CAIE FP1 2013 November — Question 8 10 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2013
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypeLine of intersection of planes
DifficultyStandard +0.3 This is a standard three-part Further Maths question on planes requiring routine techniques: finding a normal vector via cross product for the Cartesian equation, using the angle formula between planes, and finding the line of intersection by solving simultaneous equations. While it involves multiple steps and Further Maths content, each part follows well-established procedures without requiring novel insight or particularly challenging problem-solving.
Spec4.04b Plane equations: cartesian and vector forms4.04d Angles: between planes and between line and plane4.04e Line intersections: parallel, skew, or intersecting

8 The plane \(\Pi _ { 1 }\) has equation \(\mathbf { r } = \left( \begin{array} { r } 2 \\ 3 \\ - 1 \end{array} \right) + s \left( \begin{array} { l } 1 \\ 0 \\ 1 \end{array} \right) + t \left( \begin{array} { r } 1 \\ - 1 \\ - 2 \end{array} \right)\). Find a cartesian equation of \(\Pi _ { 1 }\). The plane \(\Pi _ { 2 }\) has equation \(2 x - y + z = 10\). Find the acute angle between \(\Pi _ { 1 }\) and \(\Pi _ { 2 }\). Find an equation of the line of intersection of \(\Pi _ { 1 }\) and \(\Pi _ { 2 }\), giving your answer in the form \(\mathbf { r } = \mathbf { a } + \lambda \mathbf { b }\).

Question 8:
Finding normal to Π₁:
AnswerMarks Guidance
\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 1 \\ 1 & -1 & -2 \end{vmatrix} = \mathbf{i} + 3\mathbf{j} - \mathbf{k}\)M1A1 Part mark: 3
Equation of \(\Pi_1\): \(x + 3y - z = 12\)A1
Finding angle between normals:
AnswerMarks Guidance
\(\cos\theta = \dfrac{2-3-1 }{\sqrt{11}\sqrt{6}} = \dfrac{2}{\sqrt{66}} \Rightarrow \theta = 75.7°\) or \(1.32\) rad
Direction of line of intersection:
AnswerMarks
\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 3 & -1 \\ 2 & -1 & 1 \end{vmatrix} = 2\mathbf{i} - 3\mathbf{j} - 7\mathbf{k}\)M1A1
Point common to both planes:
AnswerMarks Guidance
Point on both planes e.g. \((6,2,0)\)M1A1
\(\mathbf{r} = 6\mathbf{i} + 2\mathbf{j} + t(2\mathbf{i} - 3\mathbf{j} - 7\mathbf{k})\) (OE)A1 Part mark: 5, Total: [10]
## Question 8:

**Finding normal to Π₁:**

$\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 1 \\ 1 & -1 & -2 \end{vmatrix} = \mathbf{i} + 3\mathbf{j} - \mathbf{k}$ | M1A1 | Part mark: 3

Equation of $\Pi_1$: $x + 3y - z = 12$ | A1 |

**Finding angle between normals:**

$\cos\theta = \dfrac{|2-3-1|}{\sqrt{11}\sqrt{6}} = \dfrac{2}{\sqrt{66}} \Rightarrow \theta = 75.7°$ or $1.32$ rad | M1, A1 | Part mark: 2

**Direction of line of intersection:**

$\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 3 & -1 \\ 2 & -1 & 1 \end{vmatrix} = 2\mathbf{i} - 3\mathbf{j} - 7\mathbf{k}$ | M1A1 |

**Point common to both planes:**

Point on both planes e.g. $(6,2,0)$ | M1A1 |

$\mathbf{r} = 6\mathbf{i} + 2\mathbf{j} + t(2\mathbf{i} - 3\mathbf{j} - 7\mathbf{k})$ (OE) | A1 | Part mark: 5, Total: **[10]**

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8 The plane $\Pi _ { 1 }$ has equation $\mathbf { r } = \left( \begin{array} { r } 2 \\ 3 \\ - 1 \end{array} \right) + s \left( \begin{array} { l } 1 \\ 0 \\ 1 \end{array} \right) + t \left( \begin{array} { r } 1 \\ - 1 \\ - 2 \end{array} \right)$. Find a cartesian equation of $\Pi _ { 1 }$.

The plane $\Pi _ { 2 }$ has equation $2 x - y + z = 10$. Find the acute angle between $\Pi _ { 1 }$ and $\Pi _ { 2 }$.

Find an equation of the line of intersection of $\Pi _ { 1 }$ and $\Pi _ { 2 }$, giving your answer in the form $\mathbf { r } = \mathbf { a } + \lambda \mathbf { b }$.

\hfill \mbox{\textit{CAIE FP1 2013 Q8 [10]}}