AS October 2020 Paper 1 Q4
4.
All units in this question are in metres.
A lawn is modelled as a plane that contains the points \(L(-2, -3, -1)\), \(M(6, -2, 0)\) and \(N(2, 0, 0)\), relative to a fixed origin \(O\).
There are two posts set in the lawn.
There is a washing line between the two posts.
The washing line is modelled as a straight line through points at the top of each post with coordinates \(P(-10, 8, 2)\) and \(Q(6, 4, 3)\).
The point \(R(2, 5, 2.75)\) lies on the washing line.
Given that the shortest distance from the point \(R\) to the lawn is actually 1.5 m,
| Scheme | Marks | AO |
|---|---|---|
| Finds any two vectors \(\pm\overrightarrow{LM}, \pm\overrightarrow{LN}\) or \(\pm\overrightarrow{MN}\) \(\pm\begin{pmatrix}8\\ 1\\ 1\end{pmatrix}\) or \(\pm\begin{pmatrix}4\\ 3\\ 1\end{pmatrix}\) or \(\pm\begin{pmatrix}-4\\ 2\\ 0\end{pmatrix}\) two out of three values correct is sufficient to imply the correct method | M1 | 3.3 |
| Applies the vector equation of the plane formula \(\mathbf{r} = \mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c}\) Where \(\mathbf{a}\) is any coordinate from L, M & N and vectors \(\mathbf{b}\) and \(\mathbf{c}\) come from an attempt at finding any two vectors that lie on the plane. | M1 | 1.1b |
| A correct equation for the plane \(\mathbf{r} = \mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c}\) \(\mathbf{a} = \begin{pmatrix}-2\\ -3\\ -1\end{pmatrix}\) or \(\begin{pmatrix}6\\ -2\\ 0\end{pmatrix}\) or \(\begin{pmatrix}2\\ 0\\ 0\end{pmatrix}\) \(\mathbf{b}\) and \(\mathbf{c}\) are any two vectors from \(\pm\begin{pmatrix}8\\ 1\\ 1\end{pmatrix}\) or \(\pm\begin{pmatrix}4\\ 3\\ 1\end{pmatrix}\) or \(\pm\begin{pmatrix}-4\\ 2\\ 0\end{pmatrix}\) | A1 | 1.1b |
| (3) |
Notes
M1: Finds any two vectors \(\pm\overrightarrow{LM}, \pm\overrightarrow{LN}\) or \(\pm\overrightarrow{MN}\) by subtracting relevant vectors. Two out three values correct is sufficient to imply the correct method
M1: Applies the vector equation of the plane formula \(\mathbf{r} = \mathbf{a} + \lambda\mathbf{b} + \mu\mathbf{c}\) where \(\mathbf{a}\) is any point on the plane and the vectors \(\mathbf{b}\) and \(\mathbf{c}\) are any two from their \(\pm\overrightarrow{LM}, \pm\overrightarrow{LN}\) or \(\pm\overrightarrow{MN}\)
A1: Any correct equation for the plane. Must start with \(\mathbf{r}\) =…
| Scheme | Marks | AO |
|---|---|---|
| (i) Applies ‘their’ \(\mathbf{b}.\begin{pmatrix}1\\ 2\\ -10\end{pmatrix}\) AND ‘their’ \(\mathbf{c}.\begin{pmatrix}1\\ 2\\ -10\end{pmatrix}\) Alternative 1: Finds ‘their \(\mathbf{b}\)’ – ‘their \(\mathbf{c}\)’ or vice versa and applies the dot product with \(\begin{pmatrix}1\\ 2\\ -10\end{pmatrix}\) AND one of their \(\mathbf{b}\) or \(\mathbf{c}\) Alternative 2: Applies ‘their’ \(\mathbf{b}.\begin{pmatrix}x\\ y\\ z\end{pmatrix}\) AND ‘their’ \(\mathbf{c}.\begin{pmatrix}x\\ y\\ z\end{pmatrix}\) and solves to find values of \(x\), \(y\) and \(z\) Alternative 3: Applies the dot product between their answer to part (a) and the vector \(\begin{pmatrix}1\\ 2\\ -10\end{pmatrix}\) | M1 | 1.1b |
| Show that both dot product(s) = 0 therefore the lawn is perpendicular (main method and Alternative 1) Alternative 2: Shows results is parallel to \(\begin{pmatrix}1\\ 2\\ -10\end{pmatrix}\) therefore the lawn is perpendicular Alternative 3: Achieves the value 2 and concludes as a constant therefore the lawn is perpendicular | A1 | 2.4 |
| Outside Specification for this paper – using the cross product Finds the cross product between ‘their \(\mathbf{b}\)’ and ‘their \(\mathbf{c}\)’ and either compares with the vector \(\begin{pmatrix}1\\ 2\\ -10\end{pmatrix}\) to show parallel or applies the dot product formula with the vector \(\begin{pmatrix}1\\ 2\\ -10\end{pmatrix}\) to show parallel | M1 | 1.1b |
| Concludes parallel therefore the lawn is perpendicular | A1 | 2.4 |
| (ii) Attempts \(\begin{pmatrix}x\\ y\\ z\end{pmatrix}.\begin{pmatrix}1\\ 2\\ -10\end{pmatrix} = \mathbf{a}.\begin{pmatrix}1\\ 2\\ -10\end{pmatrix}\) where \(\mathbf{a} = \begin{pmatrix}-2\\ -3\\ -1\end{pmatrix}\) or \(\begin{pmatrix}6\\ -2\\ 0\end{pmatrix}\) or \(\begin{pmatrix}2\\ 0\\ 0\end{pmatrix}\) Allow \(\mathbf{r}.\begin{pmatrix}1\\ 2\\ -10\end{pmatrix} = \mathbf{a}.\begin{pmatrix}1\\ 2\\ -10\end{pmatrix}\) for this mark | M1 | 1.1b |
| \(x + 2y - 10z = 2\) or \(x + 2y - 10z - 2 = 0\) | A1 | 1.1b |
| (4) |
Notes
(Corrected from the printed mark scheme: in (ii) the printed scheme lists \(\mathbf{a} = \begin{pmatrix}6\\ -3\\ 0\end{pmatrix}\); the point \(M\) is \((6, -2, 0)\), as in the notes.)
