Question 5 - A-Level Maths

Edexcel FP3 2017 June — Question 5 11 marks

Exam Board	Edexcel
Module	FP3 (Further Pure Mathematics 3)
Year	2017
Session	June
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Vectors: Lines & Planes
Type	Angle between two planes
Difficulty	Standard +0.3 This is a standard Further Maths vectors question testing routine techniques: (a) angle between planes using dot product of normals, (b) finding intersection of line and plane using parametric equations, (c) finding plane equation perpendicular to two given planes using cross product. All parts follow textbook methods with straightforward arithmetic, making it slightly easier than average A-level difficulty.
Spec	4.04b Plane equations: cartesian and vector forms 4.04c Scalar product: calculate and use for angles 4.04d Angles: between planes and between line and plane 4.04f Line-plane intersection: find point

5. The plane \(\Pi _ { 1 }\) has equation \(x - 2 y - 3 z = 5\) and the plane \(\Pi _ { 2 }\) has equation \(6 x + y - 4 z = 7\)

Find, to the nearest degree, the acute angle between \(\Pi _ { 1 }\) and \(\Pi _ { 2 }\) The point \(P\) has coordinates \(( 2,3 , - 1 )\). The line \(l\) is perpendicular to \(\Pi _ { 1 }\) and passes through the point \(P\). The line \(l\) intersects \(\Pi _ { 2 }\) at the point \(Q\).
Find the coordinates of \(Q\). The plane \(\Pi _ { 3 }\) passes through the point \(Q\) and is perpendicular to \(\Pi _ { 1 }\) and \(\Pi _ { 2 }\)
Find an equation of the plane \(\Pi _ { 3 }\) in the form \(\mathbf { r } . \mathbf { n } = p\)

Show mark scheme Show mark scheme source

Question 5:

Part (a) Way 1:

Answer	Marks	Guidance
Answer/Working	Marks	Guidance Notes
\(\begin{pmatrix}1\\-2\\-3\end{pmatrix}\cdot\begin{pmatrix}6\\1\\-4\end{pmatrix}=6-2+12\)	M1	Attempts scalar product of normal vectors, allowing one slip. May be implied by value of 16
\(16=\sqrt{1^2+2^2+3^2}\sqrt{6^2+1^2+4^2}\cos\theta \Rightarrow \cos\theta=\ldots\)	M1	Complete attempt to find \(\cos\theta\)
\(\cos\theta=\frac{16}{\sqrt{14}\sqrt{53}}\Rightarrow\theta=54°\)	A1	cao and not isw. Do not allow 54.0°

Part (a) Way 2:

Answer	Marks	Guidance
Answer/Working	Marks	Guidance Notes
\(\begin{pmatrix}1\\-2\\-3\end{pmatrix}\times\begin{pmatrix}6\\1\\-4\end{pmatrix}=\begin{pmatrix}11\\-14\\13\end{pmatrix}\)	M1	Attempts cross product of normal vectors. 2 components correct if no working
\(\sqrt{11^2+14^2+13^2}=\sqrt{1^2+2^2+3^2}\sqrt{6^2+1^2+4^2}\sin\theta \Rightarrow \sin\theta=\ldots\)	M1	Complete attempt to find \(\sin\theta\)
\(\sin\theta=\frac{9\sqrt{6}}{\sqrt{14}\sqrt{53}}\Rightarrow\theta=54°\)	A1	cao and not isw. Do not allow 54.0°

Part (b):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance Notes
\(\mathbf{PQ}=\begin{pmatrix}2\\3\\-1\end{pmatrix}+\lambda\begin{pmatrix}1\\-2\\-3\end{pmatrix}\) or \(\begin{pmatrix}2+\lambda\\3-2\lambda\\-1-3\lambda\end{pmatrix}\)	M1	Attempt parametric form of PQ using point \(P\) and normal to \(\Pi_1\)
\(6(2+\lambda)+(3-2\lambda)-4(-1-3\lambda)=7 \Rightarrow \lambda=\ldots\)	M1	Substitutes parametric form into equation of \(\Pi_2\) and solves for \(\lambda\)
\(\lambda=-\frac{3}{4}\Rightarrow Q\) is \(\left(\frac{5}{4},\frac{9}{2},\frac{5}{4}\right)\)	M1A1	M1: Uses \(\lambda\) in PQ equation; A1: Correct coordinates or vector

Part (c):

Answer	Marks	Guidance
Answer/Working	Marks	Guidance Notes
\(\begin{pmatrix}1\\-2\\-3\end{pmatrix}\times\begin{pmatrix}6\\1\\-4\end{pmatrix}=\begin{pmatrix}11\\-14\\13\end{pmatrix}\)	M1A1	M1: Attempt cross product between normals; A1: Correct normal vector (any multiple)
\(\begin{pmatrix}11\\-14\\13\end{pmatrix}\cdot\begin{pmatrix}5/4\\9/2\\5/4\end{pmatrix}=\ldots\) or \(\begin{pmatrix}11\\-14\\13\end{pmatrix}\cdot\begin{pmatrix}2\\3\\-1\end{pmatrix}=\ldots\)	M1	Attempt scalar product between normal and OQ or OP. Must obtain a value
\(\mathbf{r}\cdot\begin{pmatrix}11\\-14\\13\end{pmatrix}=-33\)	A1	Any multiple e.g. \(\mathbf{r}\cdot\begin{pmatrix}11k\\-14k\\13k\end{pmatrix}=-33k\) \((k\neq0)\)

