FP3 June 2010 Q7
7. The plane \(\Pi\) has vector equation \[\mathbf{r} = 3\mathbf{i} + \mathbf{k} + \lambda(-4\mathbf{i} + \mathbf{j}) + \mu(6\mathbf{i} - 2\mathbf{j} + \mathbf{k})\]
(a) Find an equation of \(\Pi\) in the form \(\mathbf{r} \cdot \mathbf{n} = p\), where \(\mathbf{n}\) is a vector perpendicular to \(\Pi\) and \(p\) is a constant. (5)
The point \(P\) has coordinates \((6, 13, 5)\). The line \(l\) passes through \(P\) and is perpendicular to \(\Pi\). The line \(l\) intersects \(\Pi\) at the point \(N\).
(b) Show that the coordinates of \(N\) are \((3, 1, -1)\). (4)
The point \(R\) lies on \(\Pi\) and has coordinates \((1, 0, 2)\).
(c) Find the perpendicular distance from \(N\) to the line \(PR\). Give your answer to 3 significant figures. (5)
| Scheme | Marks |
|---|---|
| \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -4 & 1 & 0 \\ 6 & -2 & 1 \end{vmatrix} = \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix}\) | M1 A2(1,0) |
| \(\begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} \bullet \begin{pmatrix} 3 \\ 0 \\ 1 \end{pmatrix} = 5\) \(\mathbf{r} \bullet \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} = 5\) | M1A1 |
| (5) |
| Scheme | Marks |
|---|---|
| Equation of \(l\) is \(\mathbf{r} = \begin{pmatrix} 6 \\ 13 \\ 5 \end{pmatrix} + t\begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix}\) | M1 |
| At intersection \(\begin{pmatrix} 6 + t \\ 13 + 4t \\ 5 + 2t \end{pmatrix} \bullet \begin{pmatrix} 1 \\ 4 \\ 2 \end{pmatrix} = 5\) | M1 |
| \(\Rightarrow 6 + t + 4\left(13 + 4t\right) + 2\left(5 + 2t\right) = 5 \Rightarrow t = -3\) | M1 |
| N is \(\left(3, 1, -1\right)\) * | A1 |
| (4) |
| Scheme | Marks |
|---|---|
| \(\overrightarrow{PN} \bullet \overrightarrow{PR} = \left(-3\mathbf{i} - 12\mathbf{j} - 6\mathbf{k}\right) \bullet \left(-5\mathbf{i} - 13\mathbf{j} - 3\mathbf{k}\right) = 189\) | M1 A1ft |
| \(\sqrt{9 + 144 + 36}\sqrt{25 + 169 + 9}\cos NPR = 189\) | A1 |
| \(NX = NP\sin NPR = \sqrt{189}\sin NPR = 3.61\) | M1A1 |
| (5) | |
| (14 marks) |