FP3 June 2011 Q6
6. The plane \(P\) has equation \[\mathbf{r} = \begin{pmatrix} 3 \\ 1 \\ 2 \end{pmatrix} + \lambda\begin{pmatrix} 0 \\ 2 \\ -1 \end{pmatrix} + \mu\begin{pmatrix} 3 \\ 2 \\ 2 \end{pmatrix}\]
The line \(l\) passes through the point \(A\ (1, 3, 3)\) and meets \(P\) at \((3, 1, 2)\).
The acute angle between the plane \(P\) and the line \(l\) is \(\alpha\).
| Scheme | Marks |
|---|---|
| \(\mathbf{n} = (2\mathbf{j} - \mathbf{k}) \times (3\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) = 6\mathbf{i} - 3\mathbf{j} - 6\mathbf{k}\) o.a.e. (e.g. \(2\mathbf{i} - \mathbf{j} - 2\mathbf{k}\)) | M1 A1 |
| (2) |
Notes
M1 Cross product of the correct vectors
A1 CAO o.e.
| Scheme | Marks |
|---|---|
| Line \(l\) has direction \(2\mathbf{i} - 2\mathbf{j} - \mathbf{k}\) | B1 |
| Angle between line \(l\) and normal is given by \((\cos\beta\) or \(\sin\alpha) = \dfrac{4 + 2 + 2}{\sqrt{9}\sqrt{9}} = \dfrac{8}{9}\) | M1 A1ft |
| \(\alpha = 90 - \beta = 63\) degrees to nearest degree. | A1 awrt |
| (4) |
Notes
B1 CAO
M1 Angle between ‘\(2\mathbf{i} - \mathbf{j} - 2\mathbf{k}\)’ and \(2\mathbf{i} - 2\mathbf{j} - \mathbf{k}\), formula of correct form
1A1ft 8/9ft
2A1 CAO awrt
| Scheme | Marks |
|---|---|
| Alt 1 Plane \(P\) has equation \(\mathbf{r} \cdot (2\mathbf{i} - \mathbf{j} - 2\mathbf{k}) = 1\) | M1 A1 |
| Perpendicular distance is \(\dfrac{1 - (-7)}{\sqrt{9}} = \dfrac{8}{3}\) | M1 A1 |
| (4) | |
| (10 marks) |
Notes
(c) Alt 2
| Scheme | Marks |
|---|---|
| Parallel plane through A has equation \(\mathbf{r} \cdot \dfrac{2\mathbf{i} - \mathbf{j} - 2\mathbf{k}}{3} = \dfrac{-7}{3}\) | M1 A1 |
| Plane P has equation \(\mathbf{r} \cdot \dfrac{2\mathbf{i} - \mathbf{j} - 2\mathbf{k}}{3} = \dfrac{1}{3}\) | M1 |
| So O lies between the two and perpendicular distance is \(\dfrac{1}{3} + \dfrac{7}{3} = \dfrac{8}{3}\) | A1 |
| (4) |
(c) Alt 3
| Scheme | Marks |
|---|---|
| Distance A to \((3, 1, 2) = \sqrt{2^2 + 2^2 + 1^2} = 3\) | M1A1 |
| Perpendicular distance is ‘3’ \(\sin\alpha = 3 \times \dfrac{8}{9} = \dfrac{8}{3}\) | M1A1 |
| (4) |
(c) Alt 4
| Scheme | Marks |
|---|---|
| Finding Cartesian equation of plane P: \(2x - y - 2z - 1 = 0\) | M1 A1 |
| \(d = \dfrac{|n_1\alpha + n_2\beta + n_3\gamma + d|}{\sqrt{n_1^2 + n_2^2 + n_3^2}} = \dfrac{|2(1) - 1(3) - 2(3) - 1|}{\sqrt{2^2 + 1^2 + 2^2}} = \dfrac{8}{3}\) | M1A1 |
| (4) |
1M1 Eqn of plane using \(2\mathbf{i} - \mathbf{j} - 2\mathbf{k}\) or dist of A from O or finding length of AP
1A1 Correct equation (must have =) or A to \((3, 1, 2) = 3\)
2M1 Using correct method to find perpendicular distance
2A1 CAO