| Exam Board | OCR MEI |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2015 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Shortest distance between two skew lines |
| Difficulty | Standard +0.8 This is a substantial multi-part Further Maths question requiring cross products, distance formulas, plane equations, and geometric reasoning across 5 parts. While each individual technique is standard for FP3, the length (likely 12-15 marks total), the need to work with 3D vectors throughout, and the integration of multiple concepts (especially the cross product application for skew line distance) place this moderately above average difficulty for A-level, though it remains a structured question with clear signposting rather than requiring novel insight. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04f Line-plane intersection: find point4.04g Vector product: a x b perpendicular vector |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Any point is \(\begin{pmatrix}8\\25\\43\end{pmatrix}+\lambda\begin{pmatrix}4\\15\\25\end{pmatrix}\) | M1 | Finding AB in terms of \(\lambda\) |
| \(AB = ((8+4\lambda),(25+15\lambda),(43+25\lambda))-(2,5,4)\) \(= ((6+4\lambda),(20+15\lambda),(39+25\lambda))\) | A1 | |
| \(\sqrt{(6+4\lambda)^2+(20+15\lambda)^2+(39+25\lambda)^2}=15\) | M1 | |
| \(\Rightarrow 866\lambda^2+2598\lambda+1732=0\) | A1 | |
| \(\Rightarrow \lambda^2+3\lambda+2=0\) | ||
| \(\Rightarrow \lambda=-1,-2\) | ||
| \(\Rightarrow\) B \((4,10,18)\), C \((0,-5,-7)\) | B1, A1 | For B; For C |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(AC=[2,10,11]\) | B1 | Or any vector in plane other than BC |
| \(\mathbf{n}=\begin{pmatrix}2\\10\\11\end{pmatrix}\times\begin{pmatrix}4\\15\\25\end{pmatrix}=\begin{pmatrix}85\\-6\\-10\end{pmatrix}\) | M1 | Suitable vector product or other method for finding normal |
| \(\Rightarrow 85x-6y-10z=c\) | A1 | |
| Sub one value to give \(c\) | ||
| \(\Rightarrow 85x-6y-10z=100\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| This plane contains line BC and \(\mathbf{n}\) \(\mathbf{n'}=\begin{pmatrix}85\\-6\\-10\end{pmatrix}\times\begin{pmatrix}4\\15\\25\end{pmatrix}=\begin{pmatrix}0\\-2165\\1299\end{pmatrix}=\begin{pmatrix}0\\-5\\3\end{pmatrix}\) | M1, A1 | Appropriate vector product; Allow uncancelled vector |
| \(\Rightarrow 5y-3z=c\) | ||
| Sub one value to give \(c\) | M1 | Must be seen! |
| \(\Rightarrow 5y-3z+4=0\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{BC}=[-4,-15,-25]\), \(\mathbf{AD}=[-1,-4,-1]\) | B1 | |
| \(\mathbf{BC}\times\mathbf{AD}=\begin{vmatrix}4&15&25\\1&4&1\end{vmatrix}=[-85,21,1]\) | M1, M1 | Finding vector product; Finding magnitude, Pythagoras must be seen |
| \( | \mathbf{BC}\times\mathbf{AD} | =\sqrt{85^2+21^2+1}=\sqrt{7667}\) |
| \(AC=[2,10,11]\) | B1 | Other vectors possible |
| \(\text{Distance}=\left | \dfrac{(\mathbf{AC})\cdot(\mathbf{BC}\times\mathbf{AD})}{\sqrt{7667}}\right | \) |
| \(=\left | \dfrac{-170+210+11}{\sqrt{7667}}\right | =\dfrac{51}{\sqrt{7667}}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{AB}=[2,5,14]\), \(\mathbf{AC}=[-2,-10,-11]\), \(\mathbf{AD}=[-1,-4,-1]\) | ||
| \(\text{Volume}=\left | \tfrac{1}{6}(\mathbf{AB}\cdot(\mathbf{CA}\times\mathbf{DA}))\right | \) |
| \(\mathbf{AC}\times\mathbf{AD}=\begin{vmatrix}-2&-10&-11\\-1&-4&-1\end{vmatrix}=[-34,9,-2]\) | A1 | Vector product |
| \(=\left | \tfrac{1}{6}((2,5,14)\cdot(-34,9,-2))\right | =\left |
| \(=\dfrac{51}{6}=\dfrac{17}{2}\) | A1 |
