| Exam Board | Edexcel |
|---|---|
| Module | F3 (Further Pure Mathematics 3) |
| Year | 2021 |
| Session | October |
| Marks | 10 |
| 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 standard Further Maths vectors question requiring cross product, plane equations in multiple forms, and distance formula application. While it involves multiple parts and careful algebraic manipulation, all techniques are routine for F3 students with no novel problem-solving required—just systematic application of learned methods. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04g Vector product: a x b perpendicular vector4.04i Shortest distance: between a point and a line |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((5\mathbf{i}+\mathbf{j})\times(8\mathbf{i}-2\mathbf{j}+3\mathbf{k}) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 5 & 1 & 0 \\ 8 & -2 & 3 \end{vmatrix} = \ldots\) OR \((u\mathbf{i}+v\mathbf{j}+w\mathbf{k}).(5\mathbf{i}+\mathbf{j})=0\) and \((u\mathbf{i}+v\mathbf{j}+w\mathbf{k}).(8\mathbf{i}-2\mathbf{j}+3\mathbf{k})=0 \Rightarrow 5u+v=0\), \(8u-2v+3w=0 \Rightarrow u,v,w=\ldots\) | M1 | Any correct method to find vector perpendicular to both direction vectors. For cross product with no method shown, look for at least 2 correct components |
| \(\mathbf{n} = 3\mathbf{i}-15\mathbf{j}-18\mathbf{k}\) or \(\alpha(\mathbf{i}-5\mathbf{j}-6\mathbf{k})\) for any \(\alpha \neq 0\) | A1 | Any correct vector, scalar multiple of \(-\mathbf{i}+5\mathbf{j}+6\mathbf{k}\) (or equivalent direction) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (i) \(\mathbf{r} = (2\mathbf{i}-4\mathbf{j}+4\mathbf{k}) + s(8\mathbf{i}-2\mathbf{j}+3\mathbf{k}) + t(5\mathbf{i}+\mathbf{j})\) | B1 | Any correct equation. Must have \(\mathbf{r}=\ldots\) or column vector form |
| (ii) \((2\mathbf{i}-4\mathbf{j}+4\mathbf{k}).(3\mathbf{i}-15\mathbf{j}-18\mathbf{k}) = \ldots (= -6)\) | M1 | Uses normal vector from (a) with any point on plane (probably \((2\mathbf{i}-4\mathbf{j}+4\mathbf{k})\)) to find \(p\). Condone slips in calculation |
| So \(\mathbf{r}.(3\mathbf{i}-15\mathbf{j}-18\mathbf{k}) = -6\) oe such as \(\mathbf{r}.(-\mathbf{i}+5\mathbf{j}+6\mathbf{k}) = 2\) | A1 | Any correct equation of correct form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Distance from plane in (b) to origin: \(\dfrac{\pm 6}{\sqrt{3^2+15^2+18^2}}\) oe, e.g. \(\dfrac{2}{\sqrt{1^2+5^2+6^2}}\). Or attempts parallel plane containing \(l_1\): \(\dfrac{(\mathbf{i}+2\mathbf{j}-5\mathbf{k}).(3\mathbf{i}-15\mathbf{j}-18\mathbf{k})}{\sqrt{3^2+15^2+18^2}} = \ldots\) | M1 | Uses plane equation from (b) or parallel plane containing \(l_1\) to find distance of one plane to origin |
