Question 11 OR - A-Level Maths

CAIE FP1 2014 June — Question 11 OR

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2014
Session	June
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Vectors: Lines & Planes
Type	Plane containing line and point/vector
Difficulty	Standard +0.8 This is a substantial multi-part Further Maths question requiring vector manipulation, plane equations, perpendicular distances, and coordinate geometry. While the individual techniques are standard (checking point in plane, finding normals, using section formula), the question requires careful bookkeeping across four connected parts with multiple vectors and planes. The final perpendicular distance calculation adds computational complexity. Slightly above average difficulty for Further Maths, but not requiring exceptional insight.
Spec	1.10a Vectors in 2D: i,j notation and column vectors 1.10b Vectors in 3D: i,j,k notation 1.10c Magnitude and direction: of vectors 1.10d Vector operations: addition and scalar multiplication 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 4.04g Vector product: a x b perpendicular vector

With respect to an origin $O$, the point $A$ has position vector $4 \mathbf { i } - 2 \mathbf { j } + 2 \mathbf { k }$ and the plane $\Pi _ { 1 }$ has equation $$\mathbf { r } = ( 4 + \lambda + 3 \mu ) \mathbf { i } + ( - 2 + 7 \lambda + \mu ) \mathbf { j } + ( 2 + \lambda - \mu ) \mathbf { k } ,$$ where $\lambda$ and $\mu$ are real. The point $L$ is such that $\overrightarrow { O L } = 3 \overrightarrow { O A }$ and $\Pi _ { 2 }$ is the plane through $L$ which is parallel to $\Pi _ { 1 }$. The point $M$ is such that $\overrightarrow { A M } = 3 \overrightarrow { M L }$.

Show that $A$ is in $\Pi _ { 1 }$.
Find a vector perpendicular to $\Pi _ { 2 }$.
Find the position vector of the point $N$ in $\Pi _ { 2 }$ such that $O N$ is perpendicular to $\Pi _ { 2 }$.
Show that the position vector of $M$ is $10 \mathbf { i } - 5 \mathbf { j } + 5 \mathbf { k }$ and find the perpendicular distance of $M$ from the line through $O$ and $N$, giving your answer correct to 3 significant figures.

Show mark scheme Show mark scheme source

Question 11:

EITHER Path

Partial Fractions Setup:

Answer	Marks
$2 + \frac{A}{x-1} + \frac{B}{x+1} = \frac{2x^2-1+A(x+1)+B(x-1)}{x^2-1} = \frac{2x^2+(A+B)x+A-B-2}{x^2-1} = \frac{2x^2-x+5}{x^2-1}$	M1
$\Rightarrow A = 3$, $B = -4$	A1A1 [3]

Turning Points:

Answer	Marks
$y' = 0 \Rightarrow (x^2-1)(4x-1)-(2x^2-x+5).2x = 0 \Rightarrow x^2-14x+1=0$	M1A1
$B^2-4AC = (-14)^2 - 4 \times 1 \times 1 = 192 > 0 \Rightarrow$ two distinct turning points	M1A1 [4]

Asymptotes and Intersection:

Answer	Marks	Guidance
Asymptotes are $x=1$, $x=-1$, $y=2$	B1B1
$y=2 \Rightarrow 2x^2-x+5=2x^2-2 \Rightarrow x=7 \Rightarrow (7,2)$	M1A1	Accept if labelled on graph

Sketch:

Answer	Marks
Middle branch crossing $y$-axis at $(0,-5)$ and left branch	B1
Right branch	B1
Working to show no intersections with $x$-axis	B1 [7]

OR Path

(i):

Answer	Marks
$\mathbf{r} = 4\mathbf{i}-2\mathbf{j}+2\mathbf{k}+\lambda(\mathbf{i}+7\mathbf{j}+\mathbf{k})+\mu(3\mathbf{i}+\mathbf{j}-\mathbf{k}) \Rightarrow A$ is in $\Pi_1$	B1 [1]

(ii):

Answer	Marks
$\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 7 & 1 \\ 3 & 1 & -1 \end{vmatrix} = \begin{pmatrix}-8\\4\\-20\end{pmatrix} \sim \begin{pmatrix}2\\-1\\5\end{pmatrix}$	M1A1 [2]

(iii):

Answer	Marks
$L$ is $(12,-6,6)$	B1
$2x-y+5z = 24+6+30=60$	B1
$\mathbf{n} = t(2\mathbf{i}-\mathbf{j}+5\mathbf{k}) \Rightarrow 4t+t+25t=60 \Rightarrow t=2$	M1A1$\checkmark$
$\mathbf{n} = 4\mathbf{i}-2\mathbf{j}+10\mathbf{k}$	A1 [5]

