Question 1 - A-Level Maths

CAIE FP1 2014 June — Question 1 5 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2014
Session	June
Marks	5
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Vectors 3D & Lines
Type	Linear combinations of vectors
Difficulty	Moderate -0.5 This is a straightforward linear combination problem requiring students to express one vector in terms of others by solving a system of linear equations. While it involves three dimensions and multiple vectors, the algebraic manipulation is routine for Further Maths students and requires no geometric insight or novel problem-solving—just systematic equation solving.
Spec	4.04a Line equations: 2D and 3D, cartesian and vector forms

1 The vectors $\mathbf { a } , \mathbf { b } , \mathbf { c }$ and $\mathbf { d }$ in $\mathbb { R } ^ { 3 }$ are given by $$\mathbf { a } = \left( \begin{array} { r } 2 \\ - 1 \\ 1 \end{array} \right)$$

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Question 1:

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
$\alpha\begin{pmatrix}2\\-1\\1\end{pmatrix}+\beta\begin{pmatrix}1\\1\\1\end{pmatrix}+\gamma\begin{pmatrix}0\\1\\-1\end{pmatrix}=\mathbf{0}$	M1
$2\alpha+\beta=0$ (1), $-\alpha+\beta+\gamma=0$ (2), $\alpha+\beta-\gamma=0$ (3)
Adding (2) and (3) $\Rightarrow 2\beta=0$. In (1) $\Rightarrow \alpha=0 \Rightarrow \gamma=0$	A1
a, b and c are linearly independent $\therefore$ form basis for $\mathbb{R}^3$	A1	[3]
$l\begin{pmatrix}2\\-1\\1\end{pmatrix}+m\begin{pmatrix}1\\1\\1\end{pmatrix}+n\begin{pmatrix}0\\1\\-1\end{pmatrix}=\begin{pmatrix}3\\-2\\0\end{pmatrix}$	M1
$\Rightarrow l=2,\ m=-1,\ n=1$	A1
$\Rightarrow \mathbf{d}=2\mathbf{a}-\mathbf{b}+\mathbf{c}$		[2]
Alternative: $\begin{vmatrix}2&1&0\\-1&1&1\\1&1&-1\end{vmatrix}=2\times(-2)-0+0=-4\neq 0$	(M1A1)
Row reduce $\begin{pmatrix}2&1&0\\-1&1&1\\1&1&-1\end{pmatrix}\rightarrow\cdots\rightarrow\begin{pmatrix}2&1&0\\0&3&2\\0&0&6\end{pmatrix}$ (OE)	(M1A1)	ISW if a 4th column appears

## Question 1:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\alpha\begin{pmatrix}2\\-1\\1\end{pmatrix}+\beta\begin{pmatrix}1\\1\\1\end{pmatrix}+\gamma\begin{pmatrix}0\\1\\-1\end{pmatrix}=\mathbf{0}$ | M1 | |
| $2\alpha+\beta=0$ (1), $-\alpha+\beta+\gamma=0$ (2), $\alpha+\beta-\gamma=0$ (3) | | |
| Adding (2) and (3) $\Rightarrow 2\beta=0$. In (1) $\Rightarrow \alpha=0 \Rightarrow \gamma=0$ | A1 | |
| **a**, **b** and **c** are linearly independent $\therefore$ form basis for $\mathbb{R}^3$ | A1 | [3] |
| $l\begin{pmatrix}2\\-1\\1\end{pmatrix}+m\begin{pmatrix}1\\1\\1\end{pmatrix}+n\begin{pmatrix}0\\1\\-1\end{pmatrix}=\begin{pmatrix}3\\-2\\0\end{pmatrix}$ | M1 | |
| $\Rightarrow l=2,\ m=-1,\ n=1$ | A1 | |
| $\Rightarrow \mathbf{d}=2\mathbf{a}-\mathbf{b}+\mathbf{c}$ | | [2] |
| **Alternative:** $\begin{vmatrix}2&1&0\\-1&1&1\\1&1&-1\end{vmatrix}=2\times(-2)-0+0=-4\neq 0$ | (M1A1) | |
| Row reduce $\begin{pmatrix}2&1&0\\-1&1&1\\1&1&-1\end{pmatrix}\rightarrow\cdots\rightarrow\begin{pmatrix}2&1&0\\0&3&2\\0&0&6\end{pmatrix}$ (OE) | (M1A1) | ISW if a 4th column appears |

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1 The vectors $\mathbf { a } , \mathbf { b } , \mathbf { c }$ and $\mathbf { d }$ in $\mathbb { R } ^ { 3 }$ are given by

$$\mathbf { a } = \left( \begin{array} { r } 
2 \\
- 1 \\
1
\end{array} \right)$$

\hfill \mbox{\textit{CAIE FP1 2014 Q1 [5]}}

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Q1 5 Q2 6 Q3 7 Q4 7 Q5 8 Q6 10 Q7 10 Q8 11 Q9 10 Q10 12 Q11 EITHER Q11 OR