(In the printed scheme the two A1 alternatives are headed Alternative 1 and Alternative 2; they go with the M1 Alternatives 2 and 3, as in the notes, and are numbered that way here.)
(b)(i)
M1: Applies the dot product between their vectors b AND c with the vector \(\begin{pmatrix}1\\ 2\\ -10\end{pmatrix}\)
A1: Shows both dot products = 0 and concludes that the lawn is perpendicular to the vector \(\begin{pmatrix}1\\ 2\\ -10\end{pmatrix}\)
(b)(i) Alternative 1
M1: Applies the dot product between their vector b – c AND one of their vectors b or c with the vector \(\begin{pmatrix}1\\ 2\\ -10\end{pmatrix}\)
A1: Shows both dot products = 0 and concludes that the lawn is perpendicular to the vector \(\begin{pmatrix}1\\ 2\\ -10\end{pmatrix}\)
(b)(i) Alternative 2
M1: Applies the dot product between their vectors b and c \(\begin{pmatrix}x\\ y\\ z\end{pmatrix}\) and attempts to find values of \(x\), \(y\) and \(z\)
A1: Shows results is parallel to \(\begin{pmatrix}1\\ 2\\ -10\end{pmatrix}\) therefore the lawn is perpendicular
(b)(i) Alternative 3
M1: Applies the dot product between their answer to part (a) and the vector \(\begin{pmatrix}1\\ 2\\ -10\end{pmatrix}\)
A1: Achieves the value 2 and concludes as a constant therefore the lawn is perpendicular
(b)(i) Outside Specification for this paper – using the cross product
M1: Finds the cross product between ‘their b’ and ‘their c’ and shows parallel to \(\begin{pmatrix}1\\ 2\\ -10\end{pmatrix}\)
A1: Concludes parallel therefore the lawn is perpendicular
(b)(ii)
M1: Applies the formula \(\mathbf{r}.\mathbf{n} = \mathbf{a}.\mathbf{n}\) where \(\mathbf{n} = \begin{pmatrix}1\\ 2\\ -10\end{pmatrix}\) and \(\mathbf{a} = \begin{pmatrix}-2\\ -3\\ -1\end{pmatrix}\) or \(\begin{pmatrix}6\\ -2\\ 0\end{pmatrix}\) or \(\begin{pmatrix}2\\ 0\\ 0\end{pmatrix}\)
A1: Correct Cartesian equation of the plane
Note: If no method is shown then it must be correct to score M1 A1, if incorrect scores M0 A0. Look at part (i) to see if there is any method as long as it if used in (ii)
| Scheme | Marks | AO |
|---|---|---|
| Finds the vector \(\overrightarrow{PQ}\) or \(\overrightarrow{QP}\) and uses it as the direction vector in the formula \(\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}\) Two out three values correct is sufficient to imply the correct method | M1 | 3.3 |
| \(\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}\) where \(\mathbf{a} = \begin{pmatrix}-10\\ 8\\ 2\end{pmatrix}\) or \(\begin{pmatrix}6\\ 4\\ 3\end{pmatrix}\) and \(\mathbf{d} = \pm\begin{pmatrix}16\\ -4\\ 1\end{pmatrix}\) | A1 | 1.1b |
| (2) |
Notes
M1: Finds the vector \(\overrightarrow{PQ}\) or \(\overrightarrow{QP}\) and uses it as the direction vector in the formula. \(\mathbf{r} = \mathbf{a} + \lambda\mathbf{d}\). Two out three values correct is sufficient to imply the correct method
A1: A correct equation including \(\mathbf{r}\) =…
| Scheme | Marks | AO |
|---|---|---|
| For example: The lawn will not be flat The washing line will not be straight | B1 | 3.5b |
| (1) |
Notes
B1: States an acceptable limitation of the model for the lawn or washing line
| Scheme | Marks | AO |
|---|---|---|
| Applies the distance formula \(\dfrac{\left|(2 \times 1) + 5 \times 2 + (2.75 \times -10) - 2\right|}{\sqrt{1^2 + 2^2 + (-10)^2}}\) | M1 | 3.4 |
| = 1.71 m or 171 cm | A1 | 2.2b |
| (2) |
Notes
M1: Applies the distance formula using the point (2, 5, 2.75) and the normal vector \(\begin{pmatrix}1\\ 2\\ -10\end{pmatrix}\)
A1: 1.71 m or 171 cm
| Scheme | Marks | AO |
|---|---|---|
| Must have an answer to part (e). Compares their answer to part (e) with 1.5 m and makes an appropriate comment about the model that is consistent with their answer to part (e). If their answer to part (e) is close to 1.5 (e.g. 1.4 to 1.6) they must compare and conclude that the model therefore is good If their answer to part (e) is significantly different to 1.5 they must compare and conclude that the model therefore it is not a good model. | B1ft | 3.5a |
| (1) | ||
| (13 marks) |
Notes
B1ft: Compares their answer to part (e) with 1.5 and makes an assessment of the model with a reason with no contradictory statements.