Question 5 - A-Level Maths

CAIE Further Paper 1 2024 June — Question 5 12 marks

Exam Board	CAIE
Module	Further Paper 1 (Further Paper 1)
Year	2024
Session	June
Marks	12
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Vectors: Cross Product & Distances
Type	Equation of plane through three points
Difficulty	Standard +0.3 This is a standard Further Maths vectors question requiring routine application of cross product for plane equation, point-to-plane distance formula, and skew lines distance formula. All techniques are textbook procedures with straightforward arithmetic, making it slightly easier than average for Further Maths content.
Spec	4.04b Plane equations: cartesian and vector forms 4.04h Shortest distances: between parallel lines and between skew lines 4.04j Shortest distance: between a point and a plane

5 The points $A , B , C$ have position vectors $$2 \mathbf { i } + 2 \mathbf { j } + 4 \mathbf { k } , \quad 2 \mathbf { i } + 4 \mathbf { j } - \mathbf { k } , \quad - 3 \mathbf { i } - 3 \mathbf { j } + 4 \mathbf { k }$$ respectively, relative to the origin $O$.

Find the equation of the plane $A B C$, giving your answer in the form $a x + b y + c z = d$.
The point $D$ has position vector $2 \mathbf { i } + \mathbf { j } + 3 \mathbf { k }$.
Find the perpendicular distance from $D$ to the plane $A B C$.
Find the shortest distance between the lines $A B$ and $C D$.

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Question 5(a):

Answer	Marks	Guidance
$\overrightarrow{AB} = 2\mathbf{j} - 5\mathbf{k}$, $\overrightarrow{AC} = -5\mathbf{i} - 5\mathbf{j}$, $\overrightarrow{BC} = -5\mathbf{i} - 7\mathbf{j} + 5\mathbf{k}$	B1	Finds direction vectors of two lines in the plane
$\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 0 & 2 & -5\\ -5 & -5 & 0\end{vmatrix} = \begin{pmatrix}-25\\ 25\\ 10\end{pmatrix} \sim \begin{pmatrix}-5\\ 5\\ 2\end{pmatrix}$	M1 A1	Finds normal to the plane $ABC$
$-5(2)+5(2)+2(4) = 8 \Rightarrow -5x+5y+2z=8$	M1 A1	Substitutes point. AEF for final answer

Question 5(b):

Answer	Marks	Guidance
$\frac{8-(2\mathbf{i}+\mathbf{j}+3\mathbf{k})\cdot(-5\mathbf{i}+5\mathbf{j}+2\mathbf{k})}{\sqrt{5^2+5^2+2^2}} = \frac{7}{\sqrt{54}} = 0.953$	M1 A1	Correct formula for distance from point to plane

Question 5(c):

Answer	Marks	Guidance
$\overrightarrow{CD} = \begin{pmatrix}2\\1\\3\end{pmatrix} - \begin{pmatrix}-3\\-3\\4\end{pmatrix} = \begin{pmatrix}5\\4\\-1\end{pmatrix}$	B1
$\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 0 & 2 & -5\\ 5 & 4 & -1\end{vmatrix} = \begin{pmatrix}18\\-25\\-10\end{pmatrix}$	M1 A1	Find common perpendicular
\(\frac{1}{\sqrt{1049}}\left	\begin{pmatrix}-5\\-5\\0\end{pmatrix}\cdot\begin{pmatrix}18\\-25\\-10\end{pmatrix}\right	= \frac{35}{\sqrt{1049}} = 1.08\)

## Question 5(a):

| $\overrightarrow{AB} = 2\mathbf{j} - 5\mathbf{k}$, $\overrightarrow{AC} = -5\mathbf{i} - 5\mathbf{j}$, $\overrightarrow{BC} = -5\mathbf{i} - 7\mathbf{j} + 5\mathbf{k}$ | B1 | Finds direction vectors of two lines in the plane |
| $\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 0 & 2 & -5\\ -5 & -5 & 0\end{vmatrix} = \begin{pmatrix}-25\\ 25\\ 10\end{pmatrix} \sim \begin{pmatrix}-5\\ 5\\ 2\end{pmatrix}$ | M1 A1 | Finds normal to the plane $ABC$ |
| $-5(2)+5(2)+2(4) = 8 \Rightarrow -5x+5y+2z=8$ | M1 A1 | Substitutes point. AEF for final answer |

