Question 7 - A-Level Maths

CAIE FP1 2010 June — Question 7 10 marks

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2010
Session	June
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Vectors: Cross Product & Distances
Type	Intersection of line with plane
Difficulty	Standard +0.8 This is a multi-part Further Maths question requiring: (i) solving simultaneous equations to verify line intersection, (ii) finding a plane equation via cross product then using point-to-plane distance formula, and (iii) applying point-to-line distance formula. While the techniques are standard for FP1, the question demands careful algebraic manipulation across multiple steps and integration of several vector concepts, placing it moderately above average difficulty.
Spec	4.04a Line equations: 2D and 3D, cartesian and vector forms 4.04e Line intersections: parallel, skew, or intersecting 4.04f Line-plane intersection: find point 4.04j Shortest distance: between a point and a plane

7 The lines $l _ { 1 }$ and $l _ { 2 }$ have vector equations $$\mathbf { r } = 4 \mathbf { i } - 2 \mathbf { j } + \lambda ( 2 \mathbf { i } + \mathbf { j } - 4 \mathbf { k } ) \quad \text { and } \quad \mathbf { r } = 4 \mathbf { i } - 5 \mathbf { j } + 2 \mathbf { k } + \mu ( \mathbf { i } - \mathbf { j } - \mathbf { k } )$$ respectively.

Show that $l _ { 1 }$ and $l _ { 2 }$ intersect.
Find the perpendicular distance from the point $P$ whose position vector is $3 \mathbf { i } - 5 \mathbf { j } + 6 \mathbf { k }$ to the plane containing $l _ { 1 }$ and $l _ { 2 }$.
Find the perpendicular distance from $P$ to $l _ { 1 }$.

Show mark scheme Show mark scheme source

Answer	Marks
(i) Solves any 2 of the equations: $4 + 2\lambda = 4 + \mu, -2 + \lambda = -5 - \mu, -4\lambda = 2 - \mu$	M1A1
to obtain $\lambda = -1, \mu = -2$
Checks consistency with the third equation	A1
[3]
(ii) $\mathbf{P} = (\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}) \cdot (\mathbf{5i} + 2\mathbf{j} + 3\mathbf{k}) / \sqrt{38}$	M1A1
$= 7 / \sqrt{38} = 1.14$	A1
[3]

OR

Answer	Marks
$\mathbf{n} = -\mathbf{5i} - 2\mathbf{j} - 3\mathbf{k}$	M1
Plane is $5x + 2y + 3z = 16$	A1
$P = \frac{15 - 10 + 18 - 16}{\sqrt{5^2 + 2^2 + 3^2}} = \frac{7}{\sqrt{38}}$	A1

OR

Answer	Marks	Guidance
Plane is $5x + 2y + 3z = 16$	M1A1
Sub. general pt on perpendicular $\left(\begin{array}{c}3 + 5t \\ -5 + 2t \\ 6 + 3t\end{array}\right) \Rightarrow t = -\frac{7}{38}$
$\Rightarrow \mathbf{P} = \left(\begin{array}{c}5t \\ 2t \\ 3t\end{array}\right) = 1.14$	A1
(iii) $(\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}) \times (2\mathbf{i} + \mathbf{j} - 4\mathbf{k}) = 6\mathbf{i} + 8\mathbf{j} + 5\mathbf{k}$	B1
OR $(\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}) \times (2\mathbf{i} + \mathbf{j} - 4\mathbf{k}) = -6\mathbf{i} - 8\mathbf{j} - 5\mathbf{k}$, etc.
\(d =	6\mathbf{i} + 8\mathbf{j} + 5\mathbf{k}	/ \sqrt{21} = \sqrt{125}/21 = 2.44\)
[4]

OR

Answer	Marks
Let $Q$ be the foot of the perpendicular from $P$ to $l$, and $A$ be the known point on $l_1$
$AQ = (\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}) \cdot \frac{(2\mathbf{i} + \mathbf{j} - 4\mathbf{k})}{\sqrt{21}} = \frac{8}{\sqrt{21}}$	M1A1
$AP^2 = 1^2 + (-2)^2 + 2^2 = 9$	B1
$PQ^2 = 9 - \frac{64}{21} = \frac{125}{21} \Rightarrow PQ = \frac{5\sqrt{5}}{21}$	A1

