| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Reflection in plane |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths question on 3D vector geometry involving standard techniques: finding line-plane intersection (routine substitution), verifying a point lies on a perpendicular (straightforward check), reflecting a point in a plane (standard formula/method), and reflecting a line in a plane (requires finding reflected point and direction, then converting to Cartesian form). While it has multiple parts and uses Further Maths content, each step follows well-established procedures without requiring novel insight or particularly complex reasoning. |
| Spec | 4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04f Line-plane intersection: find point |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(((2+3\lambda)\mathbf{i}+(1+2\lambda)\mathbf{j}+(-2+\lambda)\mathbf{k})\cdot(\mathbf{i}+\mathbf{j}-2\mathbf{k})=19\) \(\Rightarrow 2+1+4+3\lambda+2\lambda-2\lambda=19 \Rightarrow \lambda = \ldots\) | M1 | Correct dot product leading to value for \(\lambda\) |
| \(\lambda = 4\) | A1 | Correct \(\lambda\) |
| \((2+3\times"4", 1+2\times"4", -2+"4")\) | M1 | Substitutes their \(\lambda\) to give coordinates |
| \((14, 9, 2)\) | A1 | Correct coordinates (allow as vector) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{AB} = 2\mathbf{i}+2\mathbf{j}-4\mathbf{k} = 2(\mathbf{i}+\mathbf{j}-2\mathbf{k})\) so is perpendicular to plane | M1 | Correct \(\overrightarrow{AB}\) and conclusion |
| Also \(B\) lies on the plane as \((4\mathbf{i}+3\mathbf{j}-6\mathbf{k})\cdot(\mathbf{i}+\mathbf{j}-2\mathbf{k})=19\) | M1 | Substitutes \(B\) into the plane equation and conclusion |
| So coordinates of \(B\) are \((4, 3, -6)\)* | A1* | Both M's scored with final conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(((2+\lambda)\mathbf{i}+(1+\lambda)\mathbf{j}+(-2-2\lambda)\mathbf{k})\cdot(\mathbf{i}+\mathbf{j}-2\mathbf{k})=19\) \(\Rightarrow 2+1+4+\lambda+\lambda+4\lambda=19 \Rightarrow \lambda=\ldots\) | M1 | Correct dot product leading to value for \(\lambda\) \((=2)\) |
| \((2+"2", 1+"2", -2-2\times"2")\) | M1 | Substitutes their \(\lambda\) to give coordinates |
| So coordinates of \(B\) are \((4, 3, -6)\)* | A1 | Both M's scored with final conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{OA'} = \overrightarrow{OA} + 2\overrightarrow{AB}\) or \(\overrightarrow{OB} + \overrightarrow{AB}\) \((2+4, 1+4, -2-8)\) or \((4+2, 3+2, -6-4)\) | M1 | Correct strategy for finding \(A'\) |
| \((6, 5, -10)\) | A1 | Correct coordinates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\pm(14\mathbf{i}+9\mathbf{j}+2\mathbf{k}-(6\mathbf{i}+5\mathbf{j}-10\mathbf{k}))\) | M1 | Correct attempt at the direction. NB require line through their \((14,9,2)\) and their \((6,5,-10)\) |
| \(\mathbf{a} = 8\mathbf{i}+4\mathbf{j}+12\mathbf{k}\) | A1 | \(\mu(8\mathbf{i}+4\mathbf{j}+12\mathbf{k})\) |
| \(\mathbf{b} = (6\mathbf{i}+5\mathbf{j}-10\mathbf{k})\times(8\mathbf{i}+4\mathbf{j}+12\mathbf{k})\) or \((14\mathbf{i}+9\mathbf{j}+2\mathbf{k})\times(8\mathbf{i}+4\mathbf{j}+12\mathbf{k})\) \(= (=100\mathbf{i}-152\mathbf{j}-16\mathbf{k})\) | dM1 | Attempt vector product of their \(6\mathbf{i}+5\mathbf{j}-10\mathbf{k}\) with their \(8\mathbf{i}+4\mathbf{j}+12\mathbf{k}\). Dependent on the previous M1 |
| \(\mathbf{r}\times(2\mathbf{i}+\mathbf{j}+3\mathbf{k}) = 25\mathbf{i}-38\mathbf{j}-4\mathbf{k}\) | A1 | \(\lambda(\mathbf{r}\times(2\mathbf{i}+\mathbf{j}+3\mathbf{k})=25\mathbf{i}-38\mathbf{j}-4\mathbf{k})\). Must be in this form for A1 and not just stating \(\mathbf{a}\) and \(\mathbf{b}\) |
