| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2022 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Line intersection with line |
| Difficulty | Standard +0.3 This is a standard line intersection problem from Further Maths P4. While it requires setting up parametric equations and solving a system, the technique is routine for this level. The question appears incomplete (missing the direction vector for l₁ and the equation for l₂), but typical problems of this type involve straightforward algebraic manipulation with no novel geometric insight required. Slightly easier than average due to being a standard textbook exercise. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10g Problem solving with vectors: in geometry4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04e Line intersections: parallel, skew, or intersecting |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(4+2\lambda=13+5\mu\), \(4-3\lambda=-1+\mu\), \(-5+6\lambda=4-3\mu\) | M1 | For writing down any 2 of these equations |
| Full method using two equations to find \(\lambda\) or \(\mu\) | M1 | Full method for finding \(\lambda\) or \(\mu\) |
| \(\lambda=2\), \(\mu=-1\) | A1 | Both correct values |
| \(-5+6\lambda=-5+12=7\), \(4-3\mu=4+3=7\), so lines intersect | B1 | Shows parameters satisfy third equation and makes a conclusion |
| \(\lambda=2\rightarrow(4+4)\mathbf{i}+(4-6)\mathbf{j}+(-5+12)\mathbf{k}\) or \(\mu=-1\rightarrow(13-5)\mathbf{i}+(-1-1)\mathbf{j}+(4+3)\mathbf{k}\) | M1 | Uses their \(\lambda\) or \(\mu\) to find \(A\) |
| \(8\mathbf{i}-2\mathbf{j}+7\mathbf{k}\) | A1 | Correct vector or coordinates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{pmatrix}2\\-3\\6\end{pmatrix}\cdot\begin{pmatrix}5\\1\\-3\end{pmatrix}=10-3-18=\sqrt{2^2+3^2+6^2}\sqrt{5^2+1^2+3^2}\cos\theta\) | M1 | Full attempt at scalar product between direction vectors |
| \(\cos\theta=\pm\dfrac{11}{7\sqrt{35}}\) | A1 | Correct magnitude for \(\cos\theta\); may be implied by e.g. \(\theta=105.4\ldots\) or \(74.6\ldots\) |
| \(\theta=74.6°\) | A1 | Awrt 74.6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \( | 2\mathbf{i}-3\mathbf{j}+6\mathbf{k} | =\sqrt{2^2+3^2+6^2}=7\) |
| \(35\div7=5\Rightarrow\lambda=5\); \(8\mathbf{i}-2\mathbf{j}+7\mathbf{k}\pm5(2\mathbf{i}-3\mathbf{j}+6\mathbf{k})\) | M1 | Correct strategy for one of the points |
| \((18,-17,37)\) or \((-2,13,-23)\) | A1 | One correct point (ignore labels) |
| \(P(18,-17,37)\) and \(Q(-2,13,-23)\) | A1 | Correct points with correct labels |
## Question 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4+2\lambda=13+5\mu$, $4-3\lambda=-1+\mu$, $-5+6\lambda=4-3\mu$ | M1 | For writing down any 2 of these equations |
| Full method using two equations to find $\lambda$ or $\mu$ | M1 | Full method for finding $\lambda$ or $\mu$ |
| $\lambda=2$, $\mu=-1$ | A1 | Both correct values |
| $-5+6\lambda=-5+12=7$, $4-3\mu=4+3=7$, so lines intersect | B1 | Shows parameters satisfy third equation and makes a conclusion |
| $\lambda=2\rightarrow(4+4)\mathbf{i}+(4-6)\mathbf{j}+(-5+12)\mathbf{k}$ or $\mu=-1\rightarrow(13-5)\mathbf{i}+(-1-1)\mathbf{j}+(4+3)\mathbf{k}$ | M1 | Uses their $\lambda$ or $\mu$ to find $A$ |
| $8\mathbf{i}-2\mathbf{j}+7\mathbf{k}$ | A1 | Correct vector or coordinates |
## Question 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}2\\-3\\6\end{pmatrix}\cdot\begin{pmatrix}5\\1\\-3\end{pmatrix}=10-3-18=\sqrt{2^2+3^2+6^2}\sqrt{5^2+1^2+3^2}\cos\theta$ | M1 | Full attempt at scalar product between direction vectors |
| $\cos\theta=\pm\dfrac{11}{7\sqrt{35}}$ | A1 | Correct magnitude for $\cos\theta$; may be implied by e.g. $\theta=105.4\ldots$ or $74.6\ldots$ |
| $\theta=74.6°$ | A1 | Awrt 74.6 |
## Question 5(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $|2\mathbf{i}-3\mathbf{j}+6\mathbf{k}|=\sqrt{2^2+3^2+6^2}=7$ | M1 | Finds magnitude of direction of $l_1$ |
| $35\div7=5\Rightarrow\lambda=5$; $8\mathbf{i}-2\mathbf{j}+7\mathbf{k}\pm5(2\mathbf{i}-3\mathbf{j}+6\mathbf{k})$ | M1 | Correct strategy for one of the points |
| $(18,-17,37)$ or $(-2,13,-23)$ | A1 | One correct point (ignore labels) |
| $P(18,-17,37)$ and $Q(-2,13,-23)$ | A1 | Correct points with correct labels |
---5. With respect to a fixed origin $O$, the lines $l _ { 1 }$ and $l _ { 2 }$ are given by the equations
$$l _ { 1 } : \mathbf { r } = \left( \begin{array} { r }
4 \\
4 \\
- 5
\end{array} \right) + \lambda \left( \begin{array} { r }
2 \\
- 3 \\
6
\end{array} \right) \quad l _ { 2 } : \mathbf { r } = \left( \begin{array} { r }
13 \\
- 1 \\
4
\end{array} \right) + \mu \left( \begin{array} { r }
5 \\
1 \\
- 3
\end{array} \right)$$
where $\lambda$ and $\mu$ are scalar parameters.
\begin{enumerate}[label=(\alph*)]
\item Show that $l _ { 1 }$ and $l _ { 2 }$ meet and find the position vector of their point of intersection $A$.
\item Find the acute angle between $l _ { 1 }$ and $l _ { 2 }$, giving your answer in degrees to one decimal place.
A circle with centre $A$ and radius 35 cuts the line $l _ { 1 }$ at the points $P$ and $Q$. Given that the $x$ coordinate of $P$ is greater than the $x$ coordinate of $Q$,
\item find the coordinates of $P$ and the coordinates of $Q$.
\section*{Question 5 continued}
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\section*{Question 5 continued}
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\section*{Question 5 continued}
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\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2022 Q5 [13]}}