| Exam Board | Edexcel |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2021 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Point on line satisfying condition |
| Difficulty | Challenging +1.2 This question requires understanding the vector equation of a line in cross-product form, parametrizing it, then using direction cosines (a standard FP1 topic) to find a specific point. While it involves multiple steps and vector manipulation, the techniques are all standard Further Maths content with no novel insight required—just systematic application of known methods. |
| Spec | 1.10c Magnitude and direction: of vectors4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Position of \(A\) is given by \(\overrightarrow{OA} = \begin{pmatrix}12+9\lambda\\16+6\lambda\\-8+2\lambda\end{pmatrix}\) | B1 | |
| \(\frac{12+9\lambda}{\sqrt{(12+9\lambda)^2+(16+6\lambda)^2+(-8+2\lambda)^2}} = \frac{3}{7}\) | M1 | |
| \(\Rightarrow 49(3(4+3\lambda))^2 = 9((12+9\lambda)^2+(16+6\lambda)^2+(-8+2\lambda)^2)\) \(\Rightarrow 2880\lambda^2+7200\lambda+2880=0\) or \(2\lambda^2+5\lambda+2=0\) | M1, A1 | |
| \(\Rightarrow (2\lambda+1)(\lambda+2)=0 \Rightarrow \lambda = ...\) | M1 | |
| Substitutes a value of \(\lambda\) to find position of \(A\), e.g. \(\overrightarrow{OA} = \begin{pmatrix}12+9(-\frac{1}{2})\\16+6(-\frac{1}{2})\\-8+2(-\frac{1}{2})\end{pmatrix}\) | M1 | |
| Coordinates of \(A\) are \(\left(\frac{15}{2}, 13, -9\right)\) only | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Direction of \(\overrightarrow{OA}\) is given by \(\mathbf{d} = \begin{pmatrix}\frac{3}{7}k\\\beta k\\\gamma k\end{pmatrix}\) or use of \(\left(\frac{3}{7}\right)^2+\beta^2+\gamma^2=1\) | B1 | |
| \(\begin{pmatrix}\frac{3}{7}k-12\\\beta k-16\\\gamma k+8\end{pmatrix}\times\begin{pmatrix}9\\6\\2\end{pmatrix}=\mathbf{0}\Rightarrow \begin{cases}6\left(\frac{3}{7}k-12\right)-9(\beta k-16)=0\\2\left(\frac{3}{7}k-12\right)-9(\gamma k+8)=0\\2(\beta k-16)-6(\gamma k+8)=0\end{cases}\) | M1 | |
| \(\Rightarrow \beta k = \frac{2}{7}k+8\) and \(\gamma k = \frac{1}{3}\left(\frac{2}{7}k-32\right)\) \(\Rightarrow \frac{9k^2}{49}+\left(\frac{2}{7}k+8\right)^2+\frac{1}{9}\left(\frac{2}{7}k-32\right)^2=k^2\Rightarrow 2k^2-7k-490=0\) | M1, A1 | |
| \(\Rightarrow (2k-35)(k+14)=0\Rightarrow k=...\) | M1 |
## Question 7:
### Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Position of $A$ is given by $\overrightarrow{OA} = \begin{pmatrix}12+9\lambda\\16+6\lambda\\-8+2\lambda\end{pmatrix}$ | B1 | |
| $\frac{12+9\lambda}{\sqrt{(12+9\lambda)^2+(16+6\lambda)^2+(-8+2\lambda)^2}} = \frac{3}{7}$ | M1 | |
| $\Rightarrow 49(3(4+3\lambda))^2 = 9((12+9\lambda)^2+(16+6\lambda)^2+(-8+2\lambda)^2)$ $\Rightarrow 2880\lambda^2+7200\lambda+2880=0$ or $2\lambda^2+5\lambda+2=0$ | M1, A1 | |
| $\Rightarrow (2\lambda+1)(\lambda+2)=0 \Rightarrow \lambda = ...$ | M1 | |
| Substitutes a value of $\lambda$ to find position of $A$, e.g. $\overrightarrow{OA} = \begin{pmatrix}12+9(-\frac{1}{2})\\16+6(-\frac{1}{2})\\-8+2(-\frac{1}{2})\end{pmatrix}$ | M1 | |
| Coordinates of $A$ are $\left(\frac{15}{2}, 13, -9\right)$ only | A1 | |
### Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Direction of $\overrightarrow{OA}$ is given by $\mathbf{d} = \begin{pmatrix}\frac{3}{7}k\\\beta k\\\gamma k\end{pmatrix}$ or use of $\left(\frac{3}{7}\right)^2+\beta^2+\gamma^2=1$ | B1 | |
| $\begin{pmatrix}\frac{3}{7}k-12\\\beta k-16\\\gamma k+8\end{pmatrix}\times\begin{pmatrix}9\\6\\2\end{pmatrix}=\mathbf{0}\Rightarrow \begin{cases}6\left(\frac{3}{7}k-12\right)-9(\beta k-16)=0\\2\left(\frac{3}{7}k-12\right)-9(\gamma k+8)=0\\2(\beta k-16)-6(\gamma k+8)=0\end{cases}$ | M1 | |
| $\Rightarrow \beta k = \frac{2}{7}k+8$ and $\gamma k = \frac{1}{3}\left(\frac{2}{7}k-32\right)$ $\Rightarrow \frac{9k^2}{49}+\left(\frac{2}{7}k+8\right)^2+\frac{1}{9}\left(\frac{2}{7}k-32\right)^2=k^2\Rightarrow 2k^2-7k-490=0$ | M1, A1 | |
| $\Rightarrow (2k-35)(k+14)=0\Rightarrow k=...$ | M1 | |\begin{enumerate}
\item With respect to a fixed origin $O$, the line $l$ has equation
\end{enumerate}
$$( \mathbf { r } - ( 12 \mathbf { i } + 16 \mathbf { j } - 8 \mathbf { k } ) ) \times ( 9 \mathbf { i } + 6 \mathbf { j } + 2 \mathbf { k } ) = \mathbf { 0 }$$
The point $A$ lies on $l$ such that the direction cosines of $\overrightarrow { O A }$ with respect to the $\mathbf { i } , \mathbf { j }$ and $\mathbf { k }$ axes are $\frac { 3 } { 7 } , \beta$ and $\gamma$.
Determine the coordinates of the point $A$.
\hfill \mbox{\textit{Edexcel FP1 2021 Q7 [7]}}