| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Line of intersection of planes |
| Difficulty | Standard +0.8 This is a Further Maths FP3 question requiring two standard techniques: finding the angle between planes using normal vectors and their dot product, then finding the line of intersection by taking the cross product of normals and finding a point on both planes. While methodical, it requires confident vector manipulation and multiple steps, placing it moderately above average difficulty. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04d Angles: between planes and between line and plane |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Normals are \(\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}\) and \(2\mathbf{i} + 2\mathbf{j} - \mathbf{k}\) | B1 | For identifying both normal vectors |
| Acute angle is \(\cos^{-1}\left(\frac{ | 2-4-2 | }{3 \times 3}\right) = 64°\) |
| M1 A1 | For completely correct process for the angle; For correct answer | |
| 4 | ||
| (ii) Direction of line is \((\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}) \times (2\mathbf{i} + 2\mathbf{j} - \mathbf{k})\), i.e. \(-2\mathbf{i} + 5\mathbf{j} + 6\mathbf{k}\) | M1 | For using vector product of normals |
| A1 | For correct vector for \(\mathbf{b}\) | |
| \(x - 2y + 2z = 1, 2x + 2y - z = 3 \Rightarrow 3x + z = 4\), so a common point is \((1, 1, 1)\), for example | M1 | For complete method to find a suitable \(\mathbf{a}\) |
| Hence line is \(\mathbf{r} = \mathbf{i} + \mathbf{j} + \mathbf{k} + t(-2\mathbf{i} + 5\mathbf{j} + 6\mathbf{k})\) | A1 | For correct equation of line (Other methods are possible) |
| 4 | ||
| 8 |
**(i)** Normals are $\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}$ and $2\mathbf{i} + 2\mathbf{j} - \mathbf{k}$ | B1 | For identifying both normal vectors
Acute angle is $\cos^{-1}\left(\frac{|2-4-2|}{3 \times 3}\right) = 64°$ | M1 | For using the scalar product of the normals
| M1 A1 | For completely correct process for the angle; For correct answer
| **4** |
**(ii)** Direction of line is $(\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}) \times (2\mathbf{i} + 2\mathbf{j} - \mathbf{k})$, i.e. $-2\mathbf{i} + 5\mathbf{j} + 6\mathbf{k}$ | M1 | For using vector product of normals
| A1 | For correct vector for $\mathbf{b}$
$x - 2y + 2z = 1, 2x + 2y - z = 3 \Rightarrow 3x + z = 4$, so a common point is $(1, 1, 1)$, for example | M1 | For complete method to find a suitable $\mathbf{a}$
Hence line is $\mathbf{r} = \mathbf{i} + \mathbf{j} + \mathbf{k} + t(-2\mathbf{i} + 5\mathbf{j} + 6\mathbf{k})$ | A1 | For correct equation of line (Other methods are possible)
| **4** |
| **8** |3 The planes $\Pi _ { 1 }$ and $\Pi _ { 2 }$ have equations $\mathbf { r } \cdot ( \mathbf { i } - 2 \mathbf { j } + 2 \mathbf { k } ) = 1$ and $\mathbf { r } \cdot ( 2 \mathbf { i } + 2 \mathbf { j } - \mathbf { k } ) = 3$ respectively. Find\\
(i) the acute angle between $\Pi _ { 1 }$ and $\Pi _ { 2 }$, correct to the nearest degree,\\
(ii) the equation of the line of intersection of $\Pi _ { 1 }$ and $\Pi _ { 2 }$, in the form $\mathbf { r } = \mathbf { a } + t \mathbf { b }$.
\hfill \mbox{\textit{OCR FP3 Q3 [8]}}