| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2018 |
| Session | June |
| Marks | 12 |
| 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 question requiring finding the line of intersection of two planes in Cartesian form (non-trivial), then finding where this line meets a third plane, and finally calculating an angle between a vector and a line. While systematic, it requires multiple coordinate geometry techniques, careful algebraic manipulation, and understanding of vector/Cartesian form conversions—significantly above average A-level difficulty but standard for Further Maths F3. |
| Spec | 1.10c Magnitude and direction: of vectors4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x=\lambda \Rightarrow y = \frac{7}{4} - \frac{3}{4}\lambda\) or \(\lambda = \frac{4y-7}{-3}\) | M1, A1 | M1: Obtains 2 equations connecting \(x,y\) or \(z\) with \(\lambda\); A1: Correct equations |
| \(z = \frac{5}{4} - \frac{1}{4}\lambda\) or \(\lambda = 5-4z\) | M1, A1 | M1: Obtains 3 equations connecting \(x,y\) or \(z\) with \(\lambda\); A1: Correct equations |
| \(\frac{x}{1} = \frac{7-4y}{3} = \frac{5-4z}{1}\ (=\lambda)\) | M1, A1 | M1: Correct use of cartesian form; A1: Correct equation (allow equivalents) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\1 & 1 & 1\\2 & 3 & -1\end{vmatrix} = \begin{pmatrix}-4\\3\\1\end{pmatrix}\) | M1, A1 | M1: Attempt vector product of normals; A1: Correct vector |
| \(x=0 \Rightarrow y+z=3,\ 3y-z=4 \Rightarrow y=\frac{7}{4},\ z=\frac{5}{4} \to \left(0,\frac{7}{4},\frac{5}{4}\right)\) | M1, A1 | M1: Attempt a point on the line; A1: Correct point (\((1,1,1)\) seen frequently) |
| \(\frac{x}{-4} = \frac{y-\frac{7}{4}}{3} = \frac{z-\frac{5}{4}}{1}\ (=\lambda)\) | M1, A1 | M1: Correct use of cartesian form; A1: Correct equation (allow equivalents). Or \(\frac{x-1}{-4}=\frac{y-1}{3}=\frac{z-1}{1}\) if \((1,1,1)\) used |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x = -\frac{4}{3}y + \frac{7}{3}\) | M1, A1 | M1: Eliminates 1 variable; A1: Correct equation |
| \(x = 5-4z\) | M1, A1 | M1: Eliminates 2nd variable; A1: Correct equation |
| \(\frac{x}{1} = -\frac{4}{3}y+\frac{7}{3} = 5-4z\) | M1, A1 | M1: Correct use of cartesian form; A1: Correct equation (allow equivalents) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(5(-4\lambda)-4\!\left(\frac{7}{4}+3\lambda\right)+4\!\left(\frac{5}{4}+\lambda\right)=12\) | M1 | Substitutes parametric form of \(L\) into \(\Pi_3\) |
| \(\lambda = -\frac{1}{2} \Rightarrow x=\ldots,\ y=\ldots,\ z=\ldots\) | dM1 | Solves for \(\lambda\) and attempts coordinates |
| \(\left(2,\frac{1}{4},\frac{3}{4}\right)\) or \(x=2,\ y=\frac{1}{4},\ z=\frac{3}{4}\) | A1 | Correct coordinates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(5x - 4\cdot\frac{3}{4}\!\left(\frac{7}{3}-x\right)+4\cdot\frac{1}{4}(5-x)=12\) | M1 | Substitutes for \(y\) and \(z\) in terms of \(x\) into \(\Pi_3\) |
| \(x=2 \Rightarrow y=\ldots,\ z=\ldots\) | dM1 | Solves for \(x\) and attempts other coordinates |
| \(\left(2,\frac{1}{4},\frac{3}{4}\right)\) or \(x=2,\ y=\frac{1}{4},\ z=\frac{3}{4}\) | A1 | Correct coordinates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\begin{pmatrix}-2\\-\frac{1}{4}\\-\frac{3}{4}\end{pmatrix}\cdot\begin{pmatrix}-4\\3\\1\end{pmatrix} = \sqrt{\frac{37}{8}}\sqrt{26}\cos\theta\) | M1 | Use scalar product between \(\pm\) their \(\overrightarrow{OA}\) and direction of their \(L\) |
| \(\frac{13}{2} = \sqrt{\frac{37}{8}}\sqrt{26}\cos\theta \Rightarrow \theta = \ldots\) | dM1 | Evaluate scalar product and complete to \(\theta=\ldots\) (or supplementary angle); check product if vectors incorrect |
| \(\theta = 53.6°\) | A1 | cao |
## Question 7:
**Given:** $\Pi_1: x+y+z=3$, $\Pi_2: 2x+3y-z=4$
### Part (a) Way 1:
| Answer | Mark | Guidance |