| \(= \pm\dfrac{2}{\sqrt{62}}\) (evaluated) or \(\mp\dfrac{21}{\sqrt{62}}\) if considering other plane | A1 | Correct distance between one plane and origin, accept \(\pm\) |
| Both \(\dfrac{\pm 6}{\sqrt{3^2+15^2+18^2}}\) and \(\dfrac{(\mathbf{i}+2\mathbf{j}-5\mathbf{k}).(3\mathbf{i}-15\mathbf{j}-18\mathbf{k})}{\sqrt{3^2+15^2+18^2}}=\ldots\) attempted | M1 | Attempts distance of both parallel planes containing \(l_1\) and \(l_2\) from origin |
| Shortest distance between lines is \(\dfrac{2}{\sqrt{62}}+\dfrac{21}{\sqrt{62}}=\ldots\) | M1 | Adds the two distances |
| \(= \dfrac{23}{\sqrt{62}}\) or \(\dfrac{23\sqrt{62}}{62}\) | A1 | Correct final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{AB} = \pm\left(({\mathbf{i}+2\mathbf{j}-5\mathbf{k}})-(2\mathbf{i}-4\mathbf{j}+4\mathbf{k})\right) = \pm(-\mathbf{i}+6\mathbf{j}-9\mathbf{k})\) | M1 A1 | |
| \(d = AB\cos\theta = \dfrac{\overrightarrow{AB}.\mathbf{n}}{ | \mathbf{n} | } = \dfrac{\pm(-\mathbf{i}+6\mathbf{j}-9\mathbf{k}).(3\mathbf{i}-15\mathbf{j}-18\mathbf{k})}{\sqrt{3^2+15^2+18^2}}\) oe |
| \(= \dfrac{\pm(-3-90+162)}{\sqrt{558}} = \dfrac{\pm 69}{\sqrt{558}} = \ldots\) | M1 | |
| \(= \dfrac{23}{\sqrt{62}}\) or \(\dfrac{23\sqrt{62}}{62}\) | A1 |
# Question 5:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(5\mathbf{i}+\mathbf{j})\times(8\mathbf{i}-2\mathbf{j}+3\mathbf{k}) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 5 & 1 & 0 \\ 8 & -2 & 3 \end{vmatrix} = \ldots$ OR $(u\mathbf{i}+v\mathbf{j}+w\mathbf{k}).(5\mathbf{i}+\mathbf{j})=0$ and $(u\mathbf{i}+v\mathbf{j}+w\mathbf{k}).(8\mathbf{i}-2\mathbf{j}+3\mathbf{k})=0 \Rightarrow 5u+v=0$, $8u-2v+3w=0 \Rightarrow u,v,w=\ldots$ | M1 | Any correct method to find vector perpendicular to both direction vectors. For cross product with no method shown, look for at least 2 correct components |
| $\mathbf{n} = 3\mathbf{i}-15\mathbf{j}-18\mathbf{k}$ or $\alpha(\mathbf{i}-5\mathbf{j}-6\mathbf{k})$ for any $\alpha \neq 0$ | A1 | Any correct vector, scalar multiple of $-\mathbf{i}+5\mathbf{j}+6\mathbf{k}$ (or equivalent direction) |
**(2 marks)**
---
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| (i) $\mathbf{r} = (2\mathbf{i}-4\mathbf{j}+4\mathbf{k}) + s(8\mathbf{i}-2\mathbf{j}+3\mathbf{k}) + t(5\mathbf{i}+\mathbf{j})$ | B1 | Any correct equation. Must have $\mathbf{r}=\ldots$ or column vector form |
| (ii) $(2\mathbf{i}-4\mathbf{j}+4\mathbf{k}).(3\mathbf{i}-15\mathbf{j}-18\mathbf{k}) = \ldots (= -6)$ | M1 | Uses normal vector from (a) with any point on plane (probably $(2\mathbf{i}-4\mathbf{j}+4\mathbf{k})$) to find $p$. Condone slips in calculation |
| So $\mathbf{r}.(3\mathbf{i}-15\mathbf{j}-18\mathbf{k}) = -6$ oe such as $\mathbf{r}.(-\mathbf{i}+5\mathbf{j}+6\mathbf{k}) = 2$ | A1 | Any correct equation of correct form |
