| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Perpendicular distance from point to line |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing standard vector techniques: normalizing a direction vector (part a), finding the closest point on a line using perpendicularity (part b), and calculating triangle area using ½|a×b| (part c). All parts follow routine procedures with no novel insight required, making it slightly easier than average. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors4.04a Line equations: 2D and 3D, cartesian and vector forms |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left\ | \begin{pmatrix}4\\4\\2\end{pmatrix}\right\ | = \sqrt{4^2+4^2+2^2}=6\) and attempts \(\frac{1}{6}\begin{pmatrix}4\\4\\2\end{pmatrix}\) |
| \(\overrightarrow{OA} = \begin{pmatrix}2/3\\2/3\\1/3\end{pmatrix}\) | A1 | Must be in vector form, not coordinate form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Position vector of \(X = \begin{pmatrix}1+4\lambda\\-10+4\lambda\\-9+2\lambda\end{pmatrix}\) | M1 | Attempt at coordinates/position vector of point \(X\) |
| \(\overrightarrow{OX}\cdot\begin{pmatrix}4\\4\\2\end{pmatrix}=0 \Rightarrow 4(1+4\lambda)+4(-10+4\lambda)+2(-9+2\lambda)=0\) | dM1 | Using \(\overrightarrow{OX}\cdot\begin{pmatrix}4\\4\\2\end{pmatrix}=0\) to set up equation in \(\lambda\) |
| \(36\lambda = 54 \Rightarrow \lambda = 1.5\) | ddM1 A1 | Dependent on both previous M marks; correct \(\lambda=1.5\) |
| \(X = (7,-4,-6)\) | A1 | Condone position vector form \(7\mathbf{i}-4\mathbf{j}-6\mathbf{k}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Finds \(OX = \sqrt{7^2+(-4)^2+(-6)^2}=\sqrt{101}\) and \(OA=1\) | M1 | Finds all elements required to calculate area |
| Area \(OXA = \frac{1}{2}\times 1\times\sqrt{101} = \frac{\sqrt{101}}{2}\) | dM1 A1 | Correct method: area \(=\frac{1}{2}\times OX\times 1\); correct answer \(\frac{\sqrt{101}}{2}\) |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left\|\begin{pmatrix}4\\4\\2\end{pmatrix}\right\| = \sqrt{4^2+4^2+2^2}=6$ and attempts $\frac{1}{6}\begin{pmatrix}4\\4\\2\end{pmatrix}$ | M1 | Correct attempt at unit vector |
| $\overrightarrow{OA} = \begin{pmatrix}2/3\\2/3\\1/3\end{pmatrix}$ | A1 | Must be in vector form, not coordinate form |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Position vector of $X = \begin{pmatrix}1+4\lambda\\-10+4\lambda\\-9+2\lambda\end{pmatrix}$ | M1 | Attempt at coordinates/position vector of point $X$ |
| $\overrightarrow{OX}\cdot\begin{pmatrix}4\\4\\2\end{pmatrix}=0 \Rightarrow 4(1+4\lambda)+4(-10+4\lambda)+2(-9+2\lambda)=0$ | dM1 | Using $\overrightarrow{OX}\cdot\begin{pmatrix}4\\4\\2\end{pmatrix}=0$ to set up equation in $\lambda$ |
| $36\lambda = 54 \Rightarrow \lambda = 1.5$ | ddM1 A1 | Dependent on both previous M marks; correct $\lambda=1.5$ |
| $X = (7,-4,-6)$ | A1 | Condone position vector form $7\mathbf{i}-4\mathbf{j}-6\mathbf{k}$ |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Finds $OX = \sqrt{7^2+(-4)^2+(-6)^2}=\sqrt{101}$ and $OA=1$ | M1 | Finds all elements required to calculate area |
| Area $OXA = \frac{1}{2}\times 1\times\sqrt{101} = \frac{\sqrt{101}}{2}$ | dM1 A1 | Correct method: area $=\frac{1}{2}\times OX\times 1$; correct answer $\frac{\sqrt{101}}{2}$ |
---\begin{enumerate}
\item Relative to a fixed origin $O$, the line $l$ has equation
\end{enumerate}
$$\mathbf { r } = \left( \begin{array} { r }
1 \\
- 10 \\
- 9
\end{array} \right) + \lambda \left( \begin{array} { l }
4 \\
4 \\
2
\end{array} \right) \quad \text { where } \lambda \text { is a scalar parameter }$$
Given that $\overrightarrow { O A }$ is a unit vector parallel to $l$,\\
(a) find $\overrightarrow { O A }$
The point $X$ lies on $l$.\\
Given that $X$ is the point on $l$ that is closest to the origin,\\
(b) find the coordinates of $X$.
The points $O , X$ and $A$ form the triangle $O X A$.\\
(c) Find the exact area of triangle $O X A$.\\
\hfill \mbox{\textit{Edexcel P4 2021 Q7 [10]}}