OR

Answer	Marks	Guidance
$\overrightarrow{PQ} = \left(\begin{array}{c}4 + 2t \\ -2 + t \\ -4t\end{array}\right) \cdot \left(\begin{array}{c}3 \\ -5 \\ -6\end{array}\right) = \left(\begin{array}{c}1 + 2t \\ 1 + 2t \\ -6 - 4t\end{array}\right) \cdot \left(\begin{array}{c}2 \\ 1 \\ -4\end{array}\right) = 0$	M1
$\Rightarrow t = -\frac{29}{21}$	A1
\(\overrightarrow{PQ} = \frac{1}{21}\left(\begin{array}{c}-37 \\ 34 \\ -10\end{array}\right) \Rightarrow	\overrightarrow{PQ}	= \frac{1}{21}\sqrt{37^2 + 34^2 + 10^2} = 2.44\)

**(i)** Solves any 2 of the equations: $4 + 2\lambda = 4 + \mu, -2 + \lambda = -5 - \mu, -4\lambda = 2 - \mu$ | M1A1 |

to obtain $\lambda = -1, \mu = -2$ |

Checks consistency with the third equation | A1 |

| [3] |

**(ii)** $\mathbf{P} = (\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}) \cdot (\mathbf{5i} + 2\mathbf{j} + 3\mathbf{k}) / \sqrt{38}$ | M1A1 |

$= 7 / \sqrt{38} = 1.14$ | A1 |

| [3] |

**OR**

$\mathbf{n} = -\mathbf{5i} - 2\mathbf{j} - 3\mathbf{k}$ | M1 |

Plane is $5x + 2y + 3z = 16$ | A1 |

$P = \frac{15 - 10 + 18 - 16}{\sqrt{5^2 + 2^2 + 3^2}} = \frac{7}{\sqrt{38}}$ | A1 |

**OR**

Plane is $5x + 2y + 3z = 16$ | M1A1 |

Sub. general pt on perpendicular $\left(\begin{array}{c}3 + 5t \\ -5 + 2t \\ 6 + 3t\end{array}\right) \Rightarrow t = -\frac{7}{38}$ |

$\Rightarrow \mathbf{P} = \left(\begin{array}{c}5t \\ 2t \\ 3t\end{array}\right) = 1.14$ | A1 |

**(iii)** $(\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}) \times (2\mathbf{i} + \mathbf{j} - 4\mathbf{k}) = 6\mathbf{i} + 8\mathbf{j} + 5\mathbf{k}$ | B1 |

OR $(\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}) \times (2\mathbf{i} + \mathbf{j} - 4\mathbf{k}) = -6\mathbf{i} - 8\mathbf{j} - 5\mathbf{k}$, etc. |

$d = |6\mathbf{i} + 8\mathbf{j} + 5\mathbf{k}| / \sqrt{21} = \sqrt{125}/21 = 2.44$ | M1A1A1 |

| [4] |

**OR**

Let $Q$ be the foot of the perpendicular from $P$ to $l$, and $A$ be the known point on $l_1$ |

$AQ = (\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}) \cdot \frac{(2\mathbf{i} + \mathbf{j} - 4\mathbf{k})}{\sqrt{21}} = \frac{8}{\sqrt{21}}$ | M1A1 |

$AP^2 = 1^2 + (-2)^2 + 2^2 = 9$ | B1 |

$PQ^2 = 9 - \frac{64}{21} = \frac{125}{21} \Rightarrow PQ = \frac{5\sqrt{5}}{21}$ | A1 |

**OR**

$\overrightarrow{PQ} = \left(\begin{array}{c}4 + 2t \\ -2 + t \\ -4t\end{array}\right) \cdot \left(\begin{array}{c}3 \\ -5 \\ -6\end{array}\right) = \left(\begin{array}{c}1 + 2t \\ 1 + 2t \\ -6 - 4t\end{array}\right) \cdot \left(\begin{array}{c}2 \\ 1 \\ -4\end{array}\right) = 0$ | M1 |

$\Rightarrow t = -\frac{29}{21}$ | A1 |

$\overrightarrow{PQ} = \frac{1}{21}\left(\begin{array}{c}-37 \\ 34 \\ -10\end{array}\right) \Rightarrow |\overrightarrow{PQ}| = \frac{1}{21}\sqrt{37^2 + 34^2 + 10^2} = 2.44$ | M1A1 |

