A2 October 2020 Paper 1 Q15
15
(a) Show that the three planes with equations\[\begin{aligned} x + \lambda y + 3z &= -12 \\ 2x + y + 5z &= -11 \\ x - 2y + 2z &= -9 \end{aligned}\]where \(\lambda\) is a constant, meet at a unique point except for one value of \(\lambda\) which is to be determined. [3]
(b) In the case \(\lambda = -2\), use matrices to find the point of intersection P of the planes, showing your method clearly. [3]
The line \(l\) has equation \(\dfrac{x - 1}{2} = \dfrac{y - 1}{-1} = \dfrac{z + 2}{-2}\).
(c) Find a vector equation of \(l\). [2]
(d) Find the shortest distance between the point P and \(l\). [4]
(e)
(i) Show that \(l\) is parallel to the plane \(x - 2y + 2z = -9\). [3]
(ii) Find the distance between \(l\) and the plane \(x - 2y + 2z = -9\). [2]
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{vmatrix} 1 & \lambda & 3 \\ 2 & 1 & 5 \\ 1 & -2 & 2 \end{vmatrix} = 1(2 + 10) - \lambda(4 - 5) + 3(-4 - 1)\) | M1 | 3.1a |
| \(= \lambda - 3\) | A1 | 1.1 |
| det \(= 0\) when \(\lambda = 3\) So unique point provided \(\lambda \neq 3\) | B1ft | 1.1 |
| [3] |
Notes
M1: calculating determinant, or full attempt to solve finding \(x\), \(y\) or \(z\) in terms of \(\lambda\)
B1ft: ft their \(\lambda\)
| Scheme | Marks | AO |
|---|---|---|
| \(\mathbf{M} = \begin{pmatrix} 1 & -2 & 3 \\ 2 & 1 & 5 \\ 1 & -2 & 2 \end{pmatrix}\) | M1 | 1.1 |
| \(\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \mathbf{M}^{-1}\begin{pmatrix} -12 \\ -11 \\ -9 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix}\) | M1 | 2.4 |
| \(\Rightarrow x = 1,\ y = 2,\ z = -3\) | A1 | 1.1 |
| [3] |
Notes
M1: matrix of coefficients or \(M^{-1}\) shown, or attempt to use row ops
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{x - 1}{2} = \dfrac{y - 1}{-1} = \dfrac{z + 2}{-2} = \lambda\) \(x = 1 + 2\lambda,\ y = 1 - \lambda,\ z = -2 - 2\lambda\) | M1 | 1.1 |
| \(\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ -1 \\ -2 \end{pmatrix}\) | A1 | 1.1 |
| [2] |
Notes
A1: oe
| Scheme | Marks | AO |
|---|---|---|
| \(\overrightarrow{\mathrm{AP}} = \begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix} - \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix}\) | B1ft | 1.1 |
| \(\overrightarrow{\mathrm{AP}} \times \mathbf{u} = \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix} \times \begin{pmatrix} 2 \\ -1 \\ -2 \end{pmatrix} = \begin{pmatrix} -3 \\ -2 \\ -2 \end{pmatrix}\) | M1 | 1.1 |
| \(d = \dfrac{\left|\overrightarrow{\mathrm{AP}} \times \mathbf{u}\right|}{|\mathbf{u}|} = \dfrac{\sqrt{17}}{\sqrt{9}} = \dfrac{\sqrt{17}}{3}\) | M1 A1 | 1.1 1.1 |
| [4] |
Notes
B1ft: ft their P; or using first principles
A1: or 1.37 or better
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\mathbf{n} = \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}\) | B1 | 2.1 |
| \(\mathbf{n}.\mathbf{u} = \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}.\begin{pmatrix} 2 \\ -1 \\ -2 \end{pmatrix} = 2 + 2 - 4 = 0\) | M1 | 2.1 |
| \(\Rightarrow\) line \(l\) is parallel to the plane | A1 | 2.2a |
| [3] | ||
| (ii) distance between \((1, 1, -2)\) and \(x - 2y + 2z = -9\) \(= \dfrac{|1 - 2 \times 1 + 2 \times -2 + 9|}{\sqrt{1^2 + (-2)^2 + 2^2}} = \dfrac{4}{3}\) | M1 A1 | 3.1a 1.1 |
| [2] |
Notes
(e)(i)
B1: soi
(e)(ii)
M1: ft position vector given in 15c; or using first principles
A1: or 1.33 or better