| Exam Board | OCR |
|---|---|
| Module | Further Pure Core AS (Further Pure Core AS) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Geometric loci and constraints |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths question requiring solid understanding of 3D vector geometry including perpendicularity, cross products, and distance calculations. While it involves several steps and techniques (shortest distance to a line, cross product, vector equations, angle calculation), each part follows standard methods without requiring novel insight. The geometric setup is moderately complex but the solution path is clear once the concepts are understood. Slightly above average difficulty due to the Further Maths context and multi-step nature, but not exceptionally challenging. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04g Vector product: a x b perpendicular vector4.04h Shortest distances: between parallel lines and between skew lines4.04i Shortest distance: between a point and a line |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Shortest distance is the length of the perpendicular from \(O\) to \(l\), so length \(OA\) | B1 | 2.2a — Can be implied by attempting to find length \(OA\) |
| \(\sqrt{\mathbf{a \cdot a}} = \sqrt{(-9)^2 + 2^2 + 6^2} = 11\) | B1 [2] | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{a} \times \mathbf{b} = (-9\mathbf{i} + 2\mathbf{j} + 6\mathbf{k}) \times (-2\mathbf{i} + \mathbf{j} + \mathbf{k}) = -4\mathbf{i} - 3\mathbf{j} - 5\mathbf{k}\) | B1 [1] | 1.1 — BC, or any non-zero multiple. Allow column vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| e.g. \(\mathbf{r} = (-9\mathbf{i} + 2\mathbf{j} + 6\mathbf{k}) + \lambda(4\mathbf{i} + 3\mathbf{j} + 5\mathbf{k})\) | B1ft [1] | 1.1 — Must be \(\mathbf{r} = \ldots\) or \(x\mathbf{i} + y\mathbf{j} + z\mathbf{k} = \ldots\) (must be an equation). Allow column vectors, allow any equivalent equation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(PAO\) is a right angled triangle | M1 | 3.1a — May be implied by diagram or attempt at trig ratio which implies right angle at \(A\) |
| \(\tan\theta = 2\) | A1 | 1.1 — Correct trig equation satisfied by \(\theta\). SC: If trying to find vector \(\overrightarrow{OP}\), B1 for using \(\overrightarrow{AP} = \lambda\begin{pmatrix}4\\3\\5\end{pmatrix}\) and finding \(\lambda = \frac{11\sqrt{2}}{5}\) oe |
| \(1.11\) rads or \(63.4°\) | A1 [3] | 1.1 — Correct answer with no working is full credit. B1 for correct use of cosine rule to find correct angle. B1 for answer of \(1.11\) (\(1.107\ldots\)) or \(63.4°\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Need \(9\mathbf{i} + 8\mathbf{j} - 12\mathbf{k} = p\mathbf{a} + \mu\mathbf{b}\) | M1 | 2.2a — Could instead consider \(p\mathbf{a} = 9\mathbf{i} + 8\mathbf{j} - 12\mathbf{k} + \mu\mathbf{b}\), this would give \(\mu = -18\) |
| (any two of) \(-9p - 2\mu = 9\), \(2p + \mu = 8\), \(6p + \mu = -12\) | M1 | 1.1 — Equating coefficients for two of \(\mathbf{i}\), \(\mathbf{j}\) and \(\mathbf{k}\). Only 2 equations necessary, \(\mu\) might be negative. No need to check consistency |
| \(p = -5\) | A1 [3] | 1.1 — BC or eliminating \(\mu\) (\(=18\)) |
# Question 3:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Shortest distance is the length of the perpendicular from $O$ to $l$, so length $OA$ | B1 | 2.2a — Can be implied by attempting to find length $OA$ |
| $\sqrt{\mathbf{a \cdot a}} = \sqrt{(-9)^2 + 2^2 + 6^2} = 11$ | B1 [2] | 1.1 |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{a} \times \mathbf{b} = (-9\mathbf{i} + 2\mathbf{j} + 6\mathbf{k}) \times (-2\mathbf{i} + \mathbf{j} + \mathbf{k}) = -4\mathbf{i} - 3\mathbf{j} - 5\mathbf{k}$ | B1 [1] | 1.1 — BC, or any non-zero multiple. Allow column vectors |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| e.g. $\mathbf{r} = (-9\mathbf{i} + 2\mathbf{j} + 6\mathbf{k}) + \lambda(4\mathbf{i} + 3\mathbf{j} + 5\mathbf{k})$ | B1ft [1] | 1.1 — Must be $\mathbf{r} = \ldots$ or $x\mathbf{i} + y\mathbf{j} + z\mathbf{k} = \ldots$ (must be an equation). Allow column vectors, allow any equivalent equation |
## Part (d)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $PAO$ is a right angled triangle | M1 | 3.1a — May be implied by diagram or attempt at trig ratio which implies right angle at $A$ |
| $\tan\theta = 2$ | A1 | 1.1 — Correct trig equation satisfied by $\theta$. SC: If trying to find vector $\overrightarrow{OP}$, B1 for using $\overrightarrow{AP} = \lambda\begin{pmatrix}4\\3\\5\end{pmatrix}$ and finding $\lambda = \frac{11\sqrt{2}}{5}$ oe |
| $1.11$ rads or $63.4°$ | A1 [3] | 1.1 — Correct answer with no working is full credit. B1 for correct use of cosine rule to find correct angle. B1 for answer of $1.11$ ($1.107\ldots$) or $63.4°$ |
## Part (e)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Need $9\mathbf{i} + 8\mathbf{j} - 12\mathbf{k} = p\mathbf{a} + \mu\mathbf{b}$ | M1 | 2.2a — Could instead consider $p\mathbf{a} = 9\mathbf{i} + 8\mathbf{j} - 12\mathbf{k} + \mu\mathbf{b}$, this would give $\mu = -18$ |
| (any two of) $-9p - 2\mu = 9$, $2p + \mu = 8$, $6p + \mu = -12$ | M1 | 1.1 — Equating coefficients for two of $\mathbf{i}$, $\mathbf{j}$ and $\mathbf{k}$. Only 2 equations necessary, $\mu$ might be negative. No need to check consistency |
| $p = -5$ | A1 [3] | 1.1 — BC or eliminating $\mu$ ($=18$) |
---3 The position vector of point $A$ is $\mathbf { a } = - 9 \mathbf { i } + 2 \mathbf { j } + 6 \mathbf { k }$.\\
The line $l$ passes through $A$ and is perpendicular to $\mathbf { a }$.
\begin{enumerate}[label=(\alph*)]
\item Determine the shortest distance between the origin, $O$, and $l$.\\
$l$ is also perpendicular to the vector $\mathbf { b }$ where $\mathbf { b } = - 2 \mathbf { i } + \mathbf { j } + \mathbf { k }$.
\item Find a vector which is perpendicular to both $\mathbf { a }$ and $\mathbf { b }$.
\item Write down an equation of $l$ in vector form.\\
$P$ is a point on $l$ such that $P A = 2 O A$.
\item Find angle $P O A$ giving your answer to 3 significant figures.\\
$C$ is a point whose position vector, $\mathbf { c }$, is given by $\mathbf { c } = p \mathbf { a }$ for some constant $p$. The line $m$ passes through $C$ and has equation $\mathbf { r } = \mathbf { c } + \mu \mathbf { b }$. The point with position vector $9 \mathbf { i } + 8 \mathbf { j } - 12 \mathbf { k }$ lies on $m$.
\item Find the value of $p$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core AS 2019 Q3 [10]}}