# Question 8:
## Part 8(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $((2+3\lambda)\mathbf{i}+(1+2\lambda)\mathbf{j}+(-2+\lambda)\mathbf{k})\cdot(\mathbf{i}+\mathbf{j}-2\mathbf{k})=19$ $\Rightarrow 2+1+4+3\lambda+2\lambda-2\lambda=19 \Rightarrow \lambda = \ldots$ | M1 | Correct dot product leading to value for $\lambda$ |
| $\lambda = 4$ | A1 | Correct $\lambda$ |
| $(2+3\times"4", 1+2\times"4", -2+"4")$ | M1 | Substitutes their $\lambda$ to give coordinates |
| $(14, 9, 2)$ | A1 | Correct coordinates (allow as vector) |
## Part 8(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{AB} = 2\mathbf{i}+2\mathbf{j}-4\mathbf{k} = 2(\mathbf{i}+\mathbf{j}-2\mathbf{k})$ so is perpendicular to plane | M1 | Correct $\overrightarrow{AB}$ and conclusion |
| Also $B$ lies on the plane as $(4\mathbf{i}+3\mathbf{j}-6\mathbf{k})\cdot(\mathbf{i}+\mathbf{j}-2\mathbf{k})=19$ | M1 | Substitutes $B$ into the plane equation and conclusion |
| So coordinates of $B$ are $(4, 3, -6)$* | A1* | Both M's scored with final conclusion |
### Alternative for 8(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $((2+\lambda)\mathbf{i}+(1+\lambda)\mathbf{j}+(-2-2\lambda)\mathbf{k})\cdot(\mathbf{i}+\mathbf{j}-2\mathbf{k})=19$ $\Rightarrow 2+1+4+\lambda+\lambda+4\lambda=19 \Rightarrow \lambda=\ldots$ | M1 | Correct dot product leading to value for $\lambda$ $(=2)$ |
| $(2+"2", 1+"2", -2-2\times"2")$ | M1 | Substitutes their $\lambda$ to give coordinates |
| So coordinates of $B$ are $(4, 3, -6)$* | A1 | Both M's scored with final conclusion |
## Part 8(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{OA'} = \overrightarrow{OA} + 2\overrightarrow{AB}$ or $\overrightarrow{OB} + \overrightarrow{AB}$ $(2+4, 1+4, -2-8)$ or $(4+2, 3+2, -6-4)$ | M1 | Correct strategy for finding $A'$ |
| $(6, 5, -10)$ | A1 | Correct coordinates |
## Part 8(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\pm(14\mathbf{i}+9\mathbf{j}+2\mathbf{k}-(6\mathbf{i}+5\mathbf{j}-10\mathbf{k}))$ | M1 | Correct attempt at the direction. **NB require line through their $(14,9,2)$ and their $(6,5,-10)$** |
| $\mathbf{a} = 8\mathbf{i}+4\mathbf{j}+12\mathbf{k}$ | A1 | $\mu(8\mathbf{i}+4\mathbf{j}+12\mathbf{k})$ |
| $\mathbf{b} = (6\mathbf{i}+5\mathbf{j}-10\mathbf{k})\times(8\mathbf{i}+4\mathbf{j}+12\mathbf{k})$ or $(14\mathbf{i}+9\mathbf{j}+2\mathbf{k})\times(8\mathbf{i}+4\mathbf{j}+12\mathbf{k})$ $= (=100\mathbf{i}-152\mathbf{j}-16\mathbf{k})$ | dM1 | Attempt vector product of their $6\mathbf{i}+5\mathbf{j}-10\mathbf{k}$ with their $8\mathbf{i}+4\mathbf{j}+12\mathbf{k}$. **Dependent on the previous M1** |
| $\mathbf{r}\times(2\mathbf{i}+\mathbf{j}+3\mathbf{k}) = 25\mathbf{i}-38\mathbf{j}-4\mathbf{k}$ | A1 | $\lambda(\mathbf{r}\times(2\mathbf{i}+\mathbf{j}+3\mathbf{k})=25\mathbf{i}-38\mathbf{j}-4\mathbf{k})$. Must be in this form for A1 and not just stating $\mathbf{a}$ and $\mathbf{b}$ |8. The line $l$ has equation
$$\mathbf { r } = ( 2 \mathbf { i } + \mathbf { j } - 2 \mathbf { k } ) + \lambda ( 3 \mathbf { i } + 2 \mathbf { j } + \mathbf { k } ) , \text { where } \lambda \text { is a scalar parameter, }$$
and the plane $\Pi$ has equation
$$\mathbf { r } . ( \mathbf { i } + \mathbf { j } - 2 \mathbf { k } ) = 19$$
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of the point of intersection of $l$ and $\Pi$.
The perpendicular to $\Pi$ from the point $A ( 2,1 , - 2 )$ meets $\Pi$ at the point $B$.
\item Verify that the coordinates of $B$ are $( 4,3 , - 6 )$.
The point $A ( 2,1 , - 2 )$ is reflected in the plane $\Pi$ to give the image point $A ^ { \prime }$.
\item Find the coordinates of the point $A ^ { \prime }$.
\item Find an equation for the line obtained by reflecting the line $l$ in the plane $\Pi$, giving your answer in the form
$$\mathbf { r } \times \mathbf { a } = \mathbf { b } ,$$
where $\mathbf { a }$ and $\mathbf { b }$ are vectors to be found.
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2014 Q8 [13]}}