CAIE FP1 2014 November — Question 10 12 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2014
SessionNovember
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Cross Product & Distances
TypeCommon perpendicular to two skew lines
DifficultyChallenging +1.3 This is a multi-part Further Maths question on skew lines requiring systematic application of vector methods: finding the common perpendicular using dot product conditions, computing shortest distance, deriving a plane equation, and finding point-to-plane distance. While it involves several steps and FM content, each part follows standard procedures without requiring novel geometric insight—harder than typical A-level but routine for FM students who know the techniques.
Spec4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04g Vector product: a x b perpendicular vector4.04i Shortest distance: between a point and a line4.04j Shortest distance: between a point and a plane
10 The line \(l _ { 1 }\) is parallel to the vector \(\mathbf { i } - 2 \mathbf { j } - 3 \mathbf { k }\) and passes through the point \(A\), whose position vector is \(3 \mathbf { i } + 3 \mathbf { j } - 4 \mathbf { k }\). The line \(l _ { 2 }\) is parallel to the vector \(- 2 \mathbf { i } + \mathbf { j } + 3 \mathbf { k }\) and passes through the point \(B\), whose position vector is \(- 3 \mathbf { i } - \mathbf { j } + 2 \mathbf { k }\). The point \(P\) on \(l _ { 1 }\) and the point \(Q\) on \(l _ { 2 }\) are such that \(P Q\) is perpendicular to both \(l _ { 1 }\) and \(l _ { 2 }\). Find
  1. the length \(P Q\),
  2. the cartesian equation of the plane \(\Pi\) containing \(P Q\) and \(l _ { 2 }\),
  3. the perpendicular distance of \(A\) from \(\Pi\).
Question 10:
Part (i):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Cross product of direction vectors \(= \begin{pmatrix}-3\\3\\-3\end{pmatrix} \sim \begin{pmatrix}1\\-1\\1\end{pmatrix}\)M1A1
\(\overrightarrow{BA} = 6\mathbf{i} + 4\mathbf{j} - 6\mathbf{k}\)B1
Shortest distance \(= \dfrac{\begin{vmatrix}\begin{pmatrix}6\\4\\-6\end{pmatrix}\cdot\begin{pmatrix}1\\-1\\1\end{pmatrix}\end{vmatrix}}{\sqrt{1^2+1^2+1^2}} = \frac{4}{\sqrt{3}}\) (= 2.31)M1A1 (5)
Part (ii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Cross product \(= \begin{pmatrix}-4\\-5\\-1\end{pmatrix} \sim \begin{pmatrix}4\\5\\1\end{pmatrix}\)M1A1
Cartesian equation of \(\Pi\): \(4x + 5y + z = -12 - 5 + 2 = -15\)M1A1 (4)
Part (iii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Distance of \(A\) from \(\Pi\): \(\dfrac{\begin{vmatrix}\begin{pmatrix}6\\4\\-6\end{pmatrix}\cdot\begin{pmatrix}4\\5\\1\end{pmatrix}\end{vmatrix}}{\sqrt{4^2+5^2+1^2}}\)M1A1
\(= \dfrac{38}{\sqrt{42}}\) (= 5.86)A1 (3) [12]
Question 11E:
Part (i):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\alpha + \beta + \gamma + \delta = -4\)B1 (1)
Part (ii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = (-4)^2 - 2\times2 = 12\)M1A1 (2)
Part (iii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}+\frac{1}{\delta} = \frac{-(-4)}{1} = 4\)M1A1 (2)
Part (iv):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{\alpha}{\beta\gamma\delta}+\frac{\beta}{\alpha\gamma\delta}+\frac{\gamma}{\alpha\beta\delta}+\frac{\delta}{\alpha\beta\gamma} = \frac{\alpha^2+\beta^2+\gamma^2+\delta^2}{\alpha\beta\gamma\delta} = \frac{12}{1} = 12\)M1A1 (2)
\(y = x+1 \Rightarrow x = y-1\); substitution and simplificationM1A1
\(2(y-1)^2 - 4(y-1)+1 = 2y^2 - 8y + 7\)A1
\(\Rightarrow x^4 + 4x^3 + 2x^2 - 4x + 1 = y^4 - 4y^2 + 4 = 0\)A1
\((y^2-2)^2 = 0 \Rightarrow y = \pm\sqrt{2}\) (twice)A1
\(\Rightarrow x = \pm\sqrt{2}-1\) (twice)M1A1 (7) [14] Some indication of four roots required for final mark
Question 11O:
Part (i):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\mathbf{Ae} = \lambda\mathbf{e}\); since \(\mathbf{A}\) is non-singular \(\mathbf{Ae} \neq 0 \Rightarrow \lambda \neq 0\) (\(\mathbf{e} \neq 0\))M1A1 (2)
Part (ii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\mathbf{Ae} = \lambda\mathbf{e} \Rightarrow \mathbf{A}^{-1}\mathbf{Ae} = \mathbf{A}^{-1}\lambda\mathbf{e}\)M1
\(\Rightarrow \mathbf{e} = \lambda\mathbf{A}^{-1}\mathbf{e} \Rightarrow \mathbf{A}^{-1}\mathbf{e} = \frac{1}{\lambda}\mathbf{e}\)A1 (2)
Eigenvalues of \(\mathbf{A}\) are \(-2, -1, 3\)B2,1,0 1 mark for any one, 2 for all three
Corresponding eigenvectors: \(\begin{pmatrix}1\\0\\0\end{pmatrix}, \begin{pmatrix}2\\1\\0\end{pmatrix}, \begin{pmatrix}-6\\25\\20\end{pmatrix}\)M1A1, A1A1 M1A1 for one, A1 for each other
\(\mathbf{P} = \begin{pmatrix}1&2&-6\\0&1&25\\0&0&20\end{pmatrix}\)B1
Eigenvalues of \(\mathbf{A}+3\mathbf{I}\) are \(1, 2, 6\)B1 Award B3 if obtained from \(\mathbf{A}+3\mathbf{I}\)
Eigenvalues of \((\mathbf{A}+3\mathbf{I})^{-1}\) are \(1, \frac{1}{2}, \frac{1}{6}\)B1
\(\mathbf{D} = \begin{pmatrix}1&0&0\\0&\frac{1}{2}&0\\0&0&\frac{1}{6}\end{pmatrix}\) (CAO)B1 (10) [14] 
# Question 10:

