FP3 June 2016 Q8
8. The plane \(\Pi_1\) has equation \[x - 5y - 2z = 3\]
The plane \(\Pi_2\) has equation \[\mathbf{r} = \mathbf{i} + 2\mathbf{j} + \mathbf{k} + \lambda(\mathbf{i} + 4\mathbf{j} + 3\mathbf{k}) + \mu(2\mathbf{i} - \mathbf{j} + \mathbf{k})\] where \(\lambda\) and \(\mu\) are scalar parameters.
| Scheme | Marks |
|---|---|
| \(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 1 & 4 & 3\\ 2 & -1 & 1\end{vmatrix} = \begin{pmatrix}7\\ 5\\ -9\end{pmatrix}\) | M1A1 |
| \(\begin{pmatrix}1\\ -5\\ -2\end{pmatrix} \bullet \begin{pmatrix}7\\ 5\\ -9\end{pmatrix}\ (= 7 - 25 + 18)\) | M1 |
| \(\begin{pmatrix}1\\ -5\\ -2\end{pmatrix} \bullet \begin{pmatrix}7\\ 5\\ -9\end{pmatrix} = 7 - 25 + 18 = 0\ \therefore\) perpendicular | A1 |
| (4) |
Notes
M1: Attempt cross product between direction vectors or any 2 vectors in the plane. If working is not shown or is unclear, 2 elements should be correct for their vectors for this mark.
A1: Correct vector
M1: Attempts \(\begin{pmatrix}1\\ -5\\ -2\end{pmatrix} \bullet\) their vector product
A1: Correctly obtains = 0 and gives a conclusion.
Note
\(\begin{pmatrix}1\\ -5\\ -2\end{pmatrix} \bullet \begin{pmatrix}7\\ 5\\ -9\end{pmatrix} = 0\ \therefore\) perpendicular scores M1A0 here.
However \(\begin{pmatrix}1\\ -5\\ -2\end{pmatrix} \bullet \begin{pmatrix}7\\ 5\\ -9\end{pmatrix} = 7 - 25 + 18 = 0\ \therefore\) perpendicular scores M1A1
BUT
If \(\begin{pmatrix}7\\ 5\\ -9\end{pmatrix}\) is incorrect then \(\begin{pmatrix}1\\ -5\\ -2\end{pmatrix} \bullet \begin{pmatrix}a\\ b\\ c\end{pmatrix} = a - 5b - 2c\) needs to be seen to score the M mark
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix}7\\ 5\\ -9\end{pmatrix} \bullet \begin{pmatrix}1\\ 2\\ 1\end{pmatrix} = 8 \Rightarrow 7x + 5y - 9z = 8\) | M1A1 |
| (2) |
Notes
M1: Uses \(\mathbf{i} + 2\mathbf{j} + \mathbf{k}\) and their vector product to find the cartesian equation of \(\Pi_2\). You may need to check their “8” if no working is shown but it must be clear that \(\mathbf{i} + 2\mathbf{j} + \mathbf{k}\) (or a point on the plane) is being used.
A1: Correct equation (any multiple or equivalent equation)
Note that part (b) is possible without part (a): e.g.
\(x = 1 + \lambda + 2\mu,\quad y = 2 + 4\lambda - \mu,\quad z = 1 + 3\lambda + \mu\)
\(\Rightarrow y + z = 3 + 7\lambda\) and \(x + 2y = 5 + 9\lambda \Rightarrow 9(y + z) - 7(x + 2y) = -8\)
\(\therefore 7x + 5y - 9z = 8\)
Score as M1: Full method leading to a Cartesian equation, A1: Correct equation
| Scheme | Marks |
|---|---|
| \(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 7 & 5 & -9\\ 1 & -5 & -2\end{vmatrix} = \begin{pmatrix}-55\\ 5\\ -40\end{pmatrix}\) | M1A1 |
| \(x = 0: (0, -\tfrac{1}{5}, -1),\quad y = 0: (-\tfrac{11}{5}, 0, -\tfrac{13}{5}),\quad z = 0: (\tfrac{11}{8}, -\tfrac{13}{40}, 0)\) Note that points on the line satisfy \((11t, -\tfrac{1}{5} - t, -1 + 8t)\) | M1A1 |
| \(\left(\mathbf{r} - \left(-\dfrac{1}{5}\mathbf{j} - \mathbf{k}\right)\right) \times (11\mathbf{i} - \mathbf{j} + 8\mathbf{k}) = \mathbf{0}\) | ddM1A1 |
| (6) | |
| (12 marks) |
Notes
M1: Attempt cross product of normal vectors.
