A2 October 2020 Paper 2 Q6
6.
\[\mathbf{M} = \begin{pmatrix}k & 5 & 7\\ 1 & 1 & 1\\ 2 & 1 & -1\end{pmatrix} \qquad \text{where } k \text{ is a constant}\]| Scheme | Marks | AO |
|---|---|---|
| \(|\mathbf{M}| = k(-1 - 1) - 5(-1 - 2) + 7(1 - 2)\) \(\{= 8 - 2k\}\) | M1 | 1.1b |
| Minors: \(\begin{pmatrix}-2 & -3 & -1\\ -12 & -k - 14 & k - 10\\ -2 & k - 7 & k - 5\end{pmatrix}\) Cofactors: \(\begin{pmatrix}-2 & 3 & -1\\ 12 & -k - 14 & 10 - k\\ -2 & 7 - k & k - 5\end{pmatrix}\) | M1 | 1.1b |
| \(\mathbf{M}^{-1} = \dfrac{1}{8 - 2k}\begin{pmatrix}-2 & 12 & -2\\ 3 & -k - 14 & 7 - k\\ -1 & 10 - k & k - 5\end{pmatrix}\) | M1 A1 | 2.1 1.1b |
| (4) |
Notes
(a)
M1: Correct method to find the determinant. Condone one sign slip
M1: A correct first step in obtaining the inverse. Could be the matrix of minors or cofactors. Condone sign slips as long as the intention is clear.
M1: Fully correct method to obtain the inverse. Attempts matrix of minors, cofactors, transposes and 1/determinant
A1: Correct matrix
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{M}^{-1} = \dfrac{1}{4}\begin{pmatrix}-2 & 12 & -2\\ 3 & -16 & 5\\ -1 & 8 & -3\end{pmatrix} \Rightarrow \begin{pmatrix}x\\ y\\ z\end{pmatrix} = \mathbf{M}^{-1}\begin{pmatrix}1\\ p\\ 2\end{pmatrix}\) Solve the equations simultaneously to achieve values for \(x\), \(y\) and \(z\) \(y + 3z = 2p - 2\) and \(4y + 8z = -1 \Rightarrow x = \ldots,\ y = \ldots,\ z = \ldots\) | M1 | 3.1a |
| \(\begin{pmatrix}x\\ y\\ z\end{pmatrix} = \begin{pmatrix}-\frac{1}{2} + 3p - 1\\[2pt] \frac{3}{4} - 4p + \frac{5}{2}\\[2pt] -\frac{1}{4} + 2p - \frac{3}{2}\end{pmatrix}\) | A1ft | 1.1b |
| \(\left(\dfrac{12p - 6}{4}, \dfrac{13 - 16p}{4}, \dfrac{8p - 7}{4}\right)\) \(\left(3p - \dfrac{3}{2}, \dfrac{13}{4} - 4p, 2p - \dfrac{7}{4}\right)\) | A1 | 2.2a |
| (3) |
Notes
(b)
M1: A complete strategy for solving the given equations e.g. multiplies the given coordinates by their inverse or solves simultaneously to achieve values for \(x\), \(y\) and \(z\)
A1ft: Correct calculation on their inverse matrix (unsimplified) or at least on correct value if solving simultaneously
A1: Correct coordinates
(c)(i)
| Scheme | Marks | AO | |||
|---|---|---|---|---|---|
| M1 | 3.1a | |||
| M1 | 1.1b | |||
| \(q = \dfrac{1}{2}\) | A1 | 1.1b |
Alternative (c)(i)
| Scheme | Marks | AO |
|---|---|---|
| Equating coefficients leading to two out of three equations and solves to find values for a and b \(4a + b = 2,\ 5a + b = 1,\ 7a + b = -1\) \(\{a = -1,\ b = 6\}\) | M1 | 3.1a |
| Forms the fourth equation involving \(q\) \(a + bq = 2\) and substitutes in the values of \(a\) and \(b\) to finds a value for \(q\) | M1 | 1.1b |
| \(q = \dfrac{1}{2}\) | A1 | 1.1b |
Alternative (c)(i)
| Scheme | Marks | AO |
|---|---|---|
| Finds a coordinate of intersection of the planes \(4x + 5y + 7z = 1\) and \(2x + y - z = 2\) e.g let \(z = 0 \Rightarrow 4x + 5y = 1\) and \(2x + y = 2 \Rightarrow y = -1,\ x = 1.5\) | M1 | 3.1a |
