CAIE Further Paper 1 2024 November — Question 7 15 marks

Exam BoardCAIE
ModuleFurther Paper 1 (Further Paper 1)
Year2024
SessionNovember
Marks15
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 vectors question requiring cross products, plane equations, and finding the common perpendicular between skew lines. Part (a) is standard (cross product for normal vector), part (b) requires finding a second plane then using dot product for angle, and part (c) involves systematic use of perpendicularity conditions with parameters. While requiring multiple techniques and careful algebra across 15 marks, these are well-practiced Further Maths procedures without requiring novel geometric insight. Slightly above average difficulty for Further Maths content.
Spec4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04d Angles: between planes and between line and plane4.04g Vector product: a x b perpendicular vector4.04h Shortest distances: between parallel lines and between skew lines
The lines \(l_1\) and \(l_2\) have equations \(\mathbf{r} = \mathbf{i} + 3\mathbf{j} - 2\mathbf{k} + \lambda(2\mathbf{i} + \mathbf{j} + \mathbf{k})\) and \(\mathbf{r} = \mathbf{i} - 2\mathbf{j} + 9\mathbf{k} + \mu(\mathbf{i} - 4\mathbf{j} + 2\mathbf{k})\) respectively. The plane \(\Pi_1\) contains \(l_1\) and is parallel to \(l_2\).
  1. Find the equation of \(\Pi_1\), giving your answer in the form \(ax + by + cz = d\). [4]
The plane \(\Pi_2\) contains \(l_2\) and the point with coordinates \((2, -1, 7)\).
  1. Find the acute angle between \(\Pi_1\) and \(\Pi_2\). [4]
The point \(P\) on \(l_1\) and the point \(Q\) on \(l_2\) are such that \(PQ\) is perpendicular to both \(l_1\) and \(l_2\).
  1. Find a vector equation for \(PQ\). [7]
Question 7:
AnswerMarks
7(a)i j k  6  −2
   
2 1 1 = −3 ~ 1
   
   
AnswerMarks Guidance
1 −4 2 −9  3 M1 A1 Finds common perpendicular. Allow one error.
−2(1)+(3)+3(−2)=−5M1 Substitutes point on l .
1
AnswerMarks Guidance
2x−y−3z=5A1 CAO.
4
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
7(b)i j k 6
 
1 −4 2 = 4
 
 
AnswerMarks Guidance
1 1 −2 5M1 A1 Finds the normal to  .
2
 2  6
    −7
−1 4 = 14 77coscos=
   
    14 77
AnswerMarks Guidance
−3 5M1 Uses dot product of normal vectors.
77.7A1 No ISW. Accept 1.36 rad.
4
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
7(c)1+2  1+   −2 
     
OP= 3+ ,OQ= −2−4 PQ= −5−4−
     
     
AnswerMarks Guidance
−2+  9+2 11+2−M1 A1 Finds PQ.
 −2  2  −2   6 
       
−5−4− 1 =0 or −5−4− =k −3
       
       
AnswerMarks Guidance
11+2− 1 11+2− −9M1 Uses that dot product of PQ with line direction
is zero, or, alternatively, PQ is a multiple of the
common perpendicular.
AnswerMarks Guidance
−6+6=0A1 Deduces one equation.
 −2   1 
   
−5−4− −4 =021+42=0
   
   
AnswerMarks Guidance
11+2−  2 A1 Deduces second equation.
 3  −1
   
=1OP= 4 or =−2OQ= 6
   
   
AnswerMarks Guidance
−1  5 M1 Solves for  or  and substitutes into OP or
OQ
−1  6 
   
r= 6 +t −3
   
   
AnswerMarks Guidance
 5  −9A1 FT OE. FT using their common perpendicular. Must
have"r=".
7
Question 7:
--- 7(a) ---
7(a) | i j k  6  −2
   
2 1 1 = −3 ~ 1
   
   
1 −4 2 −9  3  | M1 A1 | Finds common perpendicular. Allow one error.
−2(1)+(3)+3(−2)=−5 | M1 | Substitutes point on l .
1
2x−y−3z=5 | A1 | CAO.
4
Question | Answer | Marks | Guidance
--- 7(b) ---
7(b) | i j k 6
 
1 −4 2 = 4
 
 
1 1 −2 5 | M1 A1 | Finds the normal to  .
2
 2  6
    −7
−1 4 = 14 77coscos=
   
    14 77
−3 5 | M1 | Uses dot product of normal vectors.
77.7 | A1 | No ISW. Accept 1.36 rad.
4
Question | Answer | Marks | Guidance
--- 7(c) ---
7(c) | 1+2  1+   −2 
     
OP= 3+ ,OQ= −2−4 PQ= −5−4−
     
     
−2+  9+2 11+2− | M1 A1 | Finds PQ.
 −2  2  −2   6 
       
−5−4− 1 =0 or −5−4− =k −3
       
       
11+2− 1 11+2− −9 | M1 | Uses that dot product of PQ with line direction
is zero, or, alternatively, PQ is a multiple of the
common perpendicular.
−6+6=0 | A1 | Deduces one equation.
 −2   1 
   
−5−4− −4 =021+42=0
   
   
11+2−  2  | A1 | Deduces second equation.
 3  −1
   
=1OP= 4 or =−2OQ= 6
   
   
−1  5  | M1 | Solves for  or  and substitutes into OP or
OQ
−1  6 
   
r= 6 +t −3
   
   
 5  −9 | A1 FT | OE. FT using their common perpendicular. Must
have"r=".
7
The lines $l_1$ and $l_2$ have equations $\mathbf{r} = \mathbf{i} + 3\mathbf{j} - 2\mathbf{k} + \lambda(2\mathbf{i} + \mathbf{j} + \mathbf{k})$ and $\mathbf{r} = \mathbf{i} - 2\mathbf{j} + 9\mathbf{k} + \mu(\mathbf{i} - 4\mathbf{j} + 2\mathbf{k})$ respectively. The plane $\Pi_1$ contains $l_1$ and is parallel to $l_2$.

\begin{enumerate}[label=(\alph*)]
\item Find the equation of $\Pi_1$, giving your answer in the form $ax + by + cz = d$. [4]
\end{enumerate}

The plane $\Pi_2$ contains $l_2$ and the point with coordinates $(2, -1, 7)$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the acute angle between $\Pi_1$ and $\Pi_2$. [4]
\end{enumerate}

The point $P$ on $l_1$ and the point $Q$ on $l_2$ are such that $PQ$ is perpendicular to both $l_1$ and $l_2$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find a vector equation for $PQ$. [7]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 1 2024 Q7 [15]}}
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