A1: \(k(11\mathbf{i} - \mathbf{j} + 8\mathbf{k})\)
M1: Attempt point on the line (\(x\), \(y\) and \(z\)).
A1: Correct coordinates
ddM1: (\(\mathbf{r}\) – their point) \(\times\) their direction “= 0” not required for this mark. Dependent on both previous method marks.
A1: Correct equation (oe)
Alternatives for part (c) by simultaneous equations
Case 1: Eliminates \(y\) then obtains \(\mathrm{f}(x) = \mathrm{g}(y) = z\)
| Scheme | Marks |
|---|---|
| \(x - 5y - 2z = 3,\ 7x + 5y - 9z = 8 \Rightarrow 8x - 11z = 11\) | |
| \(z = \dfrac{8x - 11}{11},\ x = \dfrac{11 + 11z}{8} \Rightarrow \dfrac{11 + 11z}{8} - 5y - 2z = 3 \Rightarrow z = \dfrac{-40y - 13}{5}\) | |
| \(\dfrac{8x - 11}{11} = \dfrac{-40y - 13}{5} = z\) | M1A1 |
| \(\dfrac{x - \frac{11}{8}}{\frac{11}{8}} = \dfrac{y + \frac{13}{40}}{-\frac{1}{8}} = \dfrac{z(-0)}{(1)}\) | M1A1 |
| \(\left(\mathbf{r} - \left(\dfrac{11}{8}\mathbf{i} - \dfrac{13}{40}\mathbf{j}\right)\right) \times \left(\dfrac{11}{8}\mathbf{i} - \dfrac{1}{8}\mathbf{j} + \mathbf{k}\right) = \mathbf{0}\) | ddM1A1 |
M1: Obtains \(\mathrm{f}(x) = \mathrm{g}(y) = z\)
A1: Correct expressions
M1: Correct processing on at least one expression (not \(z\)) to enable identification of position and direction.
A1: Correct equations
ddM1: (\(\mathbf{r}\) – their point) \(\times\) their direction “= 0” not required for this mark. Dependent on both previous method marks.
A1: Correct equation (oe)
Case 2: Eliminates \(x\) then obtains \(\mathrm{f}(x) = y = \mathrm{g}(z)\)
| Scheme | Marks |
|---|---|
| \(x - 5y - 2z = 3,\ 7x + 5y - 9z = 8 \Rightarrow 40y + 5z = -13\) | |
| \(y = \dfrac{-13 - 5z}{40},\ z = \dfrac{-13 - 40y}{5} \Rightarrow x - 5y + 2\left(\dfrac{13 + 40y}{5}\right) = 3 \Rightarrow y = \dfrac{-5x - 11}{55}\) | |
| \(\dfrac{-5x - 11}{55} = y = \dfrac{-13 - 5z}{40}\) | M1A1 |
| \(\dfrac{x + \frac{11}{5}}{-11} = \dfrac{y(-0)}{(1)} = \dfrac{z + \frac{13}{5}}{-8}\) | M1A1 |
| \(\left(\mathbf{r} - \left(-\dfrac{11}{5}\mathbf{i} - \dfrac{13}{5}\mathbf{k}\right)\right) \times (-11\mathbf{i} + \mathbf{j} - 8\mathbf{k}) = \mathbf{0}\) | ddM1A1 |
M1: Obtains \(\mathrm{f}(x) = y = \mathrm{g}(z)\)
A1: Correct expressions
M1: Correct processing on at least one expression (not \(y\)) to enable identification of position and direction.
