FP3 June 2013 Q8
8. The plane \(\Pi_1\) has vector equation \[\mathbf{r} \cdot (3\mathbf{i} - 4\mathbf{j} + 2\mathbf{k}) = 5\]
(a) Find the perpendicular distance from the point \((6, 2, 12)\) to the plane \(\Pi_1\) (3)
The plane \(\Pi_2\) has vector equation \[\mathbf{r} = \lambda(2\mathbf{i} + \mathbf{j} + 5\mathbf{k}) + \mu(\mathbf{i} - \mathbf{j} - 2\mathbf{k}), \quad \text{where } \lambda \text{ and } \mu \text{ are scalar parameters.}\]
(b) Find the acute angle between \(\Pi_1\) and \(\Pi_2\) giving your answer to the nearest degree. (5)
(c) Find an equation of the line of intersection of the two planes in the form \(\mathbf{r} \times \mathbf{a} = \mathbf{b}\), where \(\mathbf{a}\) and \(\mathbf{b}\) are constant vectors. (6)
| Scheme | Marks |
|---|---|
| \((6\mathbf{i} + 2\mathbf{j} + 12\mathbf{k}) \cdot (3\mathbf{i} - 4\mathbf{j} + 2\mathbf{k}) = 34\) Attempt scalar product | M1 |
| \(\left|\dfrac{(6\mathbf{i} + 2\mathbf{j} + 12\mathbf{k}) \cdot (3\mathbf{i} - 4\mathbf{j} + 2\mathbf{k}) - 5}{\sqrt{3^2 + 4^2 + 2^2}}\right|\) Use of correct formula | M1 |
| \(\sqrt{29}\) (not \(-\sqrt{29}\)) Correct distance (Allow \(29/\sqrt{29}\)) | A1 |
| (3) |
Notes
(a) Way 2
| Scheme | Marks |
|---|---|
| \(\mathbf{r} = (6\mathbf{i} + 2\mathbf{j} + 12\mathbf{k}) + \lambda(3\mathbf{i} - 4\mathbf{j} + 2\mathbf{k})\) \(\therefore (6 + 3\lambda)3 + (2 - 4\lambda)(-4) + (12 + 2\lambda)2 = 5\) Substitutes the parametric coordinates of the line through (6, 2, 12) perpendicular to the plane into the cartesian equation. | M1 |
| \(\lambda = -1 \Rightarrow (3, 6, 10)\) or \(-3\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}\) Solves for \(\lambda\) to obtain the required point or vector. | M1 |
| \(\sqrt{29}\) Correct distance | A1 |
(a) Way 3
| Scheme | Marks |
|---|---|
| Parallel plane containing \((6, 2, 12)\) is \(\mathbf{r} \cdot (3\mathbf{i} - 4\mathbf{j} + 2\mathbf{k}) = 34\) \(\Rightarrow \dfrac{\mathbf{r} \cdot (3\mathbf{i} - 4\mathbf{j} + 2\mathbf{k})}{\sqrt{29}} = \dfrac{34}{\sqrt{29}}\) Origin to this plane is \(\dfrac{34}{\sqrt{29}}\) | M1 |
| \(\Rightarrow \dfrac{\mathbf{r} \cdot (3\mathbf{i} - 4\mathbf{j} + 2\mathbf{k})}{\sqrt{29}} = \dfrac{5}{\sqrt{29}}\) Origin to plane is \(\dfrac{5}{\sqrt{29}}\) | M1 |
| \(\dfrac{34}{\sqrt{29}} - \dfrac{5}{\sqrt{29}} = \sqrt{29}\) Correct distance | A1 |
| Scheme | Marks |
|---|---|
| \(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 1 & 5 \\ 1 & -1 & -2\end{vmatrix} = \begin{pmatrix}3 \\ 9 \\ -3\end{pmatrix}\) M1: Attempts \((2\mathbf{i} + 1\mathbf{j} + 5\mathbf{k}) \times (\mathbf{i} - \mathbf{j} - 2\mathbf{k})\) A1: Any multiple of \(\mathbf{i} + 3\mathbf{j} - \mathbf{k}\) | M1A1 |
| \((\cos\theta) = \dfrac{(3\mathbf{i} - 4\mathbf{j} + 2\mathbf{k}) \cdot (\mathbf{i} + 3\mathbf{j} - \mathbf{k})}{\sqrt{3^2 + 4^2 + 2^2}\sqrt{1^2 + 3^2 + 1^2}} \quad \left(= \dfrac{-11}{\sqrt{29}\sqrt{11}}\right)\) Attempts scalar product of normal vectors including magnitudes | M1 |
| 52 Obtains angle using arccos (dependent on previous M1) | dM1 A1 |
| Do not isw and mark the final answer e.g. 90 – 52 = 38 loses the A1 | |
| (5) |
Notes
For a cross product, if the method is unclear, 2 out of 3 components should be correct for M1
| Scheme | Marks |
|---|---|
| \(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 3 & -1 \\ 3 & -4 & 2\end{vmatrix} = \begin{pmatrix}2 \\ -5 \\ -13\end{pmatrix}\) M1: Attempt cross product of normal vectors A1: Correct vector | M1A1 |
