| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Session | Specimen |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Find reflection of point in line or plane |
| Difficulty | Challenging +1.8 This is a Further Maths Core Pure question requiring multiple sophisticated steps: finding where the line intersects the plane, reflecting the direction vector in the plane using the normal vector, and constructing the reflected line equation. While the individual techniques (line-plane intersection, vector reflection formula) are standard for Further Maths, combining them in this multi-stage geometric problem requires solid spatial reasoning and careful execution, placing it well above average difficulty. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04f Line-plane intersection: find point |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2+4\lambda - 2(4-2\lambda)-6+\lambda = 6 \Rightarrow \lambda = \ldots\) | M1 | Substitutes the parametric equation of the line into the equation of the plane and solves for \(\lambda\) |
| \(\lambda = 2 \Rightarrow\) Required point is \((2+2(4),\, 4+2(-2),\, -6+2(1))\) giving \((10, 0, -4)\) | A1 | Obtains the correct coordinates of the intersection of the line and the plane |
| \(2+t-2(4-2t)-6+t = 6 \Rightarrow t = \ldots\) | M1 | Substitutes the parametric form of the line perpendicular to the plane passing through \((2,4,-6)\) into the equation of the plane to find \(t\) |
| \(t=3\) so reflection of \((2,4,-6)\) is \((2+6(1),\, 4+6(-2),\, -6+6(1))\) | M1 | Find the reflection of \((2,4,-6)\) in the plane |
| \((8,-8,0)\) | A1 | Correct coordinates |
| \(\begin{pmatrix}10\\0\\-4\end{pmatrix} - \begin{pmatrix}8\\-8\\0\end{pmatrix} = \begin{pmatrix}2\\8\\-4\end{pmatrix}\) | M1 | Determines the direction of \(l\) by subtracting the appropriate vectors |
| \(\mathbf{r} = \begin{pmatrix}10\\0\\-4\end{pmatrix} + k\begin{pmatrix}1\\4\\-2\end{pmatrix}\) or equivalent | A1 | Correct vector equation using the correct notation |
## Question 8:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2+4\lambda - 2(4-2\lambda)-6+\lambda = 6 \Rightarrow \lambda = \ldots$ | M1 | Substitutes the parametric equation of the line into the equation of the plane and solves for $\lambda$ |
| $\lambda = 2 \Rightarrow$ Required point is $(2+2(4),\, 4+2(-2),\, -6+2(1))$ giving $(10, 0, -4)$ | A1 | Obtains the correct coordinates of the intersection of the line and the plane |
| $2+t-2(4-2t)-6+t = 6 \Rightarrow t = \ldots$ | M1 | Substitutes the parametric form of the line perpendicular to the plane passing through $(2,4,-6)$ into the equation of the plane to find $t$ |
| $t=3$ so reflection of $(2,4,-6)$ is $(2+6(1),\, 4+6(-2),\, -6+6(1))$ | M1 | Find the reflection of $(2,4,-6)$ in the plane |
| $(8,-8,0)$ | A1 | Correct coordinates |
| $\begin{pmatrix}10\\0\\-4\end{pmatrix} - \begin{pmatrix}8\\-8\\0\end{pmatrix} = \begin{pmatrix}2\\8\\-4\end{pmatrix}$ | M1 | Determines the direction of $l$ by subtracting the appropriate vectors |
| $\mathbf{r} = \begin{pmatrix}10\\0\\-4\end{pmatrix} + k\begin{pmatrix}1\\4\\-2\end{pmatrix}$ or equivalent | A1 | Correct vector equation using the correct notation |
**(7 marks)**
---\begin{enumerate}
\item The line $l _ { 1 }$ has equation $\frac { x - 2 } { 4 } = \frac { y - 4 } { - 2 } = \frac { z + 6 } { 1 }$
\end{enumerate}
The plane $\Pi$ has equation $x - 2 y + z = 6$\\
The line $l _ { 2 }$ is the reflection of the line $l _ { 1 }$ in the plane $\Pi$.\\
Find a vector equation of the line $l _ { 2 }$
\hfill \mbox{\textit{Edexcel CP1 Q8 [7]}}