| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2012 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Line intersection with plane |
| Difficulty | Standard +0.3 This is a standard Further Maths vectors question requiring routine techniques: finding a normal vector via cross product, converting plane forms, substituting a line equation into a plane, and finding a perpendicular vector. All parts are textbook exercises with no novel insight required, though it's slightly above average difficulty due to being Further Maths content and requiring multiple standard procedures. |
| Spec | 1.10c Magnitude and direction: of vectors4.04b Plane equations: cartesian and vector forms4.04f Line-plane intersection: find point |
| Answer | Marks | Guidance |
|---|---|---|
| METHOD 1 \(\mathbf{n} = [2, -1, -1] \times [2, -3, -5] = [2, 8, -4]\) | M1 | For finding vector product of 2 vectors in \(\Pi\) (or 2 scalar products = 0, with attempt to solve) |
| \(\mathbf{n} = k[1, 4, -2]\) | A1 | For correct \(\mathbf{n}\) |
| \(\Pi\) is \(r \cdot \mathbf{n} = [1, 6, 7] \cdot \mathbf{n}\) | M1 | For attempt to find equation of \(\Pi\), including cartesian equation |
| \(\Rightarrow r \cdot [1, 4, -2] = 11\) | A1 | For correct equation (allow multiples) |
| METHOD 2 \(y - z = -1 + 2u\) \(\mu = \frac{y - z + 1}{2}\) | M1 | for finding \(\lambda\) or \(\mu\) in terms of two from \(x,y,z\). |
| \(\lambda = 7 - z - 5 \frac{y - z + 1}{2}\) | M1 | For both \(\lambda\) & \(\mu\) |
| \(x = 11 + 2z - 4y\) | A1 | AEF |
| \(r(1, 4, -2) = 11\) | A1 | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \([7 + 3t, 4, 1 - t] \cdot \mathbf{n} = 11 \Rightarrow t = -2\) | M1 | For attempt to find \(t, \lambda\) and \(\mu\) by equating original equations) |
| \(\Rightarrow [1, 4, 3]\) | A1 | For correct position vector OR point |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| METHOD 1 \(\mathbf{c} = [1, 4, -2] \times [2, -1, -1]\) | M1 | For using given vector product (or 2 correct 'scalar products = 0') |
| M1 | For calculating given vector product (or 2 correct scalar products = 0, with attempt to solve) (or M1 for using vector product of \(\mathbf{c}\) with \(\mathbf{n}\) or \((2,-1,-1)\) in an equation, followed by M1 for calculating vector product and attempting to solve) | |
| \(\mathbf{c} = k[2, 1, 3]\) | A1 | For correct \(\mathbf{c}\) |
| [3] | ||
| METHOD 2 \(\mathbf{c} = [2, -3, -5] + s[2, -1, -1]\) | M1 | For \(\mathbf{c} =\) linear combination of \([2, -3, -5]\) and \([2, -1, -1]\) |
| \(\mathbf{c} \cdot [2, -1, -1] = 0 \Rightarrow 2(2 + 2s) - l(-3 - s) - l(-5 - s) = 0\) | M1 | For an equation in \(s\) from \(\mathbf{c} \cdot [2, -1, -1] = 0\) |
| \(\Rightarrow s = -2 \Rightarrow \mathbf{c} = k[2, 1, 3]\) | A1 | For correct \(\mathbf{c}\) |
**(i)**
| METHOD 1 $\mathbf{n} = [2, -1, -1] \times [2, -3, -5] = [2, 8, -4]$ | M1 | For finding vector product of 2 vectors in $\Pi$ (or 2 scalar products = 0, with attempt to solve) |
| $\mathbf{n} = k[1, 4, -2]$ | A1 | For correct $\mathbf{n}$ |
| $\Pi$ is $r \cdot \mathbf{n} = [1, 6, 7] \cdot \mathbf{n}$ | M1 | For attempt to find equation of $\Pi$, including cartesian equation |
| $\Rightarrow r \cdot [1, 4, -2] = 11$ | A1 | For correct equation (allow multiples) |
| METHOD 2 $y - z = -1 + 2u$ $\mu = \frac{y - z + 1}{2}$ | M1 | for finding $\lambda$ or $\mu$ in terms of two from $x,y,z$. |
| $\lambda = 7 - z - 5 \frac{y - z + 1}{2}$ | M1 | For both $\lambda$ & $\mu$ |
| $x = 11 + 2z - 4y$ | A1 | AEF |
| $r(1, 4, -2) = 11$ | A1 | |
| | [4] | |
**(ii)**
| $[7 + 3t, 4, 1 - t] \cdot \mathbf{n} = 11 \Rightarrow t = -2$ | M1 | For attempt to find $t, \lambda$ and $\mu$ by equating original equations) |
| $\Rightarrow [1, 4, 3]$ | A1 | For correct position vector OR point |
| | [2] | |
**(iii)**
| METHOD 1 $\mathbf{c} = [1, 4, -2] \times [2, -1, -1]$ | M1 | For using given vector product (or 2 correct 'scalar products = 0') |
| | M1 | For calculating given vector product (or 2 correct scalar products = 0, with attempt to solve) (or M1 for using vector product of $\mathbf{c}$ with $\mathbf{n}$ or $(2,-1,-1)$ in an equation, followed by M1 for calculating vector product and attempting to solve) |
| $\mathbf{c} = k[2, 1, 3]$ | A1 | For correct $\mathbf{c}$ |
| | [3] | |
| METHOD 2 $\mathbf{c} = [2, -3, -5] + s[2, -1, -1]$ | M1 | For $\mathbf{c} =$ linear combination of $[2, -3, -5]$ and $[2, -1, -1]$ |
| $\mathbf{c} \cdot [2, -1, -1] = 0 \Rightarrow 2(2 + 2s) - l(-3 - s) - l(-5 - s) = 0$ | M1 | For an equation in $s$ from $\mathbf{c} \cdot [2, -1, -1] = 0$ |
| $\Rightarrow s = -2 \Rightarrow \mathbf{c} = k[2, 1, 3]$ | A1 | For correct $\mathbf{c}$ |6 The plane $\Pi$ has equation $\mathbf { r } = \left( \begin{array} { l } 1 \\ 6 \\ 7 \end{array} \right) + \lambda \left( \begin{array} { r } 2 \\ - 1 \\ - 1 \end{array} \right) + \mu \left( \begin{array} { r } 2 \\ - 3 \\ - 5 \end{array} \right)$ and the line $l$ has equation $\mathbf { r } = \left( \begin{array} { l } 7 \\ 4 \\ 1 \end{array} \right) + t \left( \begin{array} { r } 3 \\ 0 \\ - 1 \end{array} \right)$.\\
(i) Express the equation of $\Pi$ in the form r.n $= p$.\\
(ii) Find the point of intersection of $l$ and $\Pi$.\\
(iii) The equation of $\Pi$ may be expressed in the form $\mathbf { r } = \left( \begin{array} { l } 1 \\ 6 \\ 7 \end{array} \right) + \lambda \left( \begin{array} { r } 2 \\ - 1 \\ - 1 \end{array} \right) + \mu \mathbf { c }$, where $\mathbf { c }$ is perpendicular to $\left( \begin{array} { r } 2 \\ - 1 \\ - 1 \end{array} \right)$. Find $\mathbf { c }$.
\hfill \mbox{\textit{OCR FP3 2012 Q6 [9]}}