| \(x = 0: \left(0, \dfrac{5}{2}, \dfrac{15}{2}\right),\quad y = 0: (1, 0, 1),\quad z = 0: \left(\dfrac{15}{13}, \dfrac{-5}{13}, 0\right)\) M1: Valid attempt at a point on both planes. A1: Correct coordinates May use way 3 to find a point on the line | M1A1 |
| \(\mathbf{r} \times (-2\mathbf{i} + 5\mathbf{j} + 13\mathbf{k}) = -5\mathbf{i} - 15\mathbf{j} + 5\mathbf{k}\) M1: \(\mathbf{r} \times \text{dir} = \text{pos.vector} \times \text{dir}\) (This way round) A1: Correct equation | M1A1 |
| (6) | |
| (14 marks) |
Notes
(c) Way 2
| Scheme | Marks |
|---|---|
| "\(x + 3y - z = 0\)" and \(3x - 4y + 2z = 5\) uses their cartesian form of \(\Pi_2\) and eliminate \(x\), or \(y\) or \(z\) and substitutes back to obtain two of the variables in terms of the third | M1 |
| \(\left(x = 1 - \tfrac{2}{5}y \text{ and } z = 1 + \tfrac{13}{5}y\right)\) or \(\left(y = \dfrac{5z - 5}{13} \text{ and } x = \dfrac{15 - 2z}{13}\right)\) or \(\left(y = \dfrac{5 - 5x}{2} \text{ and } z = \dfrac{15 - 13x}{2}\right)\) | A1 |
| Cartesian Equations: \(x = \dfrac{y - \frac{5}{2}}{-\frac{5}{2}} = \dfrac{z - \frac{15}{2}}{-\frac{13}{2}}\) or \(\dfrac{x - 1}{-\frac{2}{5}} = y = \dfrac{z - 1}{\frac{13}{5}}\) or \(\dfrac{x - \frac{15}{13}}{-\frac{2}{13}} = \dfrac{y + \frac{5}{13}}{\frac{5}{13}} = z\) | |
| Points and Directions: Direction can be any multiple \(\left(0, \tfrac{5}{2}, \tfrac{15}{2}\right), \mathbf{i} - \tfrac{5}{2}\mathbf{j} - \tfrac{13}{2}\mathbf{k}\) or \((1, 0, 1), -\tfrac{2}{5}\mathbf{i} + \mathbf{j} + \tfrac{13}{5}\mathbf{k}\) or \(\left(\tfrac{15}{13}, -\tfrac{5}{13}, 0\right), -\tfrac{2}{13}\mathbf{i} + \tfrac{5}{13}\mathbf{j} + \mathbf{k}\) M1: Uses their Cartesian equations correctly to obtain a point and direction A1: Correct point and direction – it may not be clear which is which – i.e. look for the correct numbers either as points or vectors | M1 A1 |
| Equation of line in required form: e.g. \(\mathbf{r} \times (-2\mathbf{i} + 5\mathbf{j} + 13\mathbf{k}) = -5\mathbf{i} - 15\mathbf{j} + 5\mathbf{k}\) Or Equivalent | M1 A1 |
| (6) |
(c) Way 3
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix}2\lambda + \mu \\ \lambda - \mu \\ 5\lambda - 2\mu\end{pmatrix} \cdot \begin{pmatrix}3 \\ -4 \\ 2\end{pmatrix} = 5 \Rightarrow 12\lambda + 3\mu = 5\) M1: Substitutes parametric form of \(\Pi_2\) into the vector equation of \(\Pi_1\) A1: Correct equation | M1A1 |
| \(\mu = \dfrac{5}{3}, \lambda = 0\) gives \(\left(\dfrac{5}{3}, -\dfrac{5}{3}, -\dfrac{10}{3}\right)\) \(\mu = 0, \lambda = \dfrac{5}{12}\) gives \(\left(\dfrac{5}{6}, \dfrac{5}{12}, \dfrac{25}{12}\right)\) Direction \(\begin{pmatrix}-2 \\ 5 \\ 13\end{pmatrix}\) M1: Finds 2 points and direction A1: Correct coordinates and direction | M1A1 |
| Equation of line in required form: e.g. \(\mathbf{r} \times (-2\mathbf{i} + 5\mathbf{j} + 13\mathbf{k}) = -5\mathbf{i} - 15\mathbf{j} + 5\mathbf{k}\) Or Equivalent | M1A1 |
(corrected from the printed mark scheme: the first point in Way 3 is printed as \(\left(\dfrac{5}{3}, -\dfrac{5}{3}, \dfrac{10}{3}\right)\); with \(\mu = \dfrac{5}{3}, \lambda = 0\) the \(z\)-coordinate is \(5\lambda - 2\mu = -\dfrac{10}{3}\))
Do not allow ‘mixed’ methods – mark the best single attempt
NB for checking, a general point on the line will be of the form: \((1 - 2\lambda, 5\lambda, 1 + 13\lambda)\)