A2 June 2024 Paper 1 Q15
15 Three planes have equations
\[\begin{aligned} x + ky + 3z &= 1, \\ 3x + 4y + 2z &= 3, \\ x + 3y - z &= -k, \end{aligned}\]
where \(k\) is a constant.
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{vmatrix} 1 & k & 3 \\ 3 & 4 & 2 \\ 1 & 3 & -1 \end{vmatrix}\) | M1 | 2.1 |
| \(\det\mathbf{M} = 1 \times (-10) - k \times (-5) + 3 \times 5\) | M1 | 2.1 |
| \(= 5k + 5\) | A1 | 1.1 |
| So planes meet at a point except when \(k = -1\) | A1 | 2.2a |
| [4] |
Notes
M1: Considering correct determinant
M1: A correct method to find the determinant, allow one slip. Must contain a sum of three terms.
A1: Could be implied by \(k = -1\) if working with \(\det\mathbf{M} = 0\)
A1: Must make it clear that the planes do meet at a point for all values of \(k\) other than \(-1\). Do not accept “solution” for “point”.
SC B2 if \(5k + 5\) found without working and a correct conclusion given.
Alternative method
| Scheme | Marks |
|---|---|
| \(x = \dfrac{-2k^2 + 15k + 17}{5k + 5},\ y = \dfrac{-7 - 7k}{5k + 5},\ z = \dfrac{3k^2 - k - 4}{5k + 5}\) | B2* |
| Undefined when \(k = -1\) | B1dep |
| So planes meet at a point except when \(k = -1\) | B1 |
| [4] |
B2*: Correctly finding \(x\), \(y\) or \(z\) in terms of \(k\) using simultaneous equations
B1: www. Must make it clear that the planes do meet at a point for all values of \(k\) other than \(-1\). Do not accept “solution” for “point”.
All three previous marks must have been awarded.
| Scheme | Marks | AO |
|---|---|---|
| M1 | 3.1a | |
| \(\mathbf{M}^{-1} = \dfrac{1}{5k + 5}\begin{pmatrix} -10 & k + 9 & 2k - 12 \\ \mathbf{5} & \mathbf{-4} & \mathbf{7} \\ 5 & k - 3 & 4 - 3k \end{pmatrix}\) | A1 M1 | 1.1 1.1 |
| A1 | 1.1 | |
| \(\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \dfrac{1}{5k + 5}\begin{pmatrix} -10 & k + 9 & 2k - 12 \\ 5 & -4 & 7 \\ 5 & k - 3 & 4 - 3k \end{pmatrix}\begin{pmatrix} 1 \\ 3 \\ -k \end{pmatrix}\) | M1 | 1.1 |
| \(y = \dfrac{-7 - 7k}{5k + 5} = -\dfrac{7}{5}\) which is independent of \(k\) | A1 | 3.2a |
| [6] |
Notes
M1: Finding at least two correct cofactors (could be in matrix of cofactors or adj \(\mathbf{M}\) or not in a matrix)
A1: All cofactors in bold correct
M1: Cofactor matrix transposed and multiplying by their \(\frac{1}{\det\mathbf{M}}\) (could be seen in later calculation)
A1: Correct inverse matrix (ignore first and third rows)
M1: soi (may only see middle row). Dependent on at least M1 scored. Do not accept \(-1\) substituted for \(k\), this is not MR.
A1: www, any values given in \(\mathbf{M}^{-1}\) and any \(x\) or \(z\) coordinates given must be correct. Statement of independence required.
Alternative method 1
| Scheme | Marks |
|---|---|
| \(5x + 10y = 3 - 2k\) | M1* A1 |
| \(4x + (9 + k)y = 1 - 3k\) | M1* A1 |
| \((5 + 5k)y = -7k - 7\) | M1dep |
| \(y = \dfrac{-7 - 7k}{5k + 5} = -\dfrac{7}{5}\) which is independent of \(k\) | A1cao |
| [6] |
NB This work may be seen in 15(a)
M1*: Eliminating \(x\) or \(z\) using two equations. Allow one slip only. Do not accept \(-1\) substituted for \(k\), this is not MR.
A1: Correct elimination
M1*: Eliminating same variable using different pair of equations. Allow one slip only. Do not accept \(-1\) substituted for \(k\), this is not MR.
A1: Correct elimination
M1dep: Eliminating correctly to leave in terms of \(y\) and \(k\) only
A1cao: www, any \(x\) or \(z\) coordinates given must be correct. Statement of independence required.
Alternative method 2
| Scheme | Marks |
|---|---|
| \(5x + 10y = 3 - 2k\) | M1* |
| \(x = \dfrac{3 - 2k - 10y}{5}\) | A1 |
| \((k - 3)y + 4z = 1 + k\) | M1* |
| \(z = \dfrac{1 + k - ky + 3y}{4}\) | A1 |
| \(\dfrac{3 - 2k - 10y}{5} + ky + \dfrac{3(1 + k - ky + 3y)}{4} = 1\) | M1dep |
| \(y = \dfrac{-7 - 7k}{5 + 5k} = -\dfrac{7}{5}\) which is independent of \(k\) | A1cao |
| [6] |
NB This work may be seen in 15(a)
M1*: Eliminating \(x\), \(y\) or \(z\). Allow one slip only. Do not accept \(-1\) substituted for \(k\), this is not MR.
A1: Finding (e.g.) \(x\) in terms of \(y\) and \(k\).
M1*: Eliminating another variable. Allow one slip only. Do not accept \(-1\) substituted for \(k\), this is not MR.
A1: Correctly
M1dep: Finding an equation for \(y\) in terms of \(k\) correctly
A1cao: www, any \(x\) or \(z\) coordinates given must be correct. Statement of independence required.