| Substitutes the values for \(x\), \(y\) and \(z\) into \(x + y + z = q\) to reach a value for \(q\) | M1 | 1.1b |
| \(q = \dfrac{1}{2}\) | A1 | 1.1b |
(c)(ii)
| Scheme | Marks | AO |
|---|---|---|
For example: \(x = \lambda \Rightarrow 3\lambda + 2y = \dfrac{5}{2},\ \lambda - 2z = \dfrac{3}{2} \Rightarrow y = \mathrm{f}(\lambda), z = \mathrm{f}(\lambda)\) \(y = \lambda \Rightarrow 3x + 2\lambda = \dfrac{5}{2},\ \lambda + 3z = -1 \Rightarrow x = \mathrm{f}(\lambda), z = \mathrm{f}(\lambda)\) \(z = \lambda \Rightarrow 3y + 9\lambda = -3,\ -6x + 12\lambda = -9 \Rightarrow x = \mathrm{f}(\lambda), y = \mathrm{f}(\lambda)\) | M1 | 3.1a |
Let \(x = \lambda\), \(\lambda = \dfrac{y - \frac{5}{4}}{-\frac{3}{2}} = \dfrac{z + \frac{3}{4}}{\frac{1}{2}}\) or \(y = \dfrac{5}{4} - \dfrac{3}{2}\lambda,\ z = -\dfrac{3}{4} + \dfrac{1}{2}\lambda\) Let \(y = \lambda\), \(\lambda = \dfrac{x - \frac{5}{6}}{-\frac{2}{3}} = \dfrac{z + \frac{1}{3}}{-\frac{1}{3}}\) or \(x = \dfrac{5}{6} - \dfrac{2}{3}\lambda,\ z = -\dfrac{1}{3} - \dfrac{1}{3}\lambda\) Let \(z = \lambda\), \(\lambda = \dfrac{x - \frac{3}{2}}{2} = \dfrac{y + 1}{-3}\) or \(x = \dfrac{3}{2} + 2\lambda,\ y = -1 - 3\lambda\) | A1 | 1.1b |
\(\mathbf{r} = \dfrac{5}{4}\mathbf{j} - \dfrac{3}{4}\mathbf{k} + t(2\mathbf{i} - 3\mathbf{j} + \mathbf{k})\) o.e. \(\mathbf{r} = \dfrac{5}{6}\mathbf{i} - \dfrac{1}{3}\mathbf{k} + t(2\mathbf{i} - 3\mathbf{j} + \mathbf{k})\) o.e. \(\mathbf{r} = \dfrac{3}{2}\mathbf{i} - \mathbf{j} + t(2\mathbf{i} - 3\mathbf{j} + \mathbf{k})\) o.e. | M1 A1 | 1.1b 2.5 |
| (7) | ||
| (14 marks) |
Alternative (c)(ii)
| Scheme | Marks | AO |
|---|---|---|
| Finds two different coordinates that lie on the line of intersection For example: setting \(x = 0 \Rightarrow \left(0, \dfrac{5}{4}, -\dfrac{3}{4}\right)\) setting \(y = 0 \Rightarrow \left(\dfrac{5}{6}, 0, -\dfrac{1}{3}\right)\) setting \(z = 0 \Rightarrow \left(\dfrac{3}{2}, -1, 0\right)\) | M1 A1 | 3.1a 1.1b |
| Finds the vector equation of the line passing through their two points | M1 | 1.1b |
\(\mathbf{r} = \dfrac{5}{4}\mathbf{j} - \dfrac{3}{4}\mathbf{k} + t(2\mathbf{i} - 3\mathbf{j} + \mathbf{k})\) o.e. \(\mathbf{r} = \dfrac{5}{6}\mathbf{i} - \dfrac{1}{3}\mathbf{k} + t(2\mathbf{i} - 3\mathbf{j} + \mathbf{k})\) o.e. \(\mathbf{r} = \dfrac{3}{2}\mathbf{i} - \mathbf{j} + t(2\mathbf{i} - 3\mathbf{j} + \mathbf{k})\) o.e. | A1 | 2.5 |
Alternative (c)(ii): Outside the spec
| Scheme | Marks | AO |
|---|---|---|
| Finds the cross product of two normal vectors and a coordinate that lies on all three planes | M1 | 3.1a |
| Correct cross product \(\begin{vmatrix}4 & 5 & 7\\ 1 & 1 & 1\end{vmatrix} = \begin{vmatrix}4 & 5 & 7\\ 2 & 1 & -1\end{vmatrix} = \begin{pmatrix}2\\ 3\\ -1\end{pmatrix}\) or \(\begin{vmatrix}1 & 1 & 1\\ 2 & 1 & -1\end{vmatrix} = \begin{pmatrix}12\\ 18\\ -3\end{pmatrix}\) | A1 | 1.1b |
