| Exam Board | CAIE |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Line of intersection of planes |
| Difficulty | Challenging +1.2 This is a standard Further Maths vectors question covering routine techniques: finding Cartesian form from vector form (cross product), finding line of intersection (solving simultaneous equations), using distance formula, and angle between line and plane. All parts follow textbook methods with no novel insight required, though the multi-part nature and Further Maths context place it slightly above average difficulty. |
| 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 plane4.04j Shortest distance: between a point and a plane |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} = \begin{pmatrix} 0 \\ 2 \\ 2 \end{pmatrix} \sim \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\) | M1 A1 | Finds common perpendicular. |
| \((-4) + (-3) = -7 \Rightarrow y + z = -7\) | M1 A1 | Substitutes point. |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| States point common to both planes e.g. \(\begin{pmatrix} -7 \\ -2 \\ -5 \end{pmatrix}\) | B1FT | |
| \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 1 & 1 \\ -5 & 3 & 5 \end{vmatrix} = \begin{pmatrix} 2 \\ -5 \\ 5 \end{pmatrix}\) | M1 A1FT | Finds direction of line. |
| \(\mathbf{r} = \begin{pmatrix} -7 \\ -2 \\ -5 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -5 \\ 5 \end{pmatrix}\) | A1 | OE. |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left\lvert \frac{a + a - 7 + 7}{\sqrt{2}} \right\rvert = \sqrt{2}\) | M1 | Uses correct formula for distance from \(A\) to \(\Pi_1\). |
| \(a = 1\) | A1 | |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left\lvert \frac{b+b}{\sqrt{2}\sqrt{(1-b)^2 + 2b^2}} \right\rvert = \frac{1}{2}\sqrt{2}\) | M1 A1 | Uses correct formula. |
| \(2b = \sqrt{(1-b)^2 + 2b^2} \Rightarrow b^2 + 2b - 1 = 0\) | M1 | Solves for \(b\). |
| \(b = -1 + \sqrt{2}\) | A1 | CAO |
| 4 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} = \begin{pmatrix} 0 \\ 2 \\ 2 \end{pmatrix} \sim \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}$ | M1 A1 | Finds common perpendicular. |
| $(-4) + (-3) = -7 \Rightarrow y + z = -7$ | M1 A1 | Substitutes point. |
| | **4** | |
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| States point common to both planes e.g. $\begin{pmatrix} -7 \\ -2 \\ -5 \end{pmatrix}$ | B1FT | |
| $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 1 & 1 \\ -5 & 3 & 5 \end{vmatrix} = \begin{pmatrix} 2 \\ -5 \\ 5 \end{pmatrix}$ | M1 A1FT | Finds direction of line. |
| $\mathbf{r} = \begin{pmatrix} -7 \\ -2 \\ -5 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -5 \\ 5 \end{pmatrix}$ | A1 | OE. |
| | **4** | |
## Question 7(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left\lvert \frac{a + a - 7 + 7}{\sqrt{2}} \right\rvert = \sqrt{2}$ | M1 | Uses correct formula for distance from $A$ to $\Pi_1$. |
| $a = 1$ | A1 | |
| | **2** | |
## Question 7(d):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left\lvert \frac{b+b}{\sqrt{2}\sqrt{(1-b)^2 + 2b^2}} \right\rvert = \frac{1}{2}\sqrt{2}$ | M1 A1 | Uses correct formula. |
| $2b = \sqrt{(1-b)^2 + 2b^2} \Rightarrow b^2 + 2b - 1 = 0$ | M1 | Solves for $b$. |
| $b = -1 + \sqrt{2}$ | A1 | CAO |
| | **4** | |7 The plane $\Pi _ { 1 }$ has equation $r = - 4 \mathbf { j } - 3 \mathbf { k } + \lambda ( \mathbf { i } - \mathbf { j } + \mathbf { k } ) + \mu ( \mathbf { i } + \mathbf { j } - \mathbf { k } )$.
\begin{enumerate}[label=(\alph*)]
\item Obtain an equation of $\Pi _ { 1 }$ in the form $\mathrm { px } + \mathrm { qy } + \mathrm { rz } = \mathrm { d }$.
\item The plane $\Pi _ { 2 }$ has equation $\mathbf { r } . ( - 5 \mathbf { i } + 3 \mathbf { j } + 5 \mathbf { k } ) = 4$.
Find a vector equation of the line of intersection of $\Pi _ { 1 }$ and $\Pi _ { 2 }$.\\
The line $l$ passes through the point $A$ with position vector $a \mathbf { i } + a \mathbf { j } + ( a - 7 ) \mathbf { k }$ and is parallel to $( 1 - b ) \mathbf { i } + b \mathbf { j } + b \mathbf { k }$, where $a$ and $b$ are positive constants.
\item Given that the perpendicular distance from $A$ to $\Pi _ { 1 }$ is $\sqrt { 2 }$, find the value of $a$.
\item Given that the obtuse angle between $l$ and $\Pi _ { 1 }$ is $\frac { 3 } { 4 } \pi$, find the exact value of $b$.\\
If you use the following page to complete the answer to any question, the question number must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 1 2023 Q7 [14]}}