## Part (i):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Cross product of direction vectors $= \begin{pmatrix}-3\\3\\-3\end{pmatrix} \sim \begin{pmatrix}1\\-1\\1\end{pmatrix}$ | M1A1 | |
| $\overrightarrow{BA} = 6\mathbf{i} + 4\mathbf{j} - 6\mathbf{k}$ | B1 | |
| Shortest distance $= \dfrac{\begin{vmatrix}\begin{pmatrix}6\\4\\-6\end{pmatrix}\cdot\begin{pmatrix}1\\-1\\1\end{pmatrix}\end{vmatrix}}{\sqrt{1^2+1^2+1^2}} = \frac{4}{\sqrt{3}}$ (= 2.31) | M1A1 (5) | |

## Part (ii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Cross product $= \begin{pmatrix}-4\\-5\\-1\end{pmatrix} \sim \begin{pmatrix}4\\5\\1\end{pmatrix}$ | M1A1 | |
| Cartesian equation of $\Pi$: $4x + 5y + z = -12 - 5 + 2 = -15$ | M1A1 (4) | |

## Part (iii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| Distance of $A$ from $\Pi$: $\dfrac{\begin{vmatrix}\begin{pmatrix}6\\4\\-6\end{pmatrix}\cdot\begin{pmatrix}4\\5\\1\end{pmatrix}\end{vmatrix}}{\sqrt{4^2+5^2+1^2}}$ | M1A1 | |
| $= \dfrac{38}{\sqrt{42}}$ (= 5.86) | A1 (3) **[12]** | |

---

# Question 11E:

## Part (i):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\alpha + \beta + \gamma + \delta = -4$ | B1 (1) | |

## Part (ii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = (-4)^2 - 2\times2 = 12$ | M1A1 (2) | |

## Part (iii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}+\frac{1}{\delta} = \frac{-(-4)}{1} = 4$ | M1A1 (2) | |

