| Exam Board | OCR |
|---|---|
| Module | Further Pure Core AS (Further Pure Core AS) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Show lines intersect and find intersection point |
| Difficulty | Standard +0.3 This is a standard Further Maths question on 3D vector lines. Part (a) requires equating components and solving simultaneous equations to find intersection—routine for FM students. Part (b) requires finding a direction vector perpendicular to both lines (cross product) and using the intersection point, which is a textbook application. While more involved than basic A-level questions, it follows standard procedures without requiring novel insight. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04e Line intersections: parallel, skew, or intersecting |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(-5 + 5\lambda = 24 + 3\mu\) & \(6 - 2\lambda = 1 + \mu\) | B1 | Forming 2 correct equations in \(\lambda\) and \(\mu\). Third equation is \(15 - 2\lambda = -5 - 4\mu\) |
| \(5\lambda - 3\mu = 29\) & \(6\lambda + 3\mu = 15\) | M1 | Attempt to solve (e.g. scaling one equation and adding; or rewriting to a standard form for solution BC). If scaling or substituting method must result in correctly eliminating one variable (other coefficients may be incorrect). |
| \(\lambda = 4\) & \(\mu = -3\) | A1 | Both |
| \(\lambda = 4\) & \(\mu = -3 \Rightarrow\) LHS \(= 15 - 2\times4 = 7\) and RHS \(= -5 - 4\times{-3} = 7 =\) LHS so all 3 equations are satisfied so \(L_1\) and \(L_2\) do intersect | A1 | Convincing justification but could be by finding the same r from both equations |
| \(\begin{pmatrix}15\\-2\\7\end{pmatrix}\) | A1 | Condone coordinates. Can be awarded even if previous A mark not awarded (i.e. if not checked third equation) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\begin{pmatrix}5\\-2\\-2\end{pmatrix} \times \begin{pmatrix}3\\1\\-4\end{pmatrix} = \begin{pmatrix}10\\14\\11\end{pmatrix}\) | B1 | Could be BC. SOI |
| \(\mathbf{r} = \begin{pmatrix}15\\-2\\7\end{pmatrix} + \nu\begin{pmatrix}10\\14\\11\end{pmatrix}\) | B1FT | FT their point of intersection from (a) and their attempt at direction vector. SOI. Condone use of \(\lambda\) or \(\mu\). No need to see "\(r =\)". Must be a recognisable attempt at a vector perpendicular to both \(L_1\) and \(L_2\). |
| \(\dfrac{x-15}{10} = \dfrac{y+2}{14} = \dfrac{z-7}{11}\) | B1FT | FT their vector equation. Correct equation here implies the other two marks. |
## Question 2(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $-5 + 5\lambda = 24 + 3\mu$ & $6 - 2\lambda = 1 + \mu$ | **B1** | Forming 2 correct equations in $\lambda$ and $\mu$. Third equation is $15 - 2\lambda = -5 - 4\mu$ |
| $5\lambda - 3\mu = 29$ & $6\lambda + 3\mu = 15$ | **M1** | Attempt to solve (e.g. scaling one equation and adding; or rewriting to a standard form for solution BC). If scaling or substituting method must result in correctly eliminating one variable (other coefficients may be incorrect). |
| $\lambda = 4$ & $\mu = -3$ | **A1** | Both |
| $\lambda = 4$ & $\mu = -3 \Rightarrow$ LHS $= 15 - 2\times4 = 7$ and RHS $= -5 - 4\times{-3} = 7 =$ LHS so all 3 equations are satisfied so $L_1$ and $L_2$ do intersect | **A1** | Convincing justification but could be by finding the same **r** from both equations |
| $\begin{pmatrix}15\\-2\\7\end{pmatrix}$ | **A1** | Condone coordinates. Can be awarded even if previous A mark not awarded (i.e. if not checked third equation) |
**[5]**
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## Question 2(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\begin{pmatrix}5\\-2\\-2\end{pmatrix} \times \begin{pmatrix}3\\1\\-4\end{pmatrix} = \begin{pmatrix}10\\14\\11\end{pmatrix}$ | **B1** | Could be BC. SOI |
| $\mathbf{r} = \begin{pmatrix}15\\-2\\7\end{pmatrix} + \nu\begin{pmatrix}10\\14\\11\end{pmatrix}$ | **B1FT** | FT their point of intersection from **(a)** and their attempt at direction vector. SOI. Condone use of $\lambda$ or $\mu$. No need to see "$r =$". Must be a recognisable attempt at a vector perpendicular to both $L_1$ and $L_2$. |
| $\dfrac{x-15}{10} = \dfrac{y+2}{14} = \dfrac{z-7}{11}$ | **B1FT** | FT their vector equation. Correct equation here implies the other two marks. |
**[3]**
---2 The lines $L _ { 1 }$ and $L _ { 2 }$ have the following equations.\\
$L _ { 1 } : \mathbf { r } = \left( \begin{array} { c } - 5 \\ 6 \\ 15 \end{array} \right) + \lambda \left( \begin{array} { c } 5 \\ - 2 \\ - 2 \end{array} \right)$\\
$L _ { 2 } : \mathbf { r } = \left( \begin{array} { c } 24 \\ 1 \\ - 5 \end{array} \right) + \mu \left( \begin{array} { c } 3 \\ 1 \\ - 4 \end{array} \right)$
\begin{enumerate}[label=(\alph*)]
\item Show that $L _ { 1 }$ and $L _ { 2 }$ intersect, giving the position vector of the point of intersection.
\item Find the equation of the line which intersects $L _ { 1 }$ and $L _ { 2 }$ and is perpendicular to both. Give your answer in cartesian form.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core AS 2023 Q2 [8]}}