| Exam Board | OCR |
|---|---|
| Module | Further Pure Core AS (Further Pure Core AS) |
| Year | 2021 |
| Session | November |
| Marks | 5 |
| 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 vector line intersection problem requiring equating components and solving simultaneous equations. While it involves more algebraic manipulation than typical A-level questions and is from Further Maths content, the method is routine and well-practiced. The 'show that' part guides students to the approach, making it slightly easier than average Further Maths questions but harder than standard A-level. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04e Line intersections: parallel, skew, or intersecting |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(8 - 2\lambda = -6 - 3\mu\) and \(-11 + 5\lambda = 11 + \mu\) | B1 | Forming 2 correct equations in \(\lambda\) and \(\mu\). Could be \(-2 + 3\lambda = 8 - \mu\) |
| Attempt to solve, e.g. scaling one equation and adding, or rewriting to standard form. Must reach equation with only one unknown. e.g. \(-25 + 13\lambda = 27\) | M1 | e.g. \(-2\lambda + 3\mu = -14\); \(5\lambda - \mu = 22\); \(3\lambda + \mu = 10\) |
| \(\lambda = 4,\ \mu = -2\) | A1 | Both |
| \(-2 + 3\times4 = 10\) and \(8 - -2 = 10\) so they do intersect | A1 | Checking consistency in 3rd equation. Both sides must be evaluated. e.g. \(8-2\times4=-6-3\times-2 \Rightarrow 0=0\). Note: \(-2+3\times4=8--2\) alone insufficient; need both sides becoming 10 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((0, 9, 10)\) | B1 | Allow as vector |
## Question 1:
### Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $8 - 2\lambda = -6 - 3\mu$ **and** $-11 + 5\lambda = 11 + \mu$ | B1 | Forming 2 correct equations in $\lambda$ and $\mu$. Could be $-2 + 3\lambda = 8 - \mu$ |
| Attempt to solve, e.g. scaling one equation and adding, or rewriting to standard form. Must reach equation with only one unknown. e.g. $-25 + 13\lambda = 27$ | M1 | e.g. $-2\lambda + 3\mu = -14$; $5\lambda - \mu = 22$; $3\lambda + \mu = 10$ |
| $\lambda = 4,\ \mu = -2$ | A1 | Both |
| $-2 + 3\times4 = 10$ and $8 - -2 = 10$ so they do intersect | A1 | Checking consistency in 3rd equation. Both sides must be evaluated. e.g. $8-2\times4=-6-3\times-2 \Rightarrow 0=0$. Note: $-2+3\times4=8--2$ alone insufficient; need both sides becoming 10 |
### Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(0, 9, 10)$ | B1 | Allow as vector |
---1 The lines $l _ { 1 }$ and $l _ { 2 }$ have the following equations.
$$\begin{aligned}
& l _ { 1 } : \mathbf { r } = \left( \begin{array} { r }
8 \\
- 11 \\
- 2
\end{array} \right) + \lambda \left( \begin{array} { r }
- 2 \\
5 \\
3
\end{array} \right) \\
& l _ { 2 } : \mathbf { r } = \left( \begin{array} { r }
- 6 \\
11 \\
8
\end{array} \right) + \mu \left( \begin{array} { r }
- 3 \\
1 \\
- 1
\end{array} \right)
\end{aligned}$$
\begin{enumerate}[label=(\alph*)]
\item Show that $l _ { 1 }$ and $l _ { 2 }$ intersect.
\item Write down the point of intersection of $l _ { 1 }$ and $l _ { 2 }$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core AS 2021 Q1 [5]}}