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## Question 5(b):

| $\frac{8-(2\mathbf{i}+\mathbf{j}+3\mathbf{k})\cdot(-5\mathbf{i}+5\mathbf{j}+2\mathbf{k})}{\sqrt{5^2+5^2+2^2}} = \frac{7}{\sqrt{54}} = 0.953$ | M1 A1 | Correct formula for distance from point to plane |

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## Question 5(c):

| $\overrightarrow{CD} = \begin{pmatrix}2\\1\\3\end{pmatrix} - \begin{pmatrix}-3\\-3\\4\end{pmatrix} = \begin{pmatrix}5\\4\\-1\end{pmatrix}$ | B1 | |
| $\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 0 & 2 & -5\\ 5 & 4 & -1\end{vmatrix} = \begin{pmatrix}18\\-25\\-10\end{pmatrix}$ | M1 A1 | Find common perpendicular |
| $\frac{1}{\sqrt{1049}}\left|\begin{pmatrix}-5\\-5\\0\end{pmatrix}\cdot\begin{pmatrix}18\\-25\\-10\end{pmatrix}\right| = \frac{35}{\sqrt{1049}} = 1.08$ | M1 A1 | Uses formula for shortest distance |

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5 The points $A , B , C$ have position vectors

$$2 \mathbf { i } + 2 \mathbf { j } + 4 \mathbf { k } , \quad 2 \mathbf { i } + 4 \mathbf { j } - \mathbf { k } , \quad - 3 \mathbf { i } - 3 \mathbf { j } + 4 \mathbf { k }$$

respectively, relative to the origin $O$.
\begin{enumerate}[label=(\alph*)]
\item Find the equation of the plane $A B C$, giving your answer in the form $a x + b y + c z = d$.\\

The point $D$ has position vector $2 \mathbf { i } + \mathbf { j } + 3 \mathbf { k }$.
\item Find the perpendicular distance from $D$ to the plane $A B C$.
\item Find the shortest distance between the lines $A B$ and $C D$.
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 1 2024 Q5 [12]}}

This paper (7 questions)

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Q1 6 Q2 6 Q3 8 Q4 13 Q5 12 Q6 15 Q7 15

Answer	Marks	Guidance
\(\overrightarrow{AB} = 2\mathbf{j} - 5\mathbf{k}\), \(\overrightarrow{AC} = -5\mathbf{i} - 5\mathbf{j}\), \(\overrightarrow{BC} = -5\mathbf{i} - 7\mathbf{j} + 5\mathbf{k}\)	B1	Finds direction vectors of two lines in the plane
\(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 0 & 2 & -5\\ -5 & -5 & 0\end{vmatrix} = \begin{pmatrix}-25\\ 25\\ 10\end{pmatrix} \sim \begin{pmatrix}-5\\ 5\\ 2\end{pmatrix}\)	M1 A1	Finds normal to the plane \(ABC\)
\(-5(2)+5(2)+2(4) = 8 \Rightarrow -5x+5y+2z=8\)	M1 A1	Substitutes point. AEF for final answer

Answer	Marks	Guidance
\(\overrightarrow{CD} = \begin{pmatrix}2\\1\\3\end{pmatrix} - \begin{pmatrix}-3\\-3\\4\end{pmatrix} = \begin{pmatrix}5\\4\\-1\end{pmatrix}\)	B1
\(\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 0 & 2 & -5\\ 5 & 4 & -1\end{vmatrix} = \begin{pmatrix}18\\-25\\-10\end{pmatrix}\)	M1 A1	Find common perpendicular
\(\frac{1}{\sqrt{1049}}\left	\begin{pmatrix}-5\\-5\\0\end{pmatrix}\cdot\begin{pmatrix}18\\-25\\-10\end{pmatrix}\right	= \frac{35}{\sqrt{1049}} = 1.08\)