| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2019 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Parallel and perpendicular lines |
| Difficulty | Standard +0.3 This is a standard A-level question on line-plane intersection requiring students to check if the direction vector is perpendicular to the normal (parallel case) or find intersection point. It involves routine dot product calculation and substitution into parametric equations—slightly easier than average as it follows a well-practiced algorithm with no novel insight required. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04e Line intersections: parallel, skew, or intersecting |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \((\mathbf{r} =)\begin{pmatrix}-2+\lambda\\5-\lambda\\4-3\lambda\end{pmatrix}\) or \(\begin{pmatrix}-2\\5\\4\end{pmatrix} + \lambda\begin{pmatrix}1\\-1\\-3\end{pmatrix}\) | M1 | Forms parametric form for the line; allow one slip |
| \(\begin{pmatrix}-2+\lambda\\5-\lambda\\4-3\lambda\end{pmatrix}\cdot\begin{pmatrix}1\\-2\\1\end{pmatrix} = -7 \Rightarrow (-2+\lambda)(1)+(5-\lambda)(-2)+(4-3\lambda)(1) = -7\) | M1 | Substitutes into equation of plane to get equation in \(\lambda\) |
| Correct equation in \(\lambda\) | A1 | Correct equation in \(\lambda\) |
| \(\Rightarrow 0\lambda - 8 = -7 \Rightarrow -8 = -7\), a contradiction so no intersection | A1ft | Simplifies to contradiction; \(\lambda\) disappears and constants unequal |
| Hence \(l\) is parallel to \(\Pi\) but not in it | A1cso | Correct deduction from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Attempts to solve simultaneously – eliminates one variable | M1 | 3.1a |
| e.g. \(y = -(x+2)+5 = -x+3 \Rightarrow x-2(-x+3)+z=-7 \Rightarrow 3x+z=-1\) | A1 | 1.1b |
| Solves reduced equations, e.g. \(-3(x+2)=z-4 \Rightarrow 3x+z=-2\) and \(3x+z=-1 \Rightarrow (3x+z)-(3x+z)=-2-(-1)\) | M1 | 1.1b |
| \(\Rightarrow 0=-1\) a contradiction so no intersection | A1ft | 2.3 |
| Hence \(l\) is parallel to \(\Pi\) but not in it. | A1cso | 3.2a |
| (5) | (5 marks) |
# Question 4:
| Working | Mark | Guidance |
|---------|------|----------|
| $(\mathbf{r} =)\begin{pmatrix}-2+\lambda\\5-\lambda\\4-3\lambda\end{pmatrix}$ or $\begin{pmatrix}-2\\5\\4\end{pmatrix} + \lambda\begin{pmatrix}1\\-1\\-3\end{pmatrix}$ | M1 | Forms parametric form for the line; allow one slip |
| $\begin{pmatrix}-2+\lambda\\5-\lambda\\4-3\lambda\end{pmatrix}\cdot\begin{pmatrix}1\\-2\\1\end{pmatrix} = -7 \Rightarrow (-2+\lambda)(1)+(5-\lambda)(-2)+(4-3\lambda)(1) = -7$ | M1 | Substitutes into equation of plane to get equation in $\lambda$ |
| Correct equation in $\lambda$ | A1 | Correct equation in $\lambda$ |
| $\Rightarrow 0\lambda - 8 = -7 \Rightarrow -8 = -7$, a contradiction so no intersection | A1ft | Simplifies to contradiction; $\lambda$ disappears and constants unequal |
| Hence $l$ is parallel to $\Pi$ but not in it | A1cso | Correct deduction from correct working |
## Alt 2:
**Line:** $\frac{x+2}{1} = \frac{y-5}{-1} = \frac{z-4}{-3}$ and plane $x - 2y + z = -7$
| Working | Mark | Guidance |
|---------|------|----------|
| Attempts to solve simultaneously – eliminates one variable | M1 | 3.1a |
| e.g. $y = -(x+2)+5 = -x+3 \Rightarrow x-2(-x+3)+z=-7 \Rightarrow 3x+z=-1$ | A1 | 1.1b |
| Solves reduced equations, e.g. $-3(x+2)=z-4 \Rightarrow 3x+z=-2$ and $3x+z=-1 \Rightarrow (3x+z)-(3x+z)=-2-(-1)$ | M1 | 1.1b |
| $\Rightarrow 0=-1$ a contradiction so no intersection | A1ft | 2.3 |
| Hence $l$ is parallel to $\Pi$ but not in it. | A1cso | 3.2a |
| **(5)** | | **(5 marks)** |
---\begin{enumerate}
\item The line $l$ has equation
\end{enumerate}
$$\frac { x + 2 } { 1 } = \frac { y - 5 } { - 1 } = \frac { z - 4 } { - 3 }$$
The plane $\Pi$ has equation
$$\mathbf { r } . ( \mathbf { i } - 2 \mathbf { j } + \mathbf { k } ) = - 7$$
Determine whether the line $l$ intersects $\Pi$ at a single point, or lies in $\Pi$, or is parallel to $\Pi$ without intersecting it.\\
(5)
\hfill \mbox{\textit{Edexcel CP AS 2019 Q4 [5]}}