| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Shortest distance between two skew lines |
| Difficulty | Standard +0.8 This is a standard Further Maths FP3 question on skew lines requiring cross product for the common perpendicular and the scalar triple product formula for shortest distance. While it involves multiple steps and vector manipulation, it's a textbook application of well-defined techniques without requiring novel insight—moderately above average difficulty due to being Further Maths content with computational complexity. |
| Spec | 4.04g Vector product: a x b perpendicular vector4.04i Shortest distance: between a point and a line |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & 3 & 2 \\ -4 & 6 & 1 \end{vmatrix} = -9\mathbf{i} - 12\mathbf{j} + 36\mathbf{k}\) | M1 | Correct attempt at vector product between \(4\mathbf{i}+3\mathbf{j}+2\mathbf{k}\) and \(-4\mathbf{i}+6\mathbf{j}+\mathbf{k}\) (if unclear, 2 components must be correct), allowing for sign error in \(y\) component |
| A1 | Any multiple of \((3\mathbf{i}+4\mathbf{j}-12\mathbf{k})\) | |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{a}_1 - \mathbf{a}_2 = \pm(2\mathbf{i}+8\mathbf{j}+\mathbf{k})\) | M1 | Attempt to subtract position vectors |
| A1 | Correct vector \(\pm(2\mathbf{i}+8\mathbf{j}+\mathbf{k})\) (allow as coordinates) | |
| \(p = \frac{\begin{pmatrix}2\\8\\1\end{pmatrix}\cdot\begin{pmatrix}-9\\-12\\36\end{pmatrix}}{\sqrt{9^2+12^2+36^2}}\) | M1 | Correct formula for distance using their vectors: \(\frac{\pm(2\mathbf{i}+8\mathbf{j}+\mathbf{k})\cdot\mathbf{"n"}}{\ |
| \(p = \frac{\pm78}{\sqrt{1521}} = \frac{\pm78}{39} = 2\) | dM1 | Correctly forms scalar product in numerator and Pythagoras in denominator. (Dependent on previous method mark) |
| A1 | \(2\) (not \(-2\)) | |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((\mathbf{i}-\mathbf{j}+\mathbf{k})\cdot(3\mathbf{i}+4\mathbf{j}-12\mathbf{k}) = -13\ (d_1)\) | M1 | Attempt scalar product between their \(\mathbf{n}\) and either position vector |
| \((3\mathbf{i}+7\mathbf{j}+2\mathbf{k})\cdot(3\mathbf{i}+4\mathbf{j}-12\mathbf{k}) = 13\ (d_2)\) | A1 | Both scalar products correct |
| \(\frac{\pm13}{\sqrt{3^2+4^2+12^2}}\ (=1)\) | M1 | Divides either scalar product by magnitude of normal vector: \(\frac{d_1 \text{ or } d_2}{\ |
| \(p = \frac{d_1}{\ | \mathbf{"n"}\ | } - \frac{d_2}{\ |
| A1 | \(2\) (not \(-2\)) | |
| (5) | ||
| Total 7 |
# Question 2:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & 3 & 2 \\ -4 & 6 & 1 \end{vmatrix} = -9\mathbf{i} - 12\mathbf{j} + 36\mathbf{k}$ | M1 | Correct attempt at vector product between $4\mathbf{i}+3\mathbf{j}+2\mathbf{k}$ and $-4\mathbf{i}+6\mathbf{j}+\mathbf{k}$ (if unclear, 2 components must be correct), allowing for sign error in $y$ component |
| | A1 | Any multiple of $(3\mathbf{i}+4\mathbf{j}-12\mathbf{k})$ |
| | **(2)** | |
## Part (b) Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{a}_1 - \mathbf{a}_2 = \pm(2\mathbf{i}+8\mathbf{j}+\mathbf{k})$ | M1 | Attempt to subtract position vectors |
| | A1 | Correct vector $\pm(2\mathbf{i}+8\mathbf{j}+\mathbf{k})$ (allow as coordinates) |
| $p = \frac{\begin{pmatrix}2\\8\\1\end{pmatrix}\cdot\begin{pmatrix}-9\\-12\\36\end{pmatrix}}{\sqrt{9^2+12^2+36^2}}$ | M1 | Correct formula for distance using their vectors: $\frac{\pm(2\mathbf{i}+8\mathbf{j}+\mathbf{k})\cdot\mathbf{"n"}}{\|\mathbf{"n"}\|}$ |
| $p = \frac{\pm78}{\sqrt{1521}} = \frac{\pm78}{39} = 2$ | dM1 | Correctly forms scalar product in numerator **and** Pythagoras in denominator. **(Dependent on previous method mark)** |
| | A1 | $2$ (**not** $-2$) |
| | **(5)** | |
## Part (b) Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(\mathbf{i}-\mathbf{j}+\mathbf{k})\cdot(3\mathbf{i}+4\mathbf{j}-12\mathbf{k}) = -13\ (d_1)$ | M1 | Attempt scalar product between their $\mathbf{n}$ and either position vector |
| $(3\mathbf{i}+7\mathbf{j}+2\mathbf{k})\cdot(3\mathbf{i}+4\mathbf{j}-12\mathbf{k}) = 13\ (d_2)$ | A1 | Both scalar products correct |
| $\frac{\pm13}{\sqrt{3^2+4^2+12^2}}\ (=1)$ | M1 | Divides either scalar product by magnitude of normal vector: $\frac{d_1 \text{ or } d_2}{\|\mathbf{"n"}\|}$ |
| $p = \frac{d_1}{\|\mathbf{"n"}\|} - \frac{d_2}{\|\mathbf{"n"}\|}$ or $2\times\frac{d_1}{\|\mathbf{"n"}\|}$ | dM1 | Correct attempt to find required distance, subtracts $\frac{d_1}{\|\mathbf{n}\|}$ and $\frac{d_2}{\|\mathbf{n}\|}$ or doubles $\frac{d_1}{\|\mathbf{n}\|}$ if $|d_1|=|d_2|$. **(Dependent on previous method mark)** |
| | A1 | $2$ (**not** $-2$) |
| | **(5)** | |
| | **Total 7** | |
---2. Two skew lines $l _ { 1 }$ and $l _ { 2 }$ have equations
$$\begin{aligned}
& l _ { 1 } : \mathbf { r } = ( \mathbf { i } - \mathbf { j } + \mathbf { k } ) + \lambda ( 4 \mathbf { i } + 3 \mathbf { j } + 2 \mathbf { k } ) \\
& l _ { 2 } : \mathbf { r } = ( 3 \mathbf { i } + 7 \mathbf { j } + 2 \mathbf { k } ) + \mu ( - 4 \mathbf { i } + 6 \mathbf { j } + \mathbf { k } )
\end{aligned}$$
respectively, where $\lambda$ and $\mu$ are real parameters.
\begin{enumerate}[label=(\alph*)]
\item Find a vector in the direction of the common perpendicular to $l _ { 1 }$ and $l _ { 2 }$
\item Find the shortest distance between these two lines.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP3 2013 Q2 [7]}}