| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2021 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Shortest distance between two skew lines |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths vectors question requiring the standard formula for distance between skew lines (involving cross product), plane equations, and angle calculations. While it involves several techniques and the parameter t adds mild algebraic complexity, each part follows well-established procedures taught in FM Pure. The cross product and distance formula are direct applications, making this moderately above average difficulty but not requiring novel insight. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04d Angles: between planes and between line and plane4.04h Shortest distances: between parallel lines and between skew lines4.04i Shortest distance: between a point and a line |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}0\\1\\t\end{pmatrix} - \begin{pmatrix}t\\1\\0\end{pmatrix} = t\begin{pmatrix}-1\\0\\1\end{pmatrix}\) | B1 | Finds vector from any point on \(l_1\) to any point on \(l_2\) |
| \(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\-2 & -1 & 0\\0 & -2 & 1\end{vmatrix} = \begin{pmatrix}-1\\2\\4\end{pmatrix}\) | M1 A1 | Finds common perpendicular |
| \(\frac{t}{\sqrt{21}}\left\vert\begin{pmatrix}-1\\0\\1\end{pmatrix}\cdot\begin{pmatrix}-1\\2\\4\end{pmatrix}\right\vert = \sqrt{21} \Rightarrow t = \frac{21}{5} = 4.2\) | M1 A1 | Uses formula for shortest distance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{r} = \frac{21}{5}\mathbf{i}+\mathbf{j}+\lambda(2\mathbf{i}+\mathbf{j})+\mu(-2\mathbf{j}+\mathbf{k})\) | B1 FT | Using their value of \(t\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}0\\-2\\1\end{pmatrix}\cdot\begin{pmatrix}5\\-6\\7\end{pmatrix} = \sqrt{5}\sqrt{110}\cos\alpha\) leading to \(\cos\alpha = \dfrac{19}{\sqrt{5}\sqrt{110}}\) | M1 A1 | Uses dot product of \(-2\mathbf{j}+\mathbf{k}\) and normal |
| Acute angle between \(l_2\) and \(\Pi_2\) is \(90° - \alpha = 54.1°\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}-1\\2\\4\end{pmatrix} \cdot \begin{pmatrix}5\\-6\\7\end{pmatrix} = \sqrt{21}\sqrt{110}\cos\alpha\) leading to \(\cos\alpha = \dfrac{11}{\sqrt{21}\sqrt{110}}\) | M1, A1 FT | Dot product using their normal to \(\Pi_1\) |
| \(76.8°\) | A1 | |
| Total: 3 |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}0\\1\\t\end{pmatrix} - \begin{pmatrix}t\\1\\0\end{pmatrix} = t\begin{pmatrix}-1\\0\\1\end{pmatrix}$ | B1 | Finds vector from any point on $l_1$ to any point on $l_2$ |
| $\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\-2 & -1 & 0\\0 & -2 & 1\end{vmatrix} = \begin{pmatrix}-1\\2\\4\end{pmatrix}$ | M1 A1 | Finds common perpendicular |
| $\frac{t}{\sqrt{21}}\left\vert\begin{pmatrix}-1\\0\\1\end{pmatrix}\cdot\begin{pmatrix}-1\\2\\4\end{pmatrix}\right\vert = \sqrt{21} \Rightarrow t = \frac{21}{5} = 4.2$ | M1 A1 | Uses formula for shortest distance |
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{r} = \frac{21}{5}\mathbf{i}+\mathbf{j}+\lambda(2\mathbf{i}+\mathbf{j})+\mu(-2\mathbf{j}+\mathbf{k})$ | B1 FT | Using their value of $t$ |
## Question 6(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}0\\-2\\1\end{pmatrix}\cdot\begin{pmatrix}5\\-6\\7\end{pmatrix} = \sqrt{5}\sqrt{110}\cos\alpha$ leading to $\cos\alpha = \dfrac{19}{\sqrt{5}\sqrt{110}}$ | M1 A1 | Uses dot product of $-2\mathbf{j}+\mathbf{k}$ and normal |
| Acute angle between $l_2$ and $\Pi_2$ is $90° - \alpha = 54.1°$ | A1 | |
## Question 6(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}-1\\2\\4\end{pmatrix} \cdot \begin{pmatrix}5\\-6\\7\end{pmatrix} = \sqrt{21}\sqrt{110}\cos\alpha$ leading to $\cos\alpha = \dfrac{11}{\sqrt{21}\sqrt{110}}$ | M1, A1 FT | Dot product using their normal to $\Pi_1$ |
| $76.8°$ | A1 | |
| **Total: 3** | | |
---6 Let $t$ be a positive constant.\\
The line $l _ { 1 }$ passes through the point with position vector $t \mathbf { i } + \mathbf { j }$ and is parallel to the vector $- 2 \mathbf { i } - \mathbf { j }$. The line $l _ { 2 }$ passes through the point with position vector $\mathbf { j } + t \mathbf { k }$ and is parallel to the vector $- 2 \mathbf { j } + \mathbf { k }$.
It is given that the shortest distance between the lines $l _ { 1 }$ and $l _ { 2 }$ is $\sqrt { \mathbf { 2 1 } }$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $t$.\\
The plane $\Pi _ { 1 }$ contains $l _ { 1 }$ and is parallel to $l _ { 2 }$.
\item Write down an equation of $\Pi _ { 1 }$, giving your answer in the form $\mathbf { r } = \mathbf { a } + \lambda \mathbf { b } + \mu \mathbf { c }$.\\
The plane $\Pi _ { 2 }$ has Cartesian equation $5 x - 6 y + 7 z = 0$.
\item Find the acute angle between $l _ { 2 }$ and $\Pi _ { 2 }$.
\item Find the acute angle between $\Pi _ { 1 }$ and $\Pi _ { 2 }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2021 Q6 [12]}}