sqrt{16+49+1}} = 6$ | dM1A1 | M1: Set $\frac{1}{3}$(their area $\times$ their distance) $= 6$. A1: Correct equation |
| $k = \frac{15}{4}$ | A1 | Cao (oe) |
| | | (4) |
| | | **Total 9** |6. The coordinates of the points $A , B$ and $C$ relative to a fixed origin $O$ are ( $1,2,3$ ), $( - 1,3,4 )$ and $( 2,1,6 )$ respectively. The plane $\Pi$ contains the points $A , B$ and $C$.
\begin{enumerate}[label=(\alph*)]
\item Find a cartesian equation of the plane $\Pi$.
The point $D$ has coordinates $( k , 4,14 )$ where $k$ is a positive constant.\\
Given that the volume of the tetrahedron $A B C D$ is 6 cubic units,
\item find the value of $k$.\\
\begin{center}
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VIIIV SIHI NI IIIIM I I O N OA & VI4V SIHI NI JIIIM ION OC & VJYV SIHI NI JIIIM ION OO \\
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\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2016 Q6 [9]}}