(iv):

Answer	Marks	Guidance
$\mathbf{m} = 4\mathbf{i}-2\mathbf{j}+2\mathbf{k}+\frac{3}{4}(8\mathbf{i}-4\mathbf{j}+4\mathbf{k}) = 10\mathbf{i}-5\mathbf{j}+5\mathbf{k}$ (AG)	B1
$M$ is $(10,-5,5) \Rightarrow \overrightarrow{NM} = 6\mathbf{i}-3\mathbf{j}-5\mathbf{k}$	B1$\checkmark$
$(6\mathbf{i}-3\mathbf{j}-5\mathbf{k})\times(2\mathbf{i}-\mathbf{j}+5\mathbf{k}) = -\mathbf{i}+2\mathbf{j}$	M1A1
Perpendicular distance is \(\dfrac{\	20(-\mathbf{i}+2\mathbf{j})\	}{\sqrt{30}} = \dfrac{20}{\sqrt{6}} = 8.16\)

# Question 11:

## EITHER Path

**Partial Fractions Setup:**
| $2 + \frac{A}{x-1} + \frac{B}{x+1} = \frac{2x^2-1+A(x+1)+B(x-1)}{x^2-1} = \frac{2x^2+(A+B)x+A-B-2}{x^2-1} = \frac{2x^2-x+5}{x^2-1}$ | M1 | |
|---|---|---|
| $\Rightarrow A = 3$, $B = -4$ | A1A1 [3] | |

**Turning Points:**
| $y' = 0 \Rightarrow (x^2-1)(4x-1)-(2x^2-x+5).2x = 0 \Rightarrow x^2-14x+1=0$ | M1A1 | |
|---|---|---|
| $B^2-4AC = (-14)^2 - 4 \times 1 \times 1 = 192 > 0 \Rightarrow$ two distinct turning points | M1A1 [4] | |

**Asymptotes and Intersection:**
| Asymptotes are $x=1$, $x=-1$, $y=2$ | B1B1 | |
|---|---|---|
| $y=2 \Rightarrow 2x^2-x+5=2x^2-2 \Rightarrow x=7 \Rightarrow (7,2)$ | M1A1 | Accept if labelled on graph |

**Sketch:**
| Middle branch crossing $y$-axis at $(0,-5)$ and left branch | B1 | |
|---|---|---|
| Right branch | B1 | |
| Working to show no intersections with $x$-axis | B1 [7] | |

---

## OR Path

**(i):**
| $\mathbf{r} = 4\mathbf{i}-2\mathbf{j}+2\mathbf{k}+\lambda(\mathbf{i}+7\mathbf{j}+\mathbf{k})+\mu(3\mathbf{i}+\mathbf{j}-\mathbf{k}) \Rightarrow A$ is in $\Pi_1$ | B1 [1] | |
|---|---|---|

**(ii):**
| $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 7 & 1 \\ 3 & 1 & -1 \end{vmatrix} = \begin{pmatrix}-8\\4\\-20\end{pmatrix} \sim \begin{pmatrix}2\\-1\\5\end{pmatrix}$ | M1A1 [2] | |
|---|---|---|

**(iii):**
| $L$ is $(12,-6,6)$ | B1 | |
|---|---|---|
| $2x-y+5z = 24+6+30=60$ | B1 | |
| $\mathbf{n} = t(2\mathbf{i}-\mathbf{j}+5\mathbf{k}) \Rightarrow 4t+t+25t=60 \Rightarrow t=2$ | M1A1$\checkmark$ | |
| $\mathbf{n} = 4\mathbf{i}-2\mathbf{j}+10\mathbf{k}$ | A1 [5] | |

**(iv):**
| $\mathbf{m} = 4\mathbf{i}-2\mathbf{j}+2\mathbf{k}+\frac{3}{4}(8\mathbf{i}-4\mathbf{j}+4\mathbf{k}) = 10\mathbf{i}-5\mathbf{j}+5\mathbf{k}$ (AG) | B1 | |
|---|---|---|
| $M$ is $(10,-5,5) \Rightarrow \overrightarrow{NM} = 6\mathbf{i}-3\mathbf{j}-5\mathbf{k}$ | B1$\checkmark$ | |
| $(6\mathbf{i}-3\mathbf{j}-5\mathbf{k})\times(2\mathbf{i}-\mathbf{j}+5\mathbf{k}) = -\mathbf{i}+2\mathbf{j}$ | M1A1 | |
| Perpendicular distance is $\dfrac{\|20(-\mathbf{i}+2\mathbf{j})\|}{\sqrt{30}} = \dfrac{20}{\sqrt{6}} = 8.16$ | M1A1 [6] | Mark various alternative methods similarly |