| Uses the point and the direction vector to find the equation of the line | M1 | 1.1b |
| \(\mathbf{r} = \dfrac{5}{4}\mathbf{j} - \dfrac{3}{4}\mathbf{k} + t(2\mathbf{i} - 3\mathbf{j} + \mathbf{k})\) o.e. | A1 | 1.1b |
| (7) |
Notes
(c)(i)
M1: Uses a correct strategy that will lead to establishing a value for \(q\). E.g. eliminating one of \(x\), \(y\) or \(z\)
M1: Solves a suitable equation to obtain a value for \(q\)
A1: Correct value
(c)(i) Alternative 1
M1: Equating coefficients leading to two out of three equations and solves to find values for a and b
M1: Solves a suitable equation to obtain a value for \(q\) using their values for \(a\) and \(b\)
(c)(i) Alternative 2
M1: Finds a coordinate of intersection of the planes \(4x + 5y + 7z = 1\) and \(2x + y - z = 2\)
M1: Substitutes the values for \(x\), \(y\) and \(z\) into \(x + y + z = q\) to reach a value for \(q\)
A1: Correct value
(ii)
M1: Uses a correct strategy to obtain the Cartesian equation of the line or the general coordinates
A1: Correct Cartesian equation or coordinates in terms of a parameter.
M1: Uses their Cartesian equation to correctly extract the position and direction to form a vector equation for the required line
A1: Correct equation (o.e.) look out for multiples of the direction vector
Alternative (ii)
M1: Finds two different coordinates that lie on the line of intersection
A1: Correct coordinates
M1: Uses their coordinates to find the vector equation of the line that passes through them.
A1: Correct equation (o.e.), look out for multiples of the direction vector. Must have \(\mathbf{r} = \ldots\)
Alternative (ii) outside spec
M1: Finds the cross product between the normal vectors of two of the planes and a coordinate that lies on all three planes. If a coordinate is found in (i) it must be used in this part to award this mark.
A1: Correct cross product
M1: Uses the coordinate and the cross product to find the equation of the line
A1: Correct equation (o.e.), look out for multiples of the direction vector. Must have \(\mathbf{r} = \ldots\)
(corrected from the printed mark scheme: in (c)(ii), \(\lambda + 3z = -1\) is printed as \(\lambda + 3z = 1\), and \(x = \dfrac{5}{6} - \dfrac{2}{3}\lambda\) is printed as \(x = \dfrac{5}{5} - \dfrac{2}{3}\lambda\))
(Note: the cross products in the “outside the spec” alternative are as printed. The correct values are \((4\mathbf{i} + 5\mathbf{j} + 7\mathbf{k}) \times (\mathbf{i} + \mathbf{j} + \mathbf{k}) = -2\mathbf{i} + 3\mathbf{j} - \mathbf{k}\), \((4\mathbf{i} + 5\mathbf{j} + 7\mathbf{k}) \times (2\mathbf{i} + \mathbf{j} - \mathbf{k}) = -12\mathbf{i} + 18\mathbf{j} - 6\mathbf{k}\) and \((\mathbf{i} + \mathbf{j} + \mathbf{k}) \times (2\mathbf{i} + \mathbf{j} - \mathbf{k}) = -2\mathbf{i} + 3\mathbf{j} - \mathbf{k}\), each a multiple of \(2\mathbf{i} - 3\mathbf{j} + \mathbf{k}\).)