AS June 2025 Paper 1 Q10
10. The plane \(\Pi_1\) has equation \(x + y - z = 3\)
The plane \(\Pi_2\) has equation \(ax + 3y + 5z = 4\) where \(a\) is an integer.
Given that \(\Pi_1\) is perpendicular to \(\Pi_2\)
The plane \(\Pi_3\) has equation \(x + by + 13z = c\) where \(b\) and \(c\) are integers.
Given that the three planes form a sheaf,
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{pmatrix}1\\ 1\\ -1\end{pmatrix} \bullet \begin{pmatrix}a\\ 3\\ 5\end{pmatrix} = a + 3 - 5 = 0 \Rightarrow a = \ldots\) | M1 | 1.1b |
| \(a = 2\) | A1 | 1.1b |
| (2) |
Notes
M1: Finds the dot product of the normal vectors, sets equal to 0 and finds a value for \(a\). Condone a sign slip.
A1: Correct value for \(a\)
| Scheme | Marks | AO |
|---|---|---|
| e.g. \(\begin{vmatrix}1 & 1 & -1\\ 2 & 3 & 5\\ 1 & b & 13\end{vmatrix} = 1(39 - 5b) - 1(26 - 5) - 1(2b - 3) = 0\) | M1 A1 | 3.1a 1.1b |
| \(b = 3\) | A1 | 1.1b |
| Uses the planes to eliminate one variable to form two equations e.g \(\left.\begin{aligned}2x + 2y - 2z &= 6\\ 2x + 3y + 5z &= 4\end{aligned}\right\} \Rightarrow -y - 7z = 2\) and \(\left.\begin{aligned}x + y - z &= 3\\ x + 3y + 13z &= c\end{aligned}\right\} \Rightarrow -2y - 14z = 3 - c\) | dM1 | 3.1a |
| Uses their equations to form and solve an equation for \(c\) e.g. \(3 - c = 4 \Rightarrow c = \ldots\) | ddM1 | 1.1b |
| \(c = -1\) cso | A1 | 1.1b |
| (6) | ||
| (8 marks) |
Notes
M1: Finds the determinant of a 3 by 3 matrix and sets = 0, this may be implied
A1: Correct determinant
A1: Correct value for \(b\)
dM1: Uses the equations of the three planes to find two equations by eliminating the same variable. Condone a slip on one coefficient or constant term
ddM1: Correctly uses their equations to form and solve an equation for \(c\)
A1: Correct value for \(c\)
Alternative 1
| Scheme | Marks | AO |
|---|---|---|
| Uses the planes to eliminate one variable to form two equations e.g \(\left.\begin{aligned}2x + 2y - 2z &= 6\\ 2x + 3y + 5z &= 4\end{aligned}\right\} \Rightarrow -y - 7z = 2\) and \(\left.\begin{aligned}x + y - z &= 3\\ x + by + 13z &= c\end{aligned}\right\} \Rightarrow (1 - b)y - 14z = 3 - c\) | M1 A1 A1 | 3.1a 1.1b 1.1b |
| \(1 - b = -2 \Rightarrow b = \ldots\) or \(3 - c = 4 \Rightarrow c = \ldots\) | dM1 | 3.1a |
| \(1 - b = -2 \Rightarrow b = \ldots\) and \(3 - c = 4 \Rightarrow c = \ldots\) | ddM1 | 1.1b |
| \(b = 3\) and \(c = -1\) cso | A1 | 1.1b |
| (6) |
M1: Uses the planes to eliminate one variable to form two equations
A1: One correct equation
A1: two correct equations
dM1: Sets up an equation for \(b\) or \(c\) by comparing coefficients and solves to find a value. Condone a slip on one coefficient or constant term
ddM1: Solves equations to find values for \(b\) and \(c\)
A1: Correct value for \(b\) and \(c\) cso
Note for 10 (b) using Alternative 1
| Eliminating \(x\) gives | Eliminating \(y\) gives | Eliminating \(z\) gives |
|---|---|---|
| \(y + 7z = -2\) | \(x - 8z = 5\) | \(7x + 8y = 19\) |
| \((b - 1)y + 14z = c - 3\) | \((b - 1)x + (-b - 13)z = 3b - c\) | \(14x + (13 + b)y = 39 + c\) |
| \((2b - 3)y + 21z = 2c - 4\) | \((2b - 3)x + (5b - 39)z = 4b - 3c\) | \(21x + (39 - 5b)y = 52 - 5c\) |
Alternative 2
| Scheme | Marks | AO |
|---|---|---|
| eliminates one variable to form two equations e.g. \(\left.\begin{aligned}2x + 2y - 2z &= 6\\ 2x + 3y + 5z &= 4\end{aligned}\right\} \Rightarrow -y - 7z = 2 \Rightarrow y = -2 - 7z\) \(x + y - z = 3 \Rightarrow x + (-2 - 7z) - z = 3 \Rightarrow x = 5 + 8z\) Or Selects values for the coordinates to find two points that lie on the line of intersection e.g. \(z = 0 \Rightarrow x + y = 3\) and \(2x + 3y = 4 \Rightarrow (5, -2, 0)\) \(z = 1 \Rightarrow x + y - 1 = 3\) and \(2x + 3y - 5 = 4 \Rightarrow (13, -9, 1)\) | M1 A1 A1 | 3.1a 1.1b 1.1b |
| \(\begin{pmatrix}x\\ y\\ z\end{pmatrix} = \begin{pmatrix}5\\ -2\\ 0\end{pmatrix} + \lambda\begin{pmatrix}8\\ -7\\ 1\end{pmatrix}\) \(\begin{pmatrix}8\\ -7\\ 1\end{pmatrix}\) is perpendicular to \(\begin{pmatrix}1\\ b\\ 13\end{pmatrix}\) leading to \(8 - 7b + 13 = 0 \Rightarrow b = \ldots\) Or \(\begin{pmatrix}13\\ -9\\ 1\end{pmatrix} - \begin{pmatrix}5\\ -2\\ 0\end{pmatrix} = \begin{pmatrix}8\\ -7\\ 1\end{pmatrix}\) then \(\begin{pmatrix}8\\ -7\\ 1\end{pmatrix}\) is perpendicular to \(\begin{pmatrix}1\\ b\\ 13\end{pmatrix}\) leading to \(8 - 7b + 13 = 0 \Rightarrow b = \ldots\) | dM1 | 3.1a |
| \((5, -2, 0)\) lies on \(\Pi_3\) leading to \(5 - 2b = c\) Uses their \(b\) to find a value for \(c\) | ddM1 | 1.1b |
| \(b = 3\ \ c = -1\) cso | A1 | 1.1b |
| (6) |
M1: Uses the planes to eliminate one variable to form two equations or substitutes in two values for one coordinate and solves to find two coordinates that lie on the line of intersection
A1: One correct equation or coordinate
A1: Two correct equations or coordinates
dM1: \(\begin{pmatrix}8\\ -7\\ 1\end{pmatrix}\) is perpendicular to \(\begin{pmatrix}1\\ b\\ 13\end{pmatrix}\) leading to \(8 - 7b + 13 = 0 \Rightarrow b = \ldots\) or uses their coordinates to find the direction vector and uses that it is perpendicular to \(\begin{pmatrix}1\\ b\\ 13\end{pmatrix}\) to find a value for \(b\)
ddM1: \((5, -2, 0)\) lies on \(\Pi_3\) leading to \(5 - 2b = c\)
A1: Correct values for \(b\) and \(c\)