|--------|------|----------|
| $x=\lambda \Rightarrow y = \frac{7}{4} - \frac{3}{4}\lambda$ or $\lambda = \frac{4y-7}{-3}$ | M1, A1 | M1: Obtains 2 equations connecting $x,y$ or $z$ with $\lambda$; A1: Correct equations |
| $z = \frac{5}{4} - \frac{1}{4}\lambda$ or $\lambda = 5-4z$ | M1, A1 | M1: Obtains 3 equations connecting $x,y$ or $z$ with $\lambda$; A1: Correct equations |
| $\frac{x}{1} = \frac{7-4y}{3} = \frac{5-4z}{1}\ (=\lambda)$ | M1, A1 | M1: Correct use of cartesian form; A1: Correct equation (allow equivalents) |
### Part (a) Way 2:
| Answer | Mark | Guidance |
|--------|------|----------|
| $\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\1 & 1 & 1\\2 & 3 & -1\end{vmatrix} = \begin{pmatrix}-4\\3\\1\end{pmatrix}$ | M1, A1 | M1: Attempt vector product of normals; A1: Correct vector |
| $x=0 \Rightarrow y+z=3,\ 3y-z=4 \Rightarrow y=\frac{7}{4},\ z=\frac{5}{4} \to \left(0,\frac{7}{4},\frac{5}{4}\right)$ | M1, A1 | M1: Attempt a point on the line; A1: Correct point ($(1,1,1)$ seen frequently) |
| $\frac{x}{-4} = \frac{y-\frac{7}{4}}{3} = \frac{z-\frac{5}{4}}{1}\ (=\lambda)$ | M1, A1 | M1: Correct use of cartesian form; A1: Correct equation (allow equivalents). Or $\frac{x-1}{-4}=\frac{y-1}{3}=\frac{z-1}{1}$ if $(1,1,1)$ used |
### Part (a) Way 3:
| Answer | Mark | Guidance |
|--------|------|----------|
| $x = -\frac{4}{3}y + \frac{7}{3}$ | M1, A1 | M1: Eliminates 1 variable; A1: Correct equation |
| $x = 5-4z$ | M1, A1 | M1: Eliminates 2nd variable; A1: Correct equation |
| $\frac{x}{1} = -\frac{4}{3}y+\frac{7}{3} = 5-4z$ | M1, A1 | M1: Correct use of cartesian form; A1: Correct equation (allow equivalents) |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $5(-4\lambda)-4\!\left(\frac{7}{4}+3\lambda\right)+4\!\left(\frac{5}{4}+\lambda\right)=12$ | M1 | Substitutes parametric form of $L$ into $\Pi_3$ |
| $\lambda = -\frac{1}{2} \Rightarrow x=\ldots,\ y=\ldots,\ z=\ldots$ | dM1 | Solves for $\lambda$ and attempts coordinates |
| $\left(2,\frac{1}{4},\frac{3}{4}\right)$ or $x=2,\ y=\frac{1}{4},\ z=\frac{3}{4}$ | A1 | Correct coordinates |
### Part (b) Way 2:
| Answer | Mark | Guidance |
|--------|------|----------|
| $5x - 4\cdot\frac{3}{4}\!\left(\frac{7}{3}-x\right)+4\cdot\frac{1}{4}(5-x)=12$ | M1 | Substitutes for $y$ and $z$ in terms of $x$ into $\Pi_3$ |
| $x=2 \Rightarrow y=\ldots,\ z=\ldots$ | dM1 | Solves for $x$ and attempts other coordinates |
| $\left(2,\frac{1}{4},\frac{3}{4}\right)$ or $x=2,\ y=\frac{1}{4},\ z=\frac{3}{4}$ | A1 | Correct coordinates |
### Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\begin{pmatrix}-2\\-\frac{1}{4}\\-\frac{3}{4}\end{pmatrix}\cdot\begin{pmatrix}-4\\3\\1\end{pmatrix} = \sqrt{\frac{37}{8}}\sqrt{26}\cos\theta$ | M1 | Use scalar product between $\pm$ their $\overrightarrow{OA}$ and direction of their $L$ |
| $\frac{13}{2} = \sqrt{\frac{37}{8}}\sqrt{26}\cos\theta \Rightarrow \theta = \ldots$ | dM1 | Evaluate scalar product and complete to $\theta=\ldots$ (or supplementary angle); check product if vectors incorrect |
| $\theta = 53.6°$ | A1 | cao |
---7. The plane $\Pi _ { 1 }$ has equation $x + y + z = 3$ and the plane $\Pi _ { 2 }$ has equation $2 x + 3 y - z = 4$ The planes $\Pi _ { 1 }$ and $\Pi _ { 2 }$ intersect in the line $L$.
\begin{enumerate}[label=(\alph*)]
\item Find a cartesian equation for the line $L$.
The plane $\Pi _ { 3 }$ has equation
$$\text { r. } \left( \begin{array} { r }
5 \\
- 4 \\
4
\end{array} \right) = 12$$
The line $L$ meets the plane $\Pi _ { 3 }$ at the point $A$.
\item Find the coordinates of $A$.
\item Find the acute angle between $\overrightarrow { O A }$ and the line $L$, where $O$ is the origin. Give your answer in degrees to one decimal place.
\end{enumerate}
\hfill \mbox{\textit{Edexcel F3 2018 Q7 [12]}}