A1: Correct equations
ddM1: (\(\mathbf{r}\) – their point) \(\times\) their direction “= 0” not required for this mark. Dependent on both previous method marks.
A1: Correct equation (oe)
Case 3: Eliminates \(z\) then obtains \(x = \mathrm{f}(y) = \mathrm{g}(z)\)
| Scheme | Marks |
|---|---|
| \(x - 5y - 2z = 3,\ 7x + 5y - 9z = 8 \Rightarrow 5x + 55y = -11\) | |
| \(x = \dfrac{-55y - 11}{5},\ y = \dfrac{-11 - 5x}{55} \Rightarrow x + 5\left(\dfrac{11 + 5x}{55}\right) - 2z = 3 \Rightarrow x = \dfrac{11z + 11}{8}\) | |
| \(x = \dfrac{-55y - 11}{5} = \dfrac{11z + 11}{8}\) | M1A1 |
| \(\dfrac{x(-0)}{(1)} = \dfrac{y + \frac{1}{5}}{-\frac{1}{11}} = \dfrac{z + 1}{\frac{8}{11}}\) | M1A1 |
| \(\left(\mathbf{r} - \left(-\dfrac{1}{5}\mathbf{j} - \mathbf{k}\right)\right) \times \left(\mathbf{i} - \dfrac{1}{11}\mathbf{j} + \dfrac{8}{11}\mathbf{k}\right) = \mathbf{0}\) | ddM1A1 |
M1: Obtains \(x = \mathrm{f}(y) = \mathrm{g}(z)\)
A1: Correct expressions
M1: Correct processing on at least one expression (not \(z\)) to enable identification of position and direction.
A1: Correct equations
ddM1: (\(\mathbf{r}\) – their point) \(\times\) their direction “= 0” not required for this mark. Dependent on both previous method marks.
A1: Correct equation (oe)
Alternatives for part (c) by parameters
Case 1: Eliminates \(x\)
| Scheme | Marks |
|---|---|
| \(x - 5y - 2z = 3,\ 7x + 5y - 9z = 8 \Rightarrow 8x - 11z = 11\) | |
| \(x = t \Rightarrow z = -1 + \dfrac{8}{11}t,\ y = -\dfrac{1}{5} - \dfrac{1}{11}t\) | M1A1 |
| \(Pos: -\dfrac{1}{5}\mathbf{j} - \mathbf{k}\quad Dir: \mathbf{i} - \dfrac{1}{11}\mathbf{j} + \dfrac{8}{11}\mathbf{k}\) | M1A1 |
| \(\left(\mathbf{r} - \left(-\dfrac{1}{5}\mathbf{j} - \mathbf{k}\right)\right) \times \left(\mathbf{i} - \dfrac{1}{11}\mathbf{j} + \dfrac{8}{11}\mathbf{k}\right) = \mathbf{0}\) | ddM1A1 |
M1: Obtains \(x\), \(y\) and \(z\) in terms of \(\lambda\)
A1: Correct expressions
M1: Uses their equations to obtain position and direction
A1: Correct position and direction
ddM1: (\(\mathbf{r}\) – their point) \(\times\) their direction “= 0” not required for this mark. Dependent on both previous method marks.
