| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 2 (Further Pure Core 2) |
| Year | 2018 |
| Session | September |
| Marks | 8 |
| Topic | Vectors: Lines & Planes |
| Type | Line intersection with line |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question testing standard 3D vector line techniques: converting between Cartesian and vector forms, finding intersection points by equating components, and using the scalar product formula for angles. All three parts follow routine procedures with no conceptual challenges or novel problem-solving required, making it slightly easier than average even for Further Maths content. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{x(+0)}{-2} = \frac{y-5}{2} = \frac{z+(-6)}{-7}\) seen or implied by \(a\) or \(d\) correct | M1 | Expressing in form from which \(a\) and \(d\) from \(r = a + \lambda d\) can be read off. Or writing \(\frac{-x}{2} = \lambda\) etc and attempting to solve for \(x, y\) and \(z\) |
| \(r = \begin{pmatrix} 0 \\ 5 \\ -6 \end{pmatrix} + \lambda \begin{pmatrix} -2 \\ 2 \\ -7 \end{pmatrix}\) or equivalent | A1 | Must be \(r = \). Must have sensible parameter name (not eg \(x, y, z\) or \(r\)). |
| Answer | Marks |
|---|---|
| Any two of \(-2\lambda = 2 + \mu\), \(5 + 2\lambda = 7 - 2\mu\) and \(-6 - 7\lambda = -1 + 4\mu\) and attempt to solve \(\lambda = -3\) or \(\mu = 4\) | M1 A1 A1 |
| (6, -1, 15) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{pmatrix} -2 \\ 2 \\ -7 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -2 \\ 4 \end{pmatrix} = -34\) | M1* | Correct method for finding (modulus of) dot product of \(\begin{pmatrix} 1 \\ -2 \\ 4 \end{pmatrix}\) and their direction vector |
| \(\cos^{-1} \frac{\pm 34}{\sqrt{57}\sqrt{21}}\) | dep*M1 | Correct process for finding moduli and correct use (ie inverse cos of \(\pm\) their dot product divided by the product of their moduli). |
| awrt 10.7° or 0.186 rads www | A1 | Acute angle only |
## (i)
$\frac{x(+0)}{-2} = \frac{y-5}{2} = \frac{z+(-6)}{-7}$ seen or implied by $a$ or $d$ correct | M1 | Expressing in form from which $a$ and $d$ from $r = a + \lambda d$ can be read off. Or writing $\frac{-x}{2} = \lambda$ etc and attempting to solve for $x, y$ and $z$
$r = \begin{pmatrix} 0 \\ 5 \\ -6 \end{pmatrix} + \lambda \begin{pmatrix} -2 \\ 2 \\ -7 \end{pmatrix}$ or equivalent | A1 | Must be $r = $. Must have sensible parameter name (not eg $x, y, z$ or $r$).
[2]
## (ii)
Any two of $-2\lambda = 2 + \mu$, $5 + 2\lambda = 7 - 2\mu$ and $-6 - 7\lambda = -1 + 4\mu$ and attempt to solve $\lambda = -3$ or $\mu = 4$ | M1 A1 A1 |
(6, -1, 15) | A1 |
[3]
## (iii)
$\begin{pmatrix} -2 \\ 2 \\ -7 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -2 \\ 4 \end{pmatrix} = -34$ | M1* | Correct method for finding (modulus of) dot product of $\begin{pmatrix} 1 \\ -2 \\ 4 \end{pmatrix}$ and their direction vector
$\cos^{-1} \frac{\pm 34}{\sqrt{57}\sqrt{21}}$ | dep*M1 | Correct process for finding moduli and correct use (ie inverse cos of $\pm$ their dot product divided by the product of their moduli).
awrt 10.7° or 0.186 rads www | A1 | Acute angle only
[3]
---Line $l_1$ has Cartesian equation
$$l_1: \frac{-x}{2} = \frac{y-5}{2} = \frac{-z-6}{7}.$$
\begin{enumerate}[label=(\roman*)]
\item Find a vector equation for $l_1$. [2]
\end{enumerate}
Line $l_2$ has vector equation
$$l_2: \mathbf{r} = \begin{pmatrix} 2 \\ 7 \\ -1 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ -2 \\ 4 \end{pmatrix}.$$
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the point of intersection of $l_1$ and $l_2$. [3]
\item Find the acute angle between $l_1$ and $l_2$. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 2 2018 Q1 [8]}}