| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 1 (Further Pure Core 1) |
| Year | 2018 |
| Session | September |
| Marks | 8 |
| Topic | Vectors: Lines & Planes |
| Type | Angle between two planes |
| Difficulty | Standard +0.3 Part (i) is a standard application of the formula for angle between planes using normal vectors (dot product and magnitudes). Part (ii) requires converting the line to parametric form and verifying it satisfies both plane equations—straightforward verification with no novel insight required. This is slightly easier than average as it's a textbook exercise testing routine techniques. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04d Angles: between planes and between line and plane |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\mathbf{n}_1 \cdot \mathbf{n}_2 = 9 = | \mathbf{n}_1 | |
| \(\Rightarrow \cos\theta = \frac{9}{\sqrt{84}}\) | A1 | soi |
| \(\Rightarrow \theta = 10.9°\) or \(0.190\) rads | A1 [3] | |
| (ii) Direction of line is \(\begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}\) | B1 [5] | direction |
| \(\mathbf{n}_1 \cdot \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix} = 0\) and \(\mathbf{n}_2 \cdot \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix} = 0\) | M1, A1 [5] | Scalar product with both normals (shown, not just stated); A1 for finding zero twice |
| Point \((0, 1, 2)\) is on \(L\): \(\Pi_1: 3 \times 0 + 2 \times 1 + 2 = 0 + 2 + 2 = 4\) ✓; \(\Pi_2: 2 \times 0 + 1 + 2 = 3\) ✓. So a point on \(L\) lies in both planes. But \(L\) is parallel to both planes so \(L\) lies in both planes. | M1, A1 [5] | A point on the line into both equations; showing it is in both planes convincingly and conclusion |
| Or Equation of line is \(\mathbf{r} = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}\) or \(\begin{pmatrix} \lambda \\ 1-\lambda \\ 2-\lambda \end{pmatrix}\) | B1 | |
| \(\Pi_1: \mathbf{r} \cdot \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} \lambda \\ 1-\lambda \\ 2-\lambda \end{pmatrix} = 3\lambda + 2 - 2\lambda + 2 - \lambda = 4\) | M1 | Substituting into equation of \(\Pi_1\) |
| \(\Pi_2: \mathbf{r} \cdot \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} \lambda \\ 1-\lambda \\ 2-\lambda \end{pmatrix} = 2\lambda + 1 - \lambda + 2 - \lambda = 3\) | M1 | Substituting into other plane equation |
| So \(L\) lies in both planes. | A1 | |
| Or \(y = 1 - x, z = 2 - x\); \(\Pi_1: 3x + 2y + z = 3x + 2(1-x) + 2 - x = 4\) | B1, M1, A1 | |
| \(\Pi_2: 2x + y + z = 2x + 1 - x + 2 - x = 3\) | M1, A1 | |
| So \(L\) lies in both planes. | A1 [5] |
**(i)** $\mathbf{n}_1 \cdot \mathbf{n}_2 = 9 = |\mathbf{n}_1||\mathbf{n}_2|\cos\theta = \sqrt{14}\sqrt{6}\cos\theta$ | M1 | $\mathbf{n}_1 = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix}$ and $\mathbf{n}_2 = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix}$
$\Rightarrow \cos\theta = \frac{9}{\sqrt{84}}$ | A1 | soi
$\Rightarrow \theta = 10.9°$ or $0.190$ rads | A1 [3]
**(ii)** Direction of line is $\begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}$ | B1 [5] | direction
$\mathbf{n}_1 \cdot \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix} = 0$ and $\mathbf{n}_2 \cdot \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix} = 0$ | M1, A1 [5] | Scalar product with both normals (shown, not just stated); A1 for finding zero twice
Point $(0, 1, 2)$ is on $L$: $\Pi_1: 3 \times 0 + 2 \times 1 + 2 = 0 + 2 + 2 = 4$ ✓; $\Pi_2: 2 \times 0 + 1 + 2 = 3$ ✓. So a point on $L$ lies in both planes. But $L$ is parallel to both planes so $L$ lies in both planes. | M1, A1 [5] | A point on the line into both equations; showing it is in both planes convincingly and conclusion
**Or** Equation of line is $\mathbf{r} = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}$ or $\begin{pmatrix} \lambda \\ 1-\lambda \\ 2-\lambda \end{pmatrix}$ | B1 |
$\Pi_1: \mathbf{r} \cdot \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} \lambda \\ 1-\lambda \\ 2-\lambda \end{pmatrix} = 3\lambda + 2 - 2\lambda + 2 - \lambda = 4$ | M1 | Substituting into equation of $\Pi_1$
$\Pi_2: \mathbf{r} \cdot \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} \lambda \\ 1-\lambda \\ 2-\lambda \end{pmatrix} = 2\lambda + 1 - \lambda + 2 - \lambda = 3$ | M1 | Substituting into other plane equation
So $L$ lies in both planes. | A1
**Or** $y = 1 - x, z = 2 - x$; $\Pi_1: 3x + 2y + z = 3x + 2(1-x) + 2 - x = 4$ | B1, M1, A1
$\Pi_2: 2x + y + z = 2x + 1 - x + 2 - x = 3$ | M1, A1
So $L$ lies in both planes. | A1 [5]
---5 Two planes, $\Pi _ { 1 }$ and $\Pi _ { 2 }$, have equations $3 x + 2 y + z = 4$ and $2 x + y + z = 3$ respectively.\\
(i) Find the acute angle between $\Pi _ { 1 }$ and $\Pi _ { 2 }$.
The line $L$ has equation $x = 1 - y = 2 - z$.\\
(ii) Show that $L$ lies in both planes.
\hfill \mbox{\textit{OCR Further Pure Core 1 2018 Q5 [8]}}