| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2023 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Lines & Planes |
| Type | Angle between line and plane |
| Difficulty | Standard +0.3 This is a standard Further Maths vectors question requiring routine techniques: cross product for normal vector, dot product formula for angle (with care to use sine not cosine for line-plane angle), and perpendicular distance formula. All are textbook methods with straightforward arithmetic, making it slightly easier than average A-level difficulty overall. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04i Shortest distance: between a point and a line |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{pmatrix}0\\3\\-2\end{pmatrix}\times\begin{pmatrix}1\\1\\2\end{pmatrix}=8\mathbf{i}-2\mathbf{j}-3\mathbf{k}\) | M1 A1 | M1: Attempts vector product of two vectors in the plane, must achieve two correct components; A1: \(\pm(8\mathbf{i}-2\mathbf{j}-3\mathbf{k})\) or multiple |
| (2) | Allow any vector notation throughout |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(l\) has direction vector \(\pm(2\mathbf{j}+2\mathbf{k})\) | B1 | Correct direction for \(l\) |
| \(\frac{\ | (8\mathbf{i}-2\mathbf{j}-3\mathbf{k})\cdot(2\mathbf{j}+2\mathbf{k})\ | }{\ |
| Acute angle between \(l\) and \(P\) \(= 90-\alpha = 90-66.23968409...\) or \(\theta = 23.76031591...\Rightarrow 24°\) | A1 | awrt 24 from correct work; mark final answer |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((\mathbf{i}+2\mathbf{j}+3\mathbf{k})\cdot(8\mathbf{i}-2\mathbf{j}-3\mathbf{k})=-5\) or \((6\mathbf{i}-3\mathbf{j}-6\mathbf{k})\cdot(8\mathbf{i}-2\mathbf{j}-3\mathbf{k})=72\) | M1 A1 | M1: scalar product of position vector of point on plane with normal; A1: \(-5\) or \(72\) (or \(5\) or \(-72\) if normal opposite direction) |
| Shortest distance \(=\left | \frac{-5-72}{\sqrt{77}}\right | =\frac{77}{\sqrt{77}}\) or \(\sqrt{77}\) |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((\mathbf{i}+2\mathbf{j}+3\mathbf{k})\cdot(8\mathbf{i}-2\mathbf{j}-3\mathbf{k})=-5\), so "\(8x-2y-3z+5=0\)" | M1 A1 | A1: \(-5\) or \(5\) if normal opposite direction |
| \(\frac{\ | (8)(6)+(−2)(−3)+(−3)(−6)+5\ | }{\sqrt{8^2+2^2+3^2}}=\frac{77}{\sqrt{77}}\) or \(\sqrt{77}\) |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Let \(Q\) be point on plane \((1,2,3)\); \(\overrightarrow{PQ}=(\mathbf{i}+2\mathbf{j}+3\mathbf{k})-(6\mathbf{i}-3\mathbf{j}-6\mathbf{k})=-5\mathbf{i}+5\mathbf{j}+9\mathbf{k}\) | M1 A1 | M1: attempts vector from given point to point on plane; A1: correct vector \((\pm)\) |
| \(\left | \overrightarrow{PQ}\cdot\hat{\mathbf{n}}\right | =\frac{\ |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(r=(6\mathbf{i}-3\mathbf{j}-6\mathbf{k})+\lambda(8\mathbf{i}-2\mathbf{j}-3\mathbf{k})\) meets plane when \(8(6+8\lambda)-2(-3-2\lambda)-3(-6-3\lambda)+5=0 \Rightarrow \lambda=-1\) | M1 A1 | M1: line through given point in direction of normal substituted into plane; \(d\) in plane equation must not be zero; A1: correct \(\lambda\) |
| \(\ | -1(8\mathbf{i}-2\mathbf{j}-3\mathbf{k})\ | =\sqrt{8^2+2^2+3^2}\); point of intersection \(=(-2,-1,-3)\); distance \(=\sqrt{(6-(-2))^2+(-3-(-1))^2+(-6-(-3))^2}\Rightarrow\sqrt{77}\) |
| (4) Total 10 | Credit for work in (b) only available for part (c) if used in (c) |
## Question 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}0\\3\\-2\end{pmatrix}\times\begin{pmatrix}1\\1\\2\end{pmatrix}=8\mathbf{i}-2\mathbf{j}-3\mathbf{k}$ | M1 A1 | M1: Attempts vector product of two vectors in the plane, must achieve two correct components; A1: $\pm(8\mathbf{i}-2\mathbf{j}-3\mathbf{k})$ or multiple |
| | **(2)** | Allow any vector notation throughout |
---
