CAIE FP1 2018 November — Question 8 10 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2018
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypeLine of intersection of planes
DifficultyStandard +0.3 This is a standard Further Maths vectors question testing routine techniques: finding a normal vector via cross product for Cartesian form, using the angle formula between planes, and finding line of intersection by solving simultaneous equations. While it requires multiple steps and is from Further Maths (inherently harder), these are well-practiced procedures without novel insight, making it slightly easier than average for FM students.
Spec4.04b Plane equations: cartesian and vector forms4.04c Scalar product: calculate and use for angles4.04d Angles: between planes and between line and plane4.04f Line-plane intersection: find point

The plane \(\Pi_1\) has equation $$\mathbf{r} = \begin{pmatrix} 5 \\ 1 \\ 0 \end{pmatrix} + s\begin{pmatrix} -4 \\ 1 \\ 3 \end{pmatrix} + t\begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}.$$
  1. Find a cartesian equation of \(\Pi_1\). [3]
The plane \(\Pi_2\) has equation \(3x + y - z = 3\).
  1. Find the acute angle between \(\Pi_1\) and \(\Pi_2\), giving your answer in degrees. [2]
  2. Find an equation of the line of intersection of \(\Pi_1\) and \(\Pi_2\), giving your answer in the form \(\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}\). [5]

Question 8:

AnswerMarks
8(i)i j k
−4 1 3 =−i+8j−4k
AnswerMarks Guidance
0 1 2M1 Finds normal to Π .
1
AnswerMarks Guidance
−x+8y−4z=3M1 A1 Uses point on plane to find cartesian equation, AEF.
3

AnswerMarks
8(ii)−1  3 
   
8 ⋅ 1 =9 11cosθ
   
   
AnswerMarks Guidance
−4 −1M1 Uses scalar product.
⇒θ =72.5°A1 Accept 1.26 rad.
2

AnswerMarks
8(iii)i j k
3 1 −1 =4i+13j+25k
AnswerMarks Guidance
−1 8 −4M1 A1 Finds direction of line of intersection.
( )
AnswerMarks Guidance
Point on both planes is, e.g. 1,1 ,1 .M1 A1 Finds point common to both planes.
r=i+j+k+λ ( 4i+13j+25k )A1 States vector equation of line.
5
AnswerMarks Guidance
QuestionAnswer Marks
Question 8:
--- 8(i) ---
8(i) | i j k
−4 1 3 =−i+8j−4k
0 1 2 | M1 | Finds normal to Π .
1
−x+8y−4z=3 | M1 A1 | Uses point on plane to find cartesian equation, AEF.
3
--- 8(ii) ---
8(ii) | −1  3 
   
8 ⋅ 1 =9 11cosθ
   
   
−4 −1 | M1 | Uses scalar product.
⇒θ =72.5° | A1 | Accept 1.26 rad.
2
--- 8(iii) ---
8(iii) | i j k
3 1 −1 =4i+13j+25k
−1 8 −4 | M1 A1 | Finds direction of line of intersection.
( )
Point on both planes is, e.g. 1,1 ,1 . | M1 A1 | Finds point common to both planes.
r=i+j+k+λ ( 4i+13j+25k ) | A1 | States vector equation of line.
5
Question | Answer | Marks | Guidance
The plane $\Pi_1$ has equation
$$\mathbf{r} = \begin{pmatrix} 5 \\ 1 \\ 0 \end{pmatrix} + s\begin{pmatrix} -4 \\ 1 \\ 3 \end{pmatrix} + t\begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix}.$$

\begin{enumerate}[label=(\roman*)]
\item Find a cartesian equation of $\Pi_1$. [3]
\end{enumerate}

The plane $\Pi_2$ has equation $3x + y - z = 3$.

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the acute angle between $\Pi_1$ and $\Pi_2$, giving your answer in degrees. [2]

\item Find an equation of the line of intersection of $\Pi_1$ and $\Pi_2$, giving your answer in the form $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE FP1 2018 Q8 [10]}}