| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 2 (Further Pure Core 2) |
| Year | 2018 |
| Session | March |
| Marks | 8 |
| Topic | Vectors: Lines & Planes |
| Type | Angle between line and plane |
| Difficulty | Standard +0.3 This is a straightforward Further Maths question testing standard 3D vector geometry techniques: angle between line and plane using dot product, intersection by substitution, and perpendicular distance formula. All three parts are routine applications of well-practiced methods with no conceptual challenges or novel problem-solving required. |
| Spec | 4.04b Plane equations: cartesian and vector forms4.04f Line-plane intersection: find point4.04j Shortest distance: between a point and a plane |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{pmatrix}2\\2\\3\end{pmatrix} \cdot \begin{pmatrix}3\\-1\\2\end{pmatrix} = 10\) | M1 | Correct process for finding the dot product of normal to the plane and direction vector of line. |
| \(\cos(90 - \theta) = \frac{10}{\sqrt{17}\sqrt{14}} (= \sin\theta)\) | M1dep* | Correct process for using dot product to find the cosine of the complement or sin of angle |
| \(= 90 - 49.6° = \text{awrt } 40.4°\) | A1 | Or awrt 0.705 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\begin{pmatrix}1\\0\\5\end{pmatrix} + \lambda \begin{pmatrix}2\\2\\3\end{pmatrix} \cdot \begin{pmatrix}3\\-1\\2\end{pmatrix} = 33\) | M1 | Correct substitution line form for r into r.n = d or \(3(1+2\lambda) - 2\lambda + 2(5 + 3\lambda) = 33\) |
| \(\lambda = 2\) | A1 | Correct substitution line form for r into r.n = d |
| \((5, 4, 11)\) | A1 [3] | Condone position vector |
| Answer | Marks | Guidance |
|---|---|---|
| \( | 3 \times 4 + (-1) \times 5 + 2 \times (-5) - 33 | \) |
| \(\sqrt{3^2 + (-1)^2 + 2^2}\) | A1 | |
| \(\frac{18\sqrt{14}}{7}\) | A1 [2] | or \(\frac{36}{\sqrt{14}}\) |
## 1(i)
$\begin{pmatrix}2\\2\\3\end{pmatrix} \cdot \begin{pmatrix}3\\-1\\2\end{pmatrix} = 10$ | M1 | Correct process for finding the dot product of normal to the plane and direction vector of line.
$\cos(90 - \theta) = \frac{10}{\sqrt{17}\sqrt{14}} (= \sin\theta)$ | M1dep* | Correct process for using dot product to find the cosine of the complement or sin of angle
$= 90 - 49.6° = \text{awrt } 40.4°$ | A1 | Or awrt 0.705
## 1(ii)
$\begin{pmatrix}1\\0\\5\end{pmatrix} + \lambda \begin{pmatrix}2\\2\\3\end{pmatrix} \cdot \begin{pmatrix}3\\-1\\2\end{pmatrix} = 33$ | M1 | Correct substitution line form for **r** into r.n = d or $3(1+2\lambda) - 2\lambda + 2(5 + 3\lambda) = 33$
$\lambda = 2$ | A1 | Correct substitution line form for **r** into r.n = d
$(5, 4, 11)$ | A1 [3] | Condone position vector
## 1(iii)
$|3 \times 4 + (-1) \times 5 + 2 \times (-5) - 33|$ | M1 | Substitution of S and the normal to the plane into the formula.
$\sqrt{3^2 + (-1)^2 + 2^2}$ | A1 |
$\frac{18\sqrt{14}}{7}$ | A1 [2] | or $\frac{36}{\sqrt{14}}$
---Plane $\Pi$ has equation $3x - y + 2z = 33$. Line $l$ has the following vector equation.
$$l: \quad \mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ 5 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 2 \\ 3 \end{pmatrix}$$
\begin{enumerate}[label=(\roman*)]
\item Find the acute angle between $\Pi$ and $l$. [3]
\item Find the coordinates of the point of intersection of $\Pi$ and $l$. [3]
\item $S$ is the point $(4, 5, -5)$. Find the shortest distance from $S$ to $\Pi$. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 2 2018 Q1 [8]}}