# Question 1:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Any point is $\begin{pmatrix}8\\25\\43\end{pmatrix}+\lambda\begin{pmatrix}4\\15\\25\end{pmatrix}$ | M1 | Finding AB in terms of $\lambda$ |
| $AB = ((8+4\lambda),(25+15\lambda),(43+25\lambda))-(2,5,4)$ $= ((6+4\lambda),(20+15\lambda),(39+25\lambda))$ | A1 | |
| $\sqrt{(6+4\lambda)^2+(20+15\lambda)^2+(39+25\lambda)^2}=15$ | M1 | |
| $\Rightarrow 866\lambda^2+2598\lambda+1732=0$ | A1 | |
| $\Rightarrow \lambda^2+3\lambda+2=0$ | | |
| $\Rightarrow \lambda=-1,-2$ | | |
| $\Rightarrow$ B $(4,10,18)$, C $(0,-5,-7)$ | B1, A1 | For B; For C | B1 can also be given for verifying $AB=15$ and showing B is on line |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $AC=[2,10,11]$ | B1 | Or any vector in plane other than BC |
| $\mathbf{n}=\begin{pmatrix}2\\10\\11\end{pmatrix}\times\begin{pmatrix}4\\15\\25\end{pmatrix}=\begin{pmatrix}85\\-6\\-10\end{pmatrix}$ | M1 | Suitable vector product or other method for finding normal |
| $\Rightarrow 85x-6y-10z=c$ | A1 | |
| Sub one value to give $c$ | | |
| $\Rightarrow 85x-6y-10z=100$ | A1 | |
## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| This plane contains line BC and $\mathbf{n}$ $\mathbf{n'}=\begin{pmatrix}85\\-6\\-10\end{pmatrix}\times\begin{pmatrix}4\\15\\25\end{pmatrix}=\begin{pmatrix}0\\-2165\\1299\end{pmatrix}=\begin{pmatrix}0\\-5\\3\end{pmatrix}$ | M1, A1 | Appropriate vector product; Allow uncancelled vector |
| $\Rightarrow 5y-3z=c$ | | |
| Sub one value to give $c$ | M1 | Must be seen! |
| $\Rightarrow 5y-3z+4=0$ | A1 | |
## Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{BC}=[-4,-15,-25]$, $\mathbf{AD}=[-1,-4,-1]$ | B1 | |
| $\mathbf{BC}\times\mathbf{AD}=\begin{vmatrix}4&15&25\\1&4&1\end{vmatrix}=[-85,21,1]$ | M1, M1 | Finding vector product; Finding magnitude, Pythagoras must be seen |
| $|\mathbf{BC}\times\mathbf{AD}|=\sqrt{85^2+21^2+1}=\sqrt{7667}$ | A1 | N.b. Answer given |
| $AC=[2,10,11]$ | B1 | Other vectors possible |
| $\text{Distance}=\left|\dfrac{(\mathbf{AC})\cdot(\mathbf{BC}\times\mathbf{AD})}{\sqrt{7667}}\right|$ | M1 | |
| $=\left|\dfrac{-170+210+11}{\sqrt{7667}}\right|=\dfrac{51}{\sqrt{7667}}$ | A1 | Accept 0.582... |
## Part (v)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{AB}=[2,5,14]$, $\mathbf{AC}=[-2,-10,-11]$, $\mathbf{AD}=[-1,-4,-1]$ | | |
| $\text{Volume}=\left|\tfrac{1}{6}(\mathbf{AB}\cdot(\mathbf{CA}\times\mathbf{DA}))\right|$ | M1 | Formula for volume |
| $\mathbf{AC}\times\mathbf{AD}=\begin{vmatrix}-2&-10&-11\\-1&-4&-1\end{vmatrix}=[-34,9,-2]$ | A1 | Vector product |
| $=\left|\tfrac{1}{6}((2,5,14)\cdot(-34,9,-2))\right|=\left|\tfrac{1}{6}(-68+45-28)\right|$ | | |
| $=\dfrac{51}{6}=\dfrac{17}{2}$ | A1 | |
---1 The point A has coordinates $( 2,5,4 )$ and the line BC has equation
$$\mathbf { r } = \left( \begin{array} { c }
8 \\
25 \\
43
\end{array} \right) + \lambda \left( \begin{array} { c }
4 \\
15 \\
25
\end{array} \right)$$
You are given that $\mathrm { AB } = \mathrm { AC } = 15$.\\
(i) Show that the coordinates of one of the points B and C are (4, 10, 18). Find the coordinates of the other point. These points are B and C respectively.\\
(ii) Find the equation of the plane ABC in cartesian form.\\
(iii) Show that the plane containing the line BC and perpendicular to the plane ABC has equation $5 y - 3 z + 4 = 0$.
The point D has coordinates $( 1,1,3 )$.\\
(iv) Show that $| \overrightarrow { B C } \times \overrightarrow { A D } | = \sqrt { 7667 }$ and hence find the shortest distance between the lines $B C$ and $A D$.\\
(v) Find the volume of the tetrahedron ABCD .
\hfill \mbox{\textit{OCR MEI FP3 2015 Q1 [24]}}