| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors: Cross Product & Distances |
| Type | Perpendicular distance from point to plane |
| Difficulty | Standard +0.3 This is a straightforward application of standard Further Maths formulas: (a) uses the perpendicular distance formula from point to plane, (b) requires computing a cross product to find the normal vector, and (c) applies the angle between planes formula. All parts are routine calculations with no problem-solving insight required, making it slightly easier than average. |
| Spec | 4.04d Angles: between planes and between line and plane4.04j Shortest distance: between a point and a plane |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\begin{pmatrix}3\\-4\\2\end{pmatrix}\cdot\begin{pmatrix}6\\2\\12\end{pmatrix} = 18-8+24\) | M1 | Attempts scalar product between normal and position vector |
| \(d = \frac{18-8+24-5}{\sqrt{3^2+4^2+2^2}}\) | M1 | Correct method for perpendicular distance |
| \(= \sqrt{29}\) | A1 | Correct distance |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\begin{pmatrix}-1\\-3\\1\end{pmatrix}\cdot\begin{pmatrix}2\\1\\5\end{pmatrix} = \ldots\) and \(\begin{pmatrix}-1\\-3\\1\end{pmatrix}\cdot\begin{pmatrix}1\\-1\\-2\end{pmatrix} = \ldots\) | M1 | Recognises need to calculate scalar product between given vector and both direction vectors |
| Both products \(= 0\); \(\therefore -\mathbf{i}-3\mathbf{j}+\mathbf{k}\) is perpendicular to \(\Pi_2\) | A1 | Obtains zero both times and makes a conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\begin{pmatrix}-1\\-3\\1\end{pmatrix}\cdot\begin{pmatrix}3\\-4\\2\end{pmatrix} = -3+12+2\) | M1 | Calculates scalar product between the two normal vectors |
| \(\sqrt{(-1)^2+(-3)^2+1^2}\sqrt{(3)^2+(-4)^2+2^2}\cos\theta = 11\), \(\Rightarrow\cos\theta = \frac{11}{\sqrt{(-1)^2+(-3)^2+1^2}\sqrt{(3)^2+(-4)^2+2^2}}\) | M1 | Applies scalar product formula to find \(\cos\theta\) |
| Angle between planes \(\theta = 52°\) | A1* | Identifies correct angle linking normal angle to angle between planes |
## Question 2(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}3\\-4\\2\end{pmatrix}\cdot\begin{pmatrix}6\\2\\12\end{pmatrix} = 18-8+24$ | M1 | Attempts scalar product between normal and position vector |
| $d = \frac{18-8+24-5}{\sqrt{3^2+4^2+2^2}}$ | M1 | Correct method for perpendicular distance |
| $= \sqrt{29}$ | A1 | Correct distance |
## Question 2(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}-1\\-3\\1\end{pmatrix}\cdot\begin{pmatrix}2\\1\\5\end{pmatrix} = \ldots$ and $\begin{pmatrix}-1\\-3\\1\end{pmatrix}\cdot\begin{pmatrix}1\\-1\\-2\end{pmatrix} = \ldots$ | M1 | Recognises need to calculate scalar product between given vector and both direction vectors |
| Both products $= 0$; $\therefore -\mathbf{i}-3\mathbf{j}+\mathbf{k}$ is perpendicular to $\Pi_2$ | A1 | Obtains zero both times and makes a conclusion |
## Question 2(c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}-1\\-3\\1\end{pmatrix}\cdot\begin{pmatrix}3\\-4\\2\end{pmatrix} = -3+12+2$ | M1 | Calculates scalar product between the two normal vectors |
| $\sqrt{(-1)^2+(-3)^2+1^2}\sqrt{(3)^2+(-4)^2+2^2}\cos\theta = 11$, $\Rightarrow\cos\theta = \frac{11}{\sqrt{(-1)^2+(-3)^2+1^2}\sqrt{(3)^2+(-4)^2+2^2}}$ | M1 | Applies scalar product formula to find $\cos\theta$ |
| Angle between planes $\theta = 52°$ | A1* | Identifies correct angle linking normal angle to angle between planes |
---\begin{enumerate}
\item The plane $\Pi _ { 1 }$ has vector equation
\end{enumerate}
$$\mathbf { r } \cdot ( 3 \mathbf { i } - 4 \mathbf { j } + 2 \mathbf { k } ) = 5$$
(a) Find the perpendicular distance from the point $( 6,2,12 )$ to the plane $\Pi _ { 1 }$
The plane $\Pi _ { 2 }$ has vector equation
$$\mathbf { r } = \lambda ( 2 \mathbf { i } + \mathbf { j } + 5 \mathbf { k } ) + \mu ( \mathbf { i } - \mathbf { j } - 2 \mathbf { k } )$$
where $\lambda$ and $\mu$ are scalar parameters.\\
(b) Show that the vector $- \mathbf { i } - 3 \mathbf { j } + \mathbf { k }$ is perpendicular to $\Pi _ { 2 }$\\
(c) Show that the acute angle between $\Pi _ { 1 }$ and $\Pi _ { 2 }$ is $52 ^ { \circ }$ to the nearest degree.
\hfill \mbox{\textit{Edexcel CP2 Q2 [8]}}