## Part (iv):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{\alpha}{\beta\gamma\delta}+\frac{\beta}{\alpha\gamma\delta}+\frac{\gamma}{\alpha\beta\delta}+\frac{\delta}{\alpha\beta\gamma} = \frac{\alpha^2+\beta^2+\gamma^2+\delta^2}{\alpha\beta\gamma\delta} = \frac{12}{1} = 12$ | M1A1 (2) | |
| $y = x+1 \Rightarrow x = y-1$; substitution and simplification | M1A1 | |
| $2(y-1)^2 - 4(y-1)+1 = 2y^2 - 8y + 7$ | A1 | |
| $\Rightarrow x^4 + 4x^3 + 2x^2 - 4x + 1 = y^4 - 4y^2 + 4 = 0$ | A1 | |
| $(y^2-2)^2 = 0 \Rightarrow y = \pm\sqrt{2}$ (twice) | A1 | |
| $\Rightarrow x = \pm\sqrt{2}-1$ (twice) | M1A1 (7) **[14]** | Some indication of four roots required for final mark |

---

# Question 11O:

## Part (i):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{Ae} = \lambda\mathbf{e}$; since $\mathbf{A}$ is non-singular $\mathbf{Ae} \neq 0 \Rightarrow \lambda \neq 0$ ($\mathbf{e} \neq 0$) | M1A1 (2) | |

## Part (ii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{Ae} = \lambda\mathbf{e} \Rightarrow \mathbf{A}^{-1}\mathbf{Ae} = \mathbf{A}^{-1}\lambda\mathbf{e}$ | M1 | |
| $\Rightarrow \mathbf{e} = \lambda\mathbf{A}^{-1}\mathbf{e} \Rightarrow \mathbf{A}^{-1}\mathbf{e} = \frac{1}{\lambda}\mathbf{e}$ | A1 (2) | |
| Eigenvalues of $\mathbf{A}$ are $-2, -1, 3$ | B2,1,0 | 1 mark for any one, 2 for all three |
| Corresponding eigenvectors: $\begin{pmatrix}1\\0\\0\end{pmatrix}, \begin{pmatrix}2\\1\\0\end{pmatrix}, \begin{pmatrix}-6\\25\\20\end{pmatrix}$ | M1A1, A1A1 | M1A1 for one, A1 for each other |
| $\mathbf{P} = \begin{pmatrix}1&2&-6\\0&1&25\\0&0&20\end{pmatrix}$ | B1 | |
| Eigenvalues of $\mathbf{A}+3\mathbf{I}$ are $1, 2, 6$ | B1 | Award B3 if obtained from $\mathbf{A}+3\mathbf{I}$ |
| Eigenvalues of $(\mathbf{A}+3\mathbf{I})^{-1}$ are $1, \frac{1}{2}, \frac{1}{6}$ | B1 | |
| $\mathbf{D} = \begin{pmatrix}1&0&0\\0&\frac{1}{2}&0\\0&0&\frac{1}{6}\end{pmatrix}$ (CAO) | B1 (10) **[14]** | |
10 The line $l _ { 1 }$ is parallel to the vector $\mathbf { i } - 2 \mathbf { j } - 3 \mathbf { k }$ and passes through the point $A$, whose position vector is $3 \mathbf { i } + 3 \mathbf { j } - 4 \mathbf { k }$. The line $l _ { 2 }$ is parallel to the vector $- 2 \mathbf { i } + \mathbf { j } + 3 \mathbf { k }$ and passes through the point $B$, whose position vector is $- 3 \mathbf { i } - \mathbf { j } + 2 \mathbf { k }$. The point $P$ on $l _ { 1 }$ and the point $Q$ on $l _ { 2 }$ are such that $P Q$ is perpendicular to both $l _ { 1 }$ and $l _ { 2 }$. Find\\
(i) the length $P Q$,\\
(ii) the cartesian equation of the plane $\Pi$ containing $P Q$ and $l _ { 2 }$,\\
(iii) the perpendicular distance of $A$ from $\Pi$.

\hfill \mbox{\textit{CAIE FP1 2014 Q10 [12]}}
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