Show LaTeX source

With respect to an origin $O$, the point $A$ has position vector $4 \mathbf { i } - 2 \mathbf { j } + 2 \mathbf { k }$ and the plane $\Pi _ { 1 }$ has equation

$$\mathbf { r } = ( 4 + \lambda + 3 \mu ) \mathbf { i } + ( - 2 + 7 \lambda + \mu ) \mathbf { j } + ( 2 + \lambda - \mu ) \mathbf { k } ,$$

where $\lambda$ and $\mu$ are real. The point $L$ is such that $\overrightarrow { O L } = 3 \overrightarrow { O A }$ and $\Pi _ { 2 }$ is the plane through $L$ which is parallel to $\Pi _ { 1 }$. The point $M$ is such that $\overrightarrow { A M } = 3 \overrightarrow { M L }$.\\
(i) Show that $A$ is in $\Pi _ { 1 }$.\\
(ii) Find a vector perpendicular to $\Pi _ { 2 }$.\\
(iii) Find the position vector of the point $N$ in $\Pi _ { 2 }$ such that $O N$ is perpendicular to $\Pi _ { 2 }$.\\
(iv) Show that the position vector of $M$ is $10 \mathbf { i } - 5 \mathbf { j } + 5 \mathbf { k }$ and find the perpendicular distance of $M$ from the line through $O$ and $N$, giving your answer correct to 3 significant figures.

\hfill \mbox{\textit{CAIE FP1 2014 Q11 OR}}

This paper (12 questions)

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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

Answer	Marks
\(2 + \frac{A}{x-1} + \frac{B}{x+1} = \frac{2x^2-1+A(x+1)+B(x-1)}{x^2-1} = \frac{2x^2+(A+B)x+A-B-2}{x^2-1} = \frac{2x^2-x+5}{x^2-1}\)	M1
\(\Rightarrow A = 3\), \(B = -4\)	A1A1 [3]

Answer	Marks
\(y' = 0 \Rightarrow (x^2-1)(4x-1)-(2x^2-x+5).2x = 0 \Rightarrow x^2-14x+1=0\)	M1A1
\(B^2-4AC = (-14)^2 - 4 \times 1 \times 1 = 192 > 0 \Rightarrow\) two distinct turning points	M1A1 [4]

Answer	Marks	Guidance
Asymptotes are \(x=1\), \(x=-1\), \(y=2\)	B1B1
\(y=2 \Rightarrow 2x^2-x+5=2x^2-2 \Rightarrow x=7 \Rightarrow (7,2)\)	M1A1	Accept if labelled on graph

Answer	Marks
Middle branch crossing \(y\)-axis at \((0,-5)\) and left branch	B1
Right branch	B1
Working to show no intersections with \(x\)-axis	B1 [7]

Answer	Marks
\(L\) is \((12,-6,6)\)	B1
\(2x-y+5z = 24+6+30=60\)	B1
\(\mathbf{n} = t(2\mathbf{i}-\mathbf{j}+5\mathbf{k}) \Rightarrow 4t+t+25t=60 \Rightarrow t=2\)	M1A1\(\checkmark\)
\(\mathbf{n} = 4\mathbf{i}-2\mathbf{j}+10\mathbf{k}\)	A1 [5]

Answer	Marks	Guidance
\(\mathbf{m} = 4\mathbf{i}-2\mathbf{j}+2\mathbf{k}+\frac{3}{4}(8\mathbf{i}-4\mathbf{j}+4\mathbf{k}) = 10\mathbf{i}-5\mathbf{j}+5\mathbf{k}\) (AG)	B1
\(M\) is \((10,-5,5) \Rightarrow \overrightarrow{NM} = 6\mathbf{i}-3\mathbf{j}-5\mathbf{k}\)	B1\(\checkmark\)
\((6\mathbf{i}-3\mathbf{j}-5\mathbf{k})\times(2\mathbf{i}-\mathbf{j}+5\mathbf{k}) = -\mathbf{i}+2\mathbf{j}\)	M1A1
Perpendicular distance is \(\dfrac{\	20(-\mathbf{i}+2\mathbf{j})\	}{\sqrt{30}} = \dfrac{20}{\sqrt{6}} = 8.16\)