| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Plane containing line and point/vector |
| Difficulty | Standard +0.3 This is a standard multi-part vectors question testing routine techniques: finding a normal vector via cross product, deriving a plane equation, finding an unknown constant, locating an intersection point, and calculating an angle. All steps follow textbook procedures with no novel insight required. While it has multiple parts (typical of 10-12 mark questions), each individual step is straightforward application of Core Pure 1 content, making it slightly easier than the average A-level question. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04d Angles: between planes and between line and plane |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \((3\mathbf{i}-10\mathbf{j}-\mathbf{k})\cdot(2\mathbf{i}+\mathbf{j}-4\mathbf{k}) = 6-10+4=0\) | M1 | Attempts scalar product between given vector and both direction vectors; calculation shown for at least one |
| \((3\mathbf{i}-10\mathbf{j}-\mathbf{k})\cdot(6\mathbf{i}+\mathbf{j}+8\mathbf{k}) = 18-10-8=0\) | ||
| So \(3\mathbf{i}-10\mathbf{j}-\mathbf{k}\) is perpendicular to \(\Pi\) | A1 | Obtains zero for both with sufficient working and concludes perpendicular; accept "normal to" as equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \((3\mathbf{i}-10\mathbf{j}-\mathbf{k})\cdot(\mathbf{i}-2\mathbf{j}+3\mathbf{k}) = 3+20-3=20\) | M1 | Attempts scalar product between normal vector and \(\mathbf{i}-2\mathbf{j}+3\mathbf{k}\); allow vector product from scratch; allow using normal form \(3x-10y-z=d\) and substituting \((1,-2,3)\) |
| \(3x-10y-z=20\) | A1 | Correct Cartesian equation; allow equivalent forms |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \((3\mathbf{i}-10\mathbf{j}-\mathbf{k})\cdot(5\mathbf{i}+p\mathbf{j}-7\mathbf{k})=\) "20" | M1 | Attempts scalar product between normal and \(5\mathbf{i}+p\mathbf{j}-7\mathbf{k}\), sets equal to 20 and solves linear equation in \(p\); alternatively substitutes \(x=5, y=p, z=-7\) into Cartesian equation |
| \(\Rightarrow 15-10p+7=\) "20" \(\Rightarrow p=\ldots\) | ||
| \(p=0.2\) (oe) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(1+2\lambda=5+6\mu,\ 3-4\lambda=-7+8\mu \Rightarrow \lambda=\ldots\) or \(\mu=\ldots\) | M1 | Complete method to find coordinates of \(A\); slip in transcribing a coordinate allowed |
| \(\mu=0.1\) (or \(\lambda=2.3\)) \(\Rightarrow A(5.6,\ 0.3,\ -6.2)\) | ||
| \(12\mathbf{i}-11\mathbf{j}+6\mathbf{k}-(5.6\mathbf{i}+0.3\mathbf{j}-6.2\mathbf{k})=6.4\mathbf{i}-11.3\mathbf{j}+12.2\mathbf{k}\) | M1 | Attempts vector \(AB\) and attempts scalar product with this and the normal vector; allow method for any value appearing afterwards |
| \((6.4\mathbf{i}-11.3\mathbf{j}+12.2\mathbf{k})\cdot(3\mathbf{i}-10\mathbf{j}-\mathbf{k})=19.2+113-12.2=120\) | ||
| \(120=\sqrt{6.4^2+11.3^2+12.2^2}\sqrt{3^2+10^2+1^2}\cos\alpha \Rightarrow \alpha=\ldots\) | M1 | Complete method to find required angle or its complement |
| or \(120=\sqrt{6.4^2+11.3^2+12.2^2}\sqrt{3^2+10^2+1^2}\sin\alpha \Rightarrow \alpha=\ldots\) | ||
| Angle between \(AB\) and plane \(\theta=40°\) (awrt) | A1 |
