| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Distance between parallel planes or line and parallel plane |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring: (a) finding a plane equation from a point and line (involving cross product of direction vectors), and (b) using the distance formula between parallel planes with algebraic manipulation. While systematic, it requires multiple FM techniques and careful algebraic work, placing it moderately above average difficulty. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(2\mathbf{i} - \mathbf{j} + 4\mathbf{k}\), \(2\mathbf{i} + \mathbf{j} - \mathbf{k}\) | B1 | 2 correct vectors lying in \(\Pi_1\) |
| \(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 4 \\ 2 & 1 & -1\end{vmatrix} = \begin{pmatrix}-3\\10\\4\end{pmatrix}\) | M1A1 | M1: Attempt normal vector using 2 vectors in \(\Pi_1\); A1: Correct normal (any multiple) |
| \(\begin{pmatrix}-3\\10\\4\end{pmatrix} \cdot \begin{pmatrix}3\\3\\-2\end{pmatrix} = -9+30-8 = 13\) | dM1 | Attempt scalar product with a point lying in the plane; dependent on previous M mark |
| \(3x - 10y - 4z = -13\) * | A1* | Correct equation |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\begin{pmatrix}-3\\10\\4\end{pmatrix}\cdot\begin{pmatrix}\alpha\\1\\1\end{pmatrix} = 14-3\alpha\) or \(\left(\begin{pmatrix}\alpha\\1\\1\end{pmatrix} - \begin{pmatrix}3\\3\\-2\end{pmatrix}\right)\cdot\begin{pmatrix}3\\-10\\-4\end{pmatrix} = 3\alpha-1\) | M1 | Attempt scalar product between \(\begin{pmatrix}\alpha\\1\\1\end{pmatrix}\) and their normal \(\begin{pmatrix}-3\\10\\4\end{pmatrix}\) or (a point in \(\Pi_1\)) |
| \(d = \left\ | \frac{3\alpha-1}{\sqrt{3^2+10^2+4^2}}\right\ | \) or \(d = \left\ |
| \(\left\ | \frac{3\alpha-1}{5\sqrt{5}}\right\ | = \frac{1}{\sqrt{5}}\) |
| \(3\alpha - 1 = \pm 5\) | A1 | Correct equations (must see \(\pm\)) |
| \(\alpha = 2, -\frac{4}{3}\) | A1, A1 | cao |
## Question 7(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $2\mathbf{i} - \mathbf{j} + 4\mathbf{k}$, $2\mathbf{i} + \mathbf{j} - \mathbf{k}$ | B1 | 2 correct vectors lying in $\Pi_1$ |
| $\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 4 \\ 2 & 1 & -1\end{vmatrix} = \begin{pmatrix}-3\\10\\4\end{pmatrix}$ | M1A1 | M1: Attempt normal vector using 2 vectors in $\Pi_1$; A1: Correct normal (any multiple) |
| $\begin{pmatrix}-3\\10\\4\end{pmatrix} \cdot \begin{pmatrix}3\\3\\-2\end{pmatrix} = -9+30-8 = 13$ | dM1 | Attempt scalar product with a point lying in the plane; dependent on previous M mark |
| $3x - 10y - 4z = -13$ * | A1* | Correct equation |
---
## Question 7(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}-3\\10\\4\end{pmatrix}\cdot\begin{pmatrix}\alpha\\1\\1\end{pmatrix} = 14-3\alpha$ or $\left(\begin{pmatrix}\alpha\\1\\1\end{pmatrix} - \begin{pmatrix}3\\3\\-2\end{pmatrix}\right)\cdot\begin{pmatrix}3\\-10\\-4\end{pmatrix} = 3\alpha-1$ | M1 | Attempt scalar product between $\begin{pmatrix}\alpha\\1\\1\end{pmatrix}$ and their normal $\begin{pmatrix}-3\\10\\4\end{pmatrix}$ or (a point in $\Pi_1$) |
| $d = \left\|\frac{3\alpha-1}{\sqrt{3^2+10^2+4^2}}\right\|$ or $d = \left\|\frac{13}{\sqrt{3^2+10^2+4^2}} - \frac{14-3\alpha}{\sqrt{3^2+10^2+4^2}}\right\|$ | dM1 | Use of correct distance method; dependent on previous M mark |
| $\left\|\frac{3\alpha-1}{5\sqrt{5}}\right\| = \frac{1}{\sqrt{5}}$ | ddM1 | Set their distance $= \frac{1}{\sqrt{5}}$; dependent on both previous M marks |
| $3\alpha - 1 = \pm 5$ | A1 | Correct equations (must see $\pm$) |
| $\alpha = 2, -\frac{4}{3}$ | A1, A1 | cao |
---\begin{enumerate}
\item The plane $\Pi _ { 1 }$ contains the point $( 3,3 , - 2 )$ and the line $\frac { x - 1 } { 2 } = \frac { y - 2 } { - 1 } = \frac { z + 1 } { 4 }$\\
(a) Show that a cartesian equation of the plane $\Pi _ { 1 }$ is
\end{enumerate}
$$3 x - 10 y - 4 z = - 13$$
The plane $\Pi _ { 2 }$ is parallel to the plane $\Pi _ { 1 }$\\
The point ( $\alpha , 1,1$ ), where $\alpha$ is a constant, lies in $\Pi _ { 2 }$\\
Given that the shortest distance between the planes $\Pi _ { 1 }$ and $\Pi _ { 2 }$ is $\frac { 1 } { \sqrt { 5 } }$\\
(b) find the possible values of $\alpha$.
\hfill \mbox{\textit{Edexcel F3 2015 Q7 [11]}}