# Question 5:

## Part (a) Way 1:

| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $\begin{pmatrix}1\\-2\\-3\end{pmatrix}\cdot\begin{pmatrix}6\\1\\-4\end{pmatrix}=6-2+12$ | M1 | Attempts scalar product of normal vectors, allowing one slip. May be implied by value of 16 |
| $16=\sqrt{1^2+2^2+3^2}\sqrt{6^2+1^2+4^2}\cos\theta \Rightarrow \cos\theta=\ldots$ | M1 | Complete attempt to find $\cos\theta$ |
| $\cos\theta=\frac{16}{\sqrt{14}\sqrt{53}}\Rightarrow\theta=54°$ | A1 | cao and not isw. Do not allow 54.0° |

## Part (a) Way 2:

| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $\begin{pmatrix}1\\-2\\-3\end{pmatrix}\times\begin{pmatrix}6\\1\\-4\end{pmatrix}=\begin{pmatrix}11\\-14\\13\end{pmatrix}$ | M1 | Attempts cross product of normal vectors. 2 components correct if no working |
| $\sqrt{11^2+14^2+13^2}=\sqrt{1^2+2^2+3^2}\sqrt{6^2+1^2+4^2}\sin\theta \Rightarrow \sin\theta=\ldots$ | M1 | Complete attempt to find $\sin\theta$ |
| $\sin\theta=\frac{9\sqrt{6}}{\sqrt{14}\sqrt{53}}\Rightarrow\theta=54°$ | A1 | cao and not isw. Do not allow 54.0° |

## Part (b):

| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $\mathbf{PQ}=\begin{pmatrix}2\\3\\-1\end{pmatrix}+\lambda\begin{pmatrix}1\\-2\\-3\end{pmatrix}$ or $\begin{pmatrix}2+\lambda\\3-2\lambda\\-1-3\lambda\end{pmatrix}$ | M1 | Attempt parametric form of PQ using point $P$ and normal to $\Pi_1$ |
| $6(2+\lambda)+(3-2\lambda)-4(-1-3\lambda)=7 \Rightarrow \lambda=\ldots$ | M1 | Substitutes parametric form into equation of $\Pi_2$ and solves for $\lambda$ |
| $\lambda=-\frac{3}{4}\Rightarrow Q$ is $\left(\frac{5}{4},\frac{9}{2},\frac{5}{4}\right)$ | M1A1 | M1: Uses $\lambda$ in PQ equation; A1: Correct coordinates or vector |

## Part (c):

| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| $\begin{pmatrix}1\\-2\\-3\end{pmatrix}\times\begin{pmatrix}6\\1\\-4\end{pmatrix}=\begin{pmatrix}11\\-14\\13\end{pmatrix}$ | M1A1 | M1: Attempt cross product between normals; A1: Correct normal vector (any multiple) |
| $\begin{pmatrix}11\\-14\\13\end{pmatrix}\cdot\begin{pmatrix}5/4\\9/2\\5/4\end{pmatrix}=\ldots$ or $\begin{pmatrix}11\\-14\\13\end{pmatrix}\cdot\begin{pmatrix}2\\3\\-1\end{pmatrix}=\ldots$ | M1 | Attempt scalar product between normal and OQ or OP. Must obtain a value |
| $\mathbf{r}\cdot\begin{pmatrix}11\\-14\\13\end{pmatrix}=-33$ | A1 | Any multiple e.g. $\mathbf{r}\cdot\begin{pmatrix}11k\\-14k\\13k\end{pmatrix}=-33k$ $(k\neq0)$ |

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5. The plane $\Pi _ { 1 }$ has equation $x - 2 y - 3 z = 5$ and the plane $\Pi _ { 2 }$ has equation $6 x + y - 4 z = 7$
\begin{enumerate}[label=(\alph*)]
\item Find, to the nearest degree, the acute angle between $\Pi _ { 1 }$ and $\Pi _ { 2 }$

The point $P$ has coordinates $( 2,3 , - 1 )$. The line $l$ is perpendicular to $\Pi _ { 1 }$ and passes through the point $P$. The line $l$ intersects $\Pi _ { 2 }$ at the point $Q$.
\item Find the coordinates of $Q$.

The plane $\Pi _ { 3 }$ passes through the point $Q$ and is perpendicular to $\Pi _ { 1 }$ and $\Pi _ { 2 }$
\item Find an equation of the plane $\Pi _ { 3 }$ in the form $\mathbf { r } . \mathbf { n } = p$
\end{enumerate}

\hfill \mbox{\textit{Edexcel FP3 2017 Q5 [11]}}

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