FP3 June 2015 Q5
5. The points \(A\), \(B\) and \(C\) have position vectors \(\begin{pmatrix}1\\ 3\\ 2\end{pmatrix},\ \begin{pmatrix}-1\\ 0\\ 1\end{pmatrix}\) and \(\begin{pmatrix}2\\ 1\\ 0\end{pmatrix}\) respectively.
The plane \(\Pi\) contains the points \(A\), \(B\) and \(C\).
| Scheme | Marks |
|---|---|
| \(\mathbf{AB} = -2\mathbf{i} - 3\mathbf{j} - \mathbf{k}\) | M1 |
| \(\mathbf{r} = \begin{pmatrix}1\\ 3\\ 2\end{pmatrix} + \lambda\begin{pmatrix}-2\\ -3\\ -1\end{pmatrix}\) or \(\left(\mathbf{r} - \begin{pmatrix}1\\ 3\\ 2\end{pmatrix}\right) \times \begin{pmatrix}-2\\ -3\\ -1\end{pmatrix} = 0\) | A1 |
| (2) |
Notes
M1: Attempt \(\pm(\mathbf{OB} - \mathbf{OA})\)
A1: Any correct vector form including the “\(\mathbf{r}\) =” and the “= 0”. “\(\mathbf{r}\) =” can be “AB =” or “\(l\) =” etc. The direction can be any multiple of that shown.
| Scheme | Marks |
|---|---|
| \(\dfrac{x - \text{"}1\text{"}}{\text{"}-2\text{"}} = \dfrac{y - \text{"}3\text{"}}{\text{"}-3\text{"}} = \dfrac{z - \text{"}2\text{"}}{\text{"}-1\text{"}}\) oe e.g. \(\dfrac{x + 1}{2} = \dfrac{y}{3} = \dfrac{z - 1}{1}\) | M1A1 |
| (2) |
Notes
M1: Correct attempt at the Cartesian form using their position and direction
A1: \(\dfrac{x - 1}{-2} = \dfrac{y - 3}{-3} = \dfrac{z - 2}{-1}\) oe
| Scheme | Marks |
|---|---|
| \(\mathbf{AB} \times \mathbf{AC} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ -2 & -3 & -1\\ 1 & -2 & -2\end{vmatrix} = \begin{pmatrix}-2\\ -3\\ -1\end{pmatrix} \times \begin{pmatrix}1\\ -2\\ -2\end{pmatrix} = \begin{pmatrix}4\\ -5\\ 7\end{pmatrix}\) \((= \mathbf{AB} \times \mathbf{BC} = \mathbf{AC} \times \mathbf{BC})\) | M1A1 |
| \(\mathbf{r}.\begin{pmatrix}4\\ -5\\ 7\end{pmatrix} = \begin{pmatrix}1\\ 3\\ 2\end{pmatrix}.\begin{pmatrix}4\\ -5\\ 7\end{pmatrix}\) i.e. \(\mathbf{r}.\begin{pmatrix}4\\ -5\\ 7\end{pmatrix} = 3\) oe e.g. \(\mathbf{r}.\begin{pmatrix}-4\\ 5\\ -7\end{pmatrix} = -3\) | dM1A1 |
| (4) |
Notes
M1: Attempts vector product of 2 vectors in the plane e.g. \(\mathbf{AB} \times \mathbf{BC}\). If there is no working, at least 2 components should be correct.
A1: Any multiple of \(4\mathbf{i} - 5\mathbf{j} + 7\mathbf{k}\)
dM1: Attempts scalar product using their normal vector and \(\mathbf{a}\), \(\mathbf{b}\) or \(\mathbf{c}\). Dependent on the previous M
A1: Correct equation (oe)
Alternatives for 5(c)
| Scheme | Marks |
|---|---|
| \(\begin{pmatrix}x\\ y\\ z\end{pmatrix} = \begin{pmatrix}1\\ 3\\ 2\end{pmatrix} + \lambda\begin{pmatrix}-2\\ -3\\ -1\end{pmatrix} + \mu\begin{pmatrix}1\\ -2\\ -2\end{pmatrix} \Rightarrow 4x - 5y + 7z = 3\) | M1A1 |
| \(4x - 5y + 7z = 3 \Rightarrow \mathbf{r}.\begin{pmatrix}4\\ -5\\ 7\end{pmatrix} = 3\) | dM1A1 |
M1: Correctly forms the parametric equation and eliminates the parameters to obtain a cartesian equation
A1: Correct cartesian equation
dM1: Converts their Cartesian equation into the form required. Dependent on the previous M
A1: Correct equation (oe)
Alternative
| Scheme | Marks |
|---|---|
| \(\left.\begin{aligned}a + 3b + 2c &= d\\ -a + c &= d\\ 2a + b &= d\end{aligned}\right\} \Rightarrow a = \dfrac{4}{3}d,\ b = -\dfrac{5}{3}d,\ c = \dfrac{7}{3}d\) | M1A1 |
| \(\dfrac{4}{3}x - \dfrac{5}{3}y + \dfrac{7}{3}z = 1 \Rightarrow \mathbf{r}.\dfrac{1}{3}\begin{pmatrix}4\\ -5\\ 7\end{pmatrix} = 1\) | dM1A1 |
M1: Substitutes to obtain 3 equations in \(a\), \(b\), \(c\) and \(d\) and solves to obtain at least one of \(a\), \(b\) or \(c\) in terms of \(d\)
A1: Correct \(a\), \(b\) and \(c\) in terms of \(d\)
dM1: Uses their cartesian equation correctly to form a vector equation Dependent on the previous M
| Scheme | Marks |
|---|---|
| \(d = \dfrac{\text{"}3\text{"}}{\left|\text{"}4\mathbf{i} - 5\mathbf{j} + 7\mathbf{k}\text{"}\right|} = \dfrac{3}{\sqrt{90}}\) | M1A1 Note B1B1 on ePEN |
| (2) | |
| (10 marks) |
Notes
M1: \(d = \dfrac{\pm\text{their } p}{|\text{their } \mathbf{n}|},\quad p \neq 0\)
A1: \(\dfrac{3}{\sqrt{90}}\) oe e.g. \(\dfrac{3}{3\sqrt{10}},\ \dfrac{1}{\sqrt{10}}\), (awrt 0.316)
Alternative
| Scheme | Marks |
|---|---|
| \(\lambda\begin{pmatrix}4\\ -5\\ 7\end{pmatrix}.\begin{pmatrix}4\\ -5\\ 7\end{pmatrix} = 3 \Rightarrow \lambda = \dfrac{1}{30} \Rightarrow d = \sqrt{\left(\dfrac{4}{30}\right)^2 + \left(\dfrac{5}{30}\right)^2 + \left(\dfrac{7}{30}\right)^2} = \dfrac{1}{\sqrt{10}}\) | M1A1 Note B1B1 on ePEN |
M1: A correct method for finding “\(\lambda\)” and attempting the length of \(\lambda\mathbf{n}\)
A1: \(\dfrac{3}{\sqrt{90}}\) oe e.g. \(\dfrac{3}{3\sqrt{10}},\ \dfrac{1}{\sqrt{10}}\), (awrt 0.316)