## Question 7:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $(3\mathbf{i}-10\mathbf{j}-\mathbf{k})\cdot(2\mathbf{i}+\mathbf{j}-4\mathbf{k}) = 6-10+4=0$ | M1 | Attempts scalar product between given vector and both direction vectors; calculation shown for at least one |
| $(3\mathbf{i}-10\mathbf{j}-\mathbf{k})\cdot(6\mathbf{i}+\mathbf{j}+8\mathbf{k}) = 18-10-8=0$ | | |
| So $3\mathbf{i}-10\mathbf{j}-\mathbf{k}$ is perpendicular to $\Pi$ | A1 | Obtains zero for both with sufficient working and concludes perpendicular; accept "normal to" as equivalent |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $(3\mathbf{i}-10\mathbf{j}-\mathbf{k})\cdot(\mathbf{i}-2\mathbf{j}+3\mathbf{k}) = 3+20-3=20$ | M1 | Attempts scalar product between normal vector and $\mathbf{i}-2\mathbf{j}+3\mathbf{k}$; allow vector product from scratch; allow using normal form $3x-10y-z=d$ and substituting $(1,-2,3)$ |
| $3x-10y-z=20$ | A1 | Correct Cartesian equation; allow equivalent forms |
### Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $(3\mathbf{i}-10\mathbf{j}-\mathbf{k})\cdot(5\mathbf{i}+p\mathbf{j}-7\mathbf{k})=$ "20" | M1 | Attempts scalar product between normal and $5\mathbf{i}+p\mathbf{j}-7\mathbf{k}$, sets equal to 20 and solves linear equation in $p$; alternatively substitutes $x=5, y=p, z=-7$ into Cartesian equation |
| $\Rightarrow 15-10p+7=$ "20" $\Rightarrow p=\ldots$ | | |
| $p=0.2$ (oe) | A1 | |
### Part (d):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $1+2\lambda=5+6\mu,\ 3-4\lambda=-7+8\mu \Rightarrow \lambda=\ldots$ or $\mu=\ldots$ | M1 | Complete method to find coordinates of $A$; slip in transcribing a coordinate allowed |
| $\mu=0.1$ (or $\lambda=2.3$) $\Rightarrow A(5.6,\ 0.3,\ -6.2)$ | | |
| $12\mathbf{i}-11\mathbf{j}+6\mathbf{k}-(5.6\mathbf{i}+0.3\mathbf{j}-6.2\mathbf{k})=6.4\mathbf{i}-11.3\mathbf{j}+12.2\mathbf{k}$ | M1 | Attempts vector $AB$ and attempts scalar product with this and the normal vector; allow method for any value appearing afterwards |
| $(6.4\mathbf{i}-11.3\mathbf{j}+12.2\mathbf{k})\cdot(3\mathbf{i}-10\mathbf{j}-\mathbf{k})=19.2+113-12.2=120$ | | |
| $120=\sqrt{6.4^2+11.3^2+12.2^2}\sqrt{3^2+10^2+1^2}\cos\alpha \Rightarrow \alpha=\ldots$ | M1 | Complete method to find required angle or its complement |
| or $120=\sqrt{6.4^2+11.3^2+12.2^2}\sqrt{3^2+10^2+1^2}\sin\alpha \Rightarrow \alpha=\ldots$ | | |
| Angle between $AB$ and plane $\theta=40°$ (awrt) | A1 | |
---\begin{enumerate}
\item The line $l _ { 1 }$ has equation
\end{enumerate}
$$\mathbf { r } = \mathbf { i } - 2 \mathbf { j } + 3 \mathbf { k } + \lambda ( 2 \mathbf { i } + \mathbf { j } - 4 \mathbf { k } )$$
and the line $l _ { 2 }$ has equation
$$\mathbf { r } = 5 \mathbf { i } + p \mathbf { j } - 7 \mathbf { k } + \mu ( 6 \mathbf { i } + \mathbf { j } + 8 \mathbf { k } )$$
where $\lambda$ and $\mu$ are scalar parameters and $p$ is a constant.\\
The plane $\Pi$ contains $l _ { 1 }$ and $l _ { 2 }$\\
(a) Show that the vector $3 \mathbf { i } - 10 \mathbf { j } - \mathbf { k }$ is perpendicular to $\Pi$\\
(b) Hence determine a Cartesian equation of $\Pi$\\
(c) Hence determine the value of $p$
Given that
\begin{itemize}
\item the lines $l _ { 1 }$ and $l _ { 2 }$ intersect at the point $A$
\item the point $B$ has coordinates $( 12 , - 11,6 )$\\
(d) determine, to the nearest degree, the acute angle between $A B$ and $\Pi$
\end{itemize}
\hfill \mbox{\textit{Edexcel CP1 2024 Q7 [10]}}