## Question 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $l$ has direction vector $\pm(2\mathbf{j}+2\mathbf{k})$ | B1 | Correct direction for $l$ |
| $\frac{\|(8\mathbf{i}-2\mathbf{j}-3\mathbf{k})\cdot(2\mathbf{j}+2\mathbf{k})\|}{\|\sqrt{8^2+2^2+3^2}\|\times\|\sqrt{2^2+2^2}\|}=\left|\frac{-10}{\sqrt{77}\times\sqrt{8}}\right|$ or $\left|\frac{-5\sqrt{154}}{154}\right|$ | M1 A1ft | M1: scalar product of normal and direction vector divided by product of magnitudes; A1ft: correct numerical expression; actual decimal 0.40291148; allow awrt 0.41, 1.16, or 1.99 if in radians |
| Acute angle between $l$ and $P$ $= 90-\alpha = 90-66.23968409...$ or $\theta = 23.76031591...\Rightarrow 24°$ | A1 | awrt 24 from correct work; mark final answer |
| | **(4)** | |
---
## Question 7(c) Way 1 (Parallel planes):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(\mathbf{i}+2\mathbf{j}+3\mathbf{k})\cdot(8\mathbf{i}-2\mathbf{j}-3\mathbf{k})=-5$ or $(6\mathbf{i}-3\mathbf{j}-6\mathbf{k})\cdot(8\mathbf{i}-2\mathbf{j}-3\mathbf{k})=72$ | M1 A1 | M1: scalar product of position vector of point on plane with normal; A1: $-5$ or $72$ (or $5$ or $-72$ if normal opposite direction) |
| Shortest distance $=\left|\frac{-5-72}{\sqrt{77}}\right|=\frac{77}{\sqrt{77}}$ or $\sqrt{77}$ | dM1 A1 | dM1: attempts both scalar products, obtains numerical expression for distance; A1: correct exact distance |
| | **(4)** | |
## Question 7(c) Way 2 (Perp. distance formula):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(\mathbf{i}+2\mathbf{j}+3\mathbf{k})\cdot(8\mathbf{i}-2\mathbf{j}-3\mathbf{k})=-5$, so "$8x-2y-3z+5=0$" | M1 A1 | A1: $-5$ or $5$ if normal opposite direction |
| $\frac{\|(8)(6)+(−2)(−3)+(−3)(−6)+5\|}{\sqrt{8^2+2^2+3^2}}=\frac{77}{\sqrt{77}}$ or $\sqrt{77}$ | dM1 A1 | dM1: uses distance formula; condone sign slip on $-5$ but $d$ must not be zero; A1: correct exact distance |
| | **(4)** | |
## Question 7(c) Way 3 (Projection/resolving):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $Q$ be point on plane $(1,2,3)$; $\overrightarrow{PQ}=(\mathbf{i}+2\mathbf{j}+3\mathbf{k})-(6\mathbf{i}-3\mathbf{j}-6\mathbf{k})=-5\mathbf{i}+5\mathbf{j}+9\mathbf{k}$ | M1 A1 | M1: attempts vector from given point to point on plane; A1: correct vector $(\pm)$ |
| $\left|\overrightarrow{PQ}\cdot\hat{\mathbf{n}}\right|=\frac{\|(-5\mathbf{i}+5\mathbf{j}+9\mathbf{k})\cdot(8\mathbf{i}-2\mathbf{j}-3\mathbf{k})\|}{\sqrt{8^2+2^2+3^2}}=\frac{77}{\sqrt{77}}$ or $\sqrt{77}$ | dM1 A1 | dM1: uses formula; A1: correct exact distance |
| | **(4)** | |
## Question 7(c) Way 4 (Line through point):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $r=(6\mathbf{i}-3\mathbf{j}-6\mathbf{k})+\lambda(8\mathbf{i}-2\mathbf{j}-3\mathbf{k})$ meets plane when $8(6+8\lambda)-2(-3-2\lambda)-3(-6-3\lambda)+5=0 \Rightarrow \lambda=-1$ | M1 A1 | M1: line through given point in direction of normal substituted into plane; $d$ in plane equation must not be zero; A1: correct $\lambda$ |
| $\|-1(8\mathbf{i}-2\mathbf{j}-3\mathbf{k})\|=\sqrt{8^2+2^2+3^2}$; point of intersection $=(-2,-1,-3)$; distance $=\sqrt{(6-(-2))^2+(-3-(-1))^2+(-6-(-3))^2}\Rightarrow\sqrt{77}$ | dM1 A1 | dM1: attempts $|\lambda\mathbf{n}|$ or finds point on plane and distance to given point; A1: correct exact distance |
| | **(4)** Total 10 | Credit for work in (b) only available for part (c) if used in (c) |
---\begin{enumerate}
\item The plane $\Pi$ has equation
\end{enumerate}
$$\mathbf { r } = \left( \begin{array} { l }
1 \\
2 \\
3
\end{array} \right) + \lambda \left( \begin{array} { r }
0 \\
3 \\
- 2
\end{array} \right) + \mu \left( \begin{array} { l }
1 \\
1 \\
2
\end{array} \right)$$
where $\lambda$ and $\mu$ are scalar parameters.\\
(a) Determine a vector perpendicular to $\Pi$
The line $l$ meets $\Pi$ at the point ( $1,2,3$ ) and passes through the point ( $1,0,1$ )\\
(b) Determine the size of the acute angle between $\Pi$ and $l$
Give your answer to the nearest degree.\\
(c) Determine the shortest distance between $\Pi$ and the point $( 6 , - 3 , - 6 )$
\hfill \mbox{\textit{Edexcel F3 2023 Q7 [10]}}