A1: Correct equation (oe)
Case 2: Eliminates \(y\)
| Scheme | Marks |
|---|---|
| \(x - 5y - 2z = 3,\ 7x + 5y - 9z = 8 \Rightarrow 40y + 5z = -13\) | |
| \(y = t \Rightarrow z = -\dfrac{13}{5} - 8t,\ x = -\dfrac{11}{5} - 11t\) | M1A1 |
| \(Pos: -\dfrac{11}{5}\mathbf{i} - \dfrac{13}{5}\mathbf{k}\quad Dir: -11\mathbf{i} + \mathbf{j} - 8\mathbf{k}\) | M1A1 |
| \(\left(\mathbf{r} - \left(-\dfrac{11}{5}\mathbf{i} - \dfrac{13}{5}\mathbf{k}\right)\right) \times (-11\mathbf{i} + \mathbf{j} - 8\mathbf{k}) = \mathbf{0}\) | ddM1A1 |
(Corrected from the printed mark scheme: the eliminated equation is printed as \(40y + 15z = -13\); eliminating \(x\) gives \(40y + 5z = -13\), as in Case 2 of the simultaneous equations method above. The next lines are printed as \(y = -\dfrac{1}{5} - 11t\) and \(Dir: -\dfrac{11}{5}\mathbf{i} + \mathbf{j} - 8\mathbf{k}\); they should read \(x = -\dfrac{11}{5} - 11t\) and \(Dir: -11\mathbf{i} + \mathbf{j} - 8\mathbf{k}\), as used in the final equation.)
M1: Obtains \(x\), \(y\) and \(z\) in terms of \(\lambda\)
A1: Correct expressions
M1: Uses their equations to obtain position and direction
A1: Correct position and direction
ddM1: (\(\mathbf{r}\) – their point) \(\times\) their direction “= 0” not required for this mark. Dependent on both previous method marks.
A1: Correct equation (oe)
Case 3: Eliminates \(z\)
| Scheme | Marks |
|---|---|
| \(x - 5y - 2z = 3,\ 7x + 5y - 9z = 8 \Rightarrow 8x - 11z = 11\) | |
| \(z = t \Rightarrow x = \dfrac{11}{8} + \dfrac{11}{8}t,\ y = -\dfrac{13}{40} - \dfrac{1}{8}t\) | M1A1 |
| \(Pos: \dfrac{11}{8}\mathbf{i} - \dfrac{13}{40}\mathbf{j}\quad Dir: \dfrac{11}{8}\mathbf{i} - \dfrac{1}{8}\mathbf{j} + \mathbf{k}\) | M1A1 |
| \(\left(\mathbf{r} - \left(\dfrac{11}{8}\mathbf{i} - \dfrac{13}{40}\mathbf{j}\right)\right) \times \left(\dfrac{11}{8}\mathbf{i} - \dfrac{1}{8}\mathbf{j} + \mathbf{k}\right) = \mathbf{0}\) | ddM1A1 |
M1: Obtains \(x\), \(y\) and \(z\) in terms of \(\lambda\)
A1: Correct expressions
M1: Uses their equations to obtain position and direction
A1: Correct position and direction
ddM1: (\(\mathbf{r}\) – their point) \(\times\) their direction “= 0” not required for this mark. Dependent on both previous method marks.
A1: Correct equation (oe)
Alternative for part (c) by finding 2 points on the line
| Scheme | Marks |
|---|---|
| \(x = 0: (0, -\tfrac{1}{5}, -1),\quad y = 0: (-\tfrac{11}{5}, 0, -\tfrac{13}{5}),\quad z = 0: (\tfrac{11}{8}, -\tfrac{13}{40}, 0)\) M1: Attempts two points on the line A1: Two correct coordinates | M1A1 |
| Dir: \(-\dfrac{1}{5}\mathbf{j} - \mathbf{k} - \left(-\dfrac{11}{5}\mathbf{i} - \dfrac{13}{5}\mathbf{k}\right) = \dfrac{11}{5}\mathbf{i} - \dfrac{1}{5}\mathbf{j} + \dfrac{8}{5}\mathbf{k}\) | M1A1 |
| \(\left(\mathbf{r} - \left(-\dfrac{1}{5}\mathbf{j} - \mathbf{k}\right)\right) \times \left(\dfrac{11}{5}\mathbf{i} - \dfrac{1}{5}\mathbf{j} + \dfrac{8}{5}\mathbf{k}\right) = \mathbf{0}\) | ddM1A1 |
M1: Subtracts to obtain direction
A1: Correct direction
ddM1: (\(\mathbf{r}\) – their point) \(\times\) their direction “= 0” not required for this mark. Dependent on both previous method marks.
A1: Correct equation (oe)