| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Line intersection with plane |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring conversion of line equation to parametric form, substitution into plane equation, angle calculation using dot product of direction vector and normal (requiring knowledge that angle with plane = 90° - angle with normal), and finally solving a distance-from-plane equation involving parameter manipulation. While each technique is standard for FP3, the combination across three parts with the final part requiring algebraic manipulation of the distance formula makes this moderately challenging, above average difficulty. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04d Angles: between planes and between line and plane4.04f Line-plane intersection: find point4.04j Shortest distance: between a point and a plane |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x=2t+1,\ y=5t-1,\ z=t+2\) | B1 | Parameterise. Or B1 for \(y\) and \(z\) correctly in terms of \(x\) e.g. \(2y=5x-7\), \(2z=x+3\) |
| \((2t+1)+2(5t-1)-2(t+2)=5\) | Substitute into plane. Then M1 for full simultaneous equations method. | |
| \(\Rightarrow 10t=10 \Rightarrow t=1\) | M1 | Solve |
| Intersect at \((3, 4, 3)\) | A1 | cao. Accept vector form |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\cos\left(\frac{1}{2}\pi - \theta\right) = \dfrac{\begin{pmatrix}2\\5\\1\end{pmatrix}\cdot\begin{pmatrix}1\\2\\-2\end{pmatrix}}{\left\ | \begin{pmatrix}2\\5\\1\end{pmatrix}\right\ | \left\ |
| \(\theta = 0.654\) | A1 | or \(37.5°\) |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| If \(P\) is point of intersection and \(Q\) is required point, \(\overrightarrow{PQ} = \lambda\begin{pmatrix}2\\5\\1\end{pmatrix}\) so \(\sin\theta = \dfrac{2}{PQ} = \dfrac{2}{ | \lambda | \sqrt{30}}\) |
| \(\dfrac{10}{3\sqrt{30}} = \dfrac{2}{ | \lambda | \sqrt{30}}\) |
| \(\lambda = \pm\dfrac{3}{5}\) | A1 | |
| points have position vectors \(\begin{pmatrix}3\\4\\3\end{pmatrix} \pm \dfrac{3}{5}\begin{pmatrix}2\\5\\1\end{pmatrix}\) | *M1 | Dep on 1st M1 |
| points at \((1.8, 1, 2.4)\) and \((4.2, 7, 3.6)\) | A1 | cao |
| Alternative: Distance \(= \dfrac{ | 2t+1+2(5t-1)-2(t+2)-5 | }{\sqrt{1^2+2^2+2^2}}=2\) |
| \(\Rightarrow t=0.4\) or \(1.6\) | *M1, A1 | Solve. At least one value found |
| \((1.8, 1, 2.4)\) and \((4.2, 7, 3.6)\) | A1 | |
| [5] |
# Question 6:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x=2t+1,\ y=5t-1,\ z=t+2$ | B1 | Parameterise. Or B1 for $y$ and $z$ correctly in terms of $x$ e.g. $2y=5x-7$, $2z=x+3$ |
| $(2t+1)+2(5t-1)-2(t+2)=5$ | | Substitute into plane. Then M1 for full simultaneous equations method. |
| $\Rightarrow 10t=10 \Rightarrow t=1$ | M1 | Solve |
| Intersect at $(3, 4, 3)$ | A1 | cao. Accept vector form |
| | **[3]** | |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cos\left(\frac{1}{2}\pi - \theta\right) = \dfrac{\begin{pmatrix}2\\5\\1\end{pmatrix}\cdot\begin{pmatrix}1\\2\\-2\end{pmatrix}}{\left\|\begin{pmatrix}2\\5\\1\end{pmatrix}\right\|\left\|\begin{pmatrix}1\\2\\-2\end{pmatrix}\right\|} = \dfrac{10}{3\sqrt{30}}$ | M1A1 | Attempt to find angle or its complement |
| $\theta = 0.654$ | A1 | or $37.5°$ |
| | **[3]** | |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| If $P$ is point of intersection and $Q$ is required point, $\overrightarrow{PQ} = \lambda\begin{pmatrix}2\\5\\1\end{pmatrix}$ so $\sin\theta = \dfrac{2}{PQ} = \dfrac{2}{|\lambda|\sqrt{30}}$ | M1* | Or $\overrightarrow{PQ}\cdot\hat{\mathbf{n}} = \pm 2$ where $\mathbf{n}=\begin{pmatrix}1\\2\\-2\end{pmatrix}$. Use $\overrightarrow{PQ}$ with right angled triangle or consider component of $\overrightarrow{PQ}$ in direction of normal vector. |
| $\dfrac{10}{3\sqrt{30}} = \dfrac{2}{|\lambda|\sqrt{30}}$ | M1 | Valid method to set up equation in $\lambda$ alone. (May work from general point on original equation) |
| $\lambda = \pm\dfrac{3}{5}$ | A1 | |
| points have position vectors $\begin{pmatrix}3\\4\\3\end{pmatrix} \pm \dfrac{3}{5}\begin{pmatrix}2\\5\\1\end{pmatrix}$ | *M1 | Dep on 1st M1 |
| points at $(1.8, 1, 2.4)$ and $(4.2, 7, 3.6)$ | A1 | cao |
| **Alternative:** Distance $= \dfrac{|2t+1+2(5t-1)-2(t+2)-5|}{\sqrt{1^2+2^2+2^2}}=2$ | M1*, A1 | Zero if formula used without substitution in of parametric form. |
| $\Rightarrow t=0.4$ or $1.6$ | *M1, A1 | Solve. At least one value found |
| $(1.8, 1, 2.4)$ and $(4.2, 7, 3.6)$ | A1 | |
| | **[5]** | |
---6 The plane $\Pi$ has equation $x + 2 y - 2 z = 5$. The line $l$ has equation $\frac { x - 1 } { 2 } = \frac { y + 1 } { 5 } = \frac { z - 2 } { 1 }$.\\
(i) Find the coordinates of the point of intersection of $l$ with the plane $\Pi$.\\
(ii) Calculate the acute angle between $l$ and $\Pi$.\\
(iii) Find the coordinates of the two points on the line $l$ such that the distance of each point from the plane $\Pi$ is 2 .
\hfill \mbox{\textit{OCR FP3 2013 Q6 [11]}}