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7 The lines $l _ { 1 }$ and $l _ { 2 }$ have vector equations

$$\mathbf { r } = 4 \mathbf { i } - 2 \mathbf { j } + \lambda ( 2 \mathbf { i } + \mathbf { j } - 4 \mathbf { k } ) \quad \text { and } \quad \mathbf { r } = 4 \mathbf { i } - 5 \mathbf { j } + 2 \mathbf { k } + \mu ( \mathbf { i } - \mathbf { j } - \mathbf { k } )$$

respectively.\\
(i) Show that $l _ { 1 }$ and $l _ { 2 }$ intersect.\\
(ii) Find the perpendicular distance from the point $P$ whose position vector is $3 \mathbf { i } - 5 \mathbf { j } + 6 \mathbf { k }$ to the plane containing $l _ { 1 }$ and $l _ { 2 }$.\\
(iii) Find the perpendicular distance from $P$ to $l _ { 1 }$.

\hfill \mbox{\textit{CAIE FP1 2010 Q7 [10]}}

This paper (12 questions)

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Q1 5 Q2 7 Q3 7 Q4 7 Q5 9 Q6 9 Q7 10 Q8 10 Q9 11 Q10 11 Q11 EITHER Q11 OR

Answer	Marks
(i) Solves any 2 of the equations: \(4 + 2\lambda = 4 + \mu, -2 + \lambda = -5 - \mu, -4\lambda = 2 - \mu\)	M1A1
to obtain \(\lambda = -1, \mu = -2\)
Checks consistency with the third equation	A1
[3]
(ii) \(\mathbf{P} = (\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}) \cdot (\mathbf{5i} + 2\mathbf{j} + 3\mathbf{k}) / \sqrt{38}\)	M1A1
\(= 7 / \sqrt{38} = 1.14\)	A1
[3]

Answer	Marks
\(\mathbf{n} = -\mathbf{5i} - 2\mathbf{j} - 3\mathbf{k}\)	M1
Plane is \(5x + 2y + 3z = 16\)	A1
\(P = \frac{15 - 10 + 18 - 16}{\sqrt{5^2 + 2^2 + 3^2}} = \frac{7}{\sqrt{38}}\)	A1

Answer	Marks	Guidance
Plane is \(5x + 2y + 3z = 16\)	M1A1
Sub. general pt on perpendicular \(\left(\begin{array}{c}3 + 5t \\ -5 + 2t \\ 6 + 3t\end{array}\right) \Rightarrow t = -\frac{7}{38}\)
\(\Rightarrow \mathbf{P} = \left(\begin{array}{c}5t \\ 2t \\ 3t\end{array}\right) = 1.14\)	A1
(iii) \((\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}) \times (2\mathbf{i} + \mathbf{j} - 4\mathbf{k}) = 6\mathbf{i} + 8\mathbf{j} + 5\mathbf{k}\)	B1
OR \((\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}) \times (2\mathbf{i} + \mathbf{j} - 4\mathbf{k}) = -6\mathbf{i} - 8\mathbf{j} - 5\mathbf{k}\), etc.
\(d =	6\mathbf{i} + 8\mathbf{j} + 5\mathbf{k}	/ \sqrt{21} = \sqrt{125}/21 = 2.44\)
[4]

Answer	Marks
Let \(Q\) be the foot of the perpendicular from \(P\) to \(l\), and \(A\) be the known point on \(l_1\)
\(AQ = (\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}) \cdot \frac{(2\mathbf{i} + \mathbf{j} - 4\mathbf{k})}{\sqrt{21}} = \frac{8}{\sqrt{21}}\)	M1A1
\(AP^2 = 1^2 + (-2)^2 + 2^2 = 9\)	B1
\(PQ^2 = 9 - \frac{64}{21} = \frac{125}{21} \Rightarrow PQ = \frac{5\sqrt{5}}{21}\)	A1

Answer	Marks	Guidance
\(\overrightarrow{PQ} = \left(\begin{array}{c}4 + 2t \\ -2 + t \\ -4t\end{array}\right) \cdot \left(\begin{array}{c}3 \\ -5 \\ -6\end{array}\right) = \left(\begin{array}{c}1 + 2t \\ 1 + 2t \\ -6 - 4t\end{array}\right) \cdot \left(\begin{array}{c}2 \\ 1 \\ -4\end{array}\right) = 0\)	M1
\(\Rightarrow t = -\frac{29}{21}\)	A1
\(\overrightarrow{PQ} = \frac{1}{21}\left(\begin{array}{c}-37 \\ 34 \\ -10\end{array}\right) \Rightarrow	\overrightarrow{PQ}	= \frac{1}{21}\sqrt{37^2 + 34^2 + 10^2} = 2.44\)