**(3 marks)**
---
## Part (c) — Way 1
| Answer/Working | Mark | Guidance |
|---|---|---|
| Distance from plane in (b) to origin: $\dfrac{\pm 6}{\sqrt{3^2+15^2+18^2}}$ oe, e.g. $\dfrac{2}{\sqrt{1^2+5^2+6^2}}$. Or attempts parallel plane containing $l_1$: $\dfrac{(\mathbf{i}+2\mathbf{j}-5\mathbf{k}).(3\mathbf{i}-15\mathbf{j}-18\mathbf{k})}{\sqrt{3^2+15^2+18^2}} = \ldots$ | M1 | Uses plane equation from (b) or parallel plane containing $l_1$ to find distance of one plane to origin |
| $= \pm\dfrac{2}{\sqrt{62}}$ (evaluated) or $\mp\dfrac{21}{\sqrt{62}}$ if considering other plane | A1 | Correct distance between one plane and origin, accept $\pm$ |
| Both $\dfrac{\pm 6}{\sqrt{3^2+15^2+18^2}}$ and $\dfrac{(\mathbf{i}+2\mathbf{j}-5\mathbf{k}).(3\mathbf{i}-15\mathbf{j}-18\mathbf{k})}{\sqrt{3^2+15^2+18^2}}=\ldots$ attempted | M1 | Attempts distance of both parallel planes containing $l_1$ and $l_2$ from origin |
| Shortest distance between lines is $\dfrac{2}{\sqrt{62}}+\dfrac{21}{\sqrt{62}}=\ldots$ | M1 | Adds the two distances |
| $= \dfrac{23}{\sqrt{62}}$ or $\dfrac{23\sqrt{62}}{62}$ | A1 | Correct final answer |
**(5 marks)**
---
## Part (c) — Way 2
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{AB} = \pm\left(({\mathbf{i}+2\mathbf{j}-5\mathbf{k}})-(2\mathbf{i}-4\mathbf{j}+4\mathbf{k})\right) = \pm(-\mathbf{i}+6\mathbf{j}-9\mathbf{k})$ | M1 A1 | |
| $d = AB\cos\theta = \dfrac{\overrightarrow{AB}.\mathbf{n}}{|\mathbf{n}|} = \dfrac{\pm(-\mathbf{i}+6\mathbf{j}-9\mathbf{k}).(3\mathbf{i}-15\mathbf{j}-18\mathbf{k})}{\sqrt{3^2+15^2+18^2}}$ oe | M1 | |
| $= \dfrac{\pm(-3-90+162)}{\sqrt{558}} = \dfrac{\pm 69}{\sqrt{558}} = \ldots$ | M1 | |
| $= \dfrac{23}{\sqrt{62}}$ or $\dfrac{23\sqrt{62}}{62}$ | A1 | |
**(5 marks) — Total: 10 marks**\begin{enumerate}
\item The skew lines $l _ { 1 }$ and $l _ { 2 }$ have equations
\end{enumerate}
$$l _ { 1 } : \mathbf { r } = ( \mathbf { i } + 2 \mathbf { j } - 5 \mathbf { k } ) + \lambda ( 5 \mathbf { i } + \mathbf { j } )$$
and
$$l _ { 2 } : \mathbf { r } = ( 2 \mathbf { i } - 4 \mathbf { j } + 4 \mathbf { k } ) + \mu ( 8 \mathbf { i } - 2 \mathbf { j } + 3 \mathbf { k } )$$
where $\lambda$ and $\mu$ are scalar parameters.\\
(a) Determine a vector that is perpendicular to both $l _ { 1 }$ and $l _ { 2 }$\\
(b) Determine an equation of the plane parallel to $l _ { 1 }$ that contains $l _ { 2 }$\\
(i) in the form $\mathbf { r } = \mathbf { a } + s \mathbf { b } + t \mathbf { c }$\\
(ii) in the form r.n $= p$\\
(c) Determine the shortest distance between $l _ { 1 }$ and $l _ { 2 }$
Give your answer in simplest form.
\hfill \mbox{\textit{Edexcel F3 2021 Q5 [10]}}