CAIE FP1 2019 November — Question 6 9 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2019
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypePerpendicular distance point to line
DifficultyChallenging +1.2 This is a standard Further Maths vectors question requiring the skew lines distance formula and plane equation from normal vector. Part (i) involves computing direction vectors, their cross product, and applying the formula—methodical but routine for FM students. Part (ii) requires identifying that the common perpendicular direction gives a normal to the plane, then forming the cartesian equation. While multi-step with 9 marks total, these are textbook techniques without novel insight, making it moderately above average difficulty.
Spec4.04b Plane equations: cartesian and vector forms4.04h Shortest distances: between parallel lines and between skew lines
With \(O\) as the origin, the points \(A\), \(B\), \(C\) have position vectors $$\mathbf{i} - \mathbf{j}, \quad 2\mathbf{i} + \mathbf{j} + 7\mathbf{k}, \quad \mathbf{i} - \mathbf{j} + \mathbf{k}$$ respectively.
  1. Find the shortest distance between the lines \(OC\) and \(AB\). [5]
  2. Find the cartesian equation of the plane containing the line \(OC\) and the common perpendicular of the lines \(OC\) and \(AB\). [4]
Question 6:
AnswerMarks
6(i)1
(cid:74)(cid:74)(cid:74)(cid:71)
 
AB= 2
 
 
7
AnswerMarks
 B1
i j k −9  3 
(cid:74)(cid:74)(cid:74)(cid:71) (cid:74)(cid:74)(cid:74)(cid:71)
   
OC×AB= 1 −1 1 = −6 =t 2
   
   
1 2 7 3 −1
AnswerMarks Guidance
   M1 A1 Finds direction of common perpendicular.
 1   3 
   
−1 ⋅ 2
   
   
 0   −1  = 1 =0.267
AnswerMarks Guidance
32 +22 +12 14M1 A1 Uses formula for shortest distance.
5
AnswerMarks
6(ii)i j k −1
 
n= 1 −1 1 =t 4
 
 
3 2 −1 5
AnswerMarks Guidance
 M1 A1 Finds normal to plane.
( )+4 ( )+5 ( )=0
AnswerMarks Guidance
− 0 0 0M1 Uses point on plane.
−x+4y+5z =0A1 AEF
4
AnswerMarks Guidance
QuestionAnswer Marks 
Question 6:
--- 6(i) ---
6(i) | 1
(cid:74)(cid:74)(cid:74)(cid:71)
 
AB= 2
 
 
7
  | B1
i j k −9  3 
(cid:74)(cid:74)(cid:74)(cid:71) (cid:74)(cid:74)(cid:74)(cid:71)
   
OC×AB= 1 −1 1 = −6 =t 2
   
   
1 2 7 3 −1
    | M1 A1 | Finds direction of common perpendicular.
 1   3 
   
−1 ⋅ 2
   
   
 0   −1  = 1 =0.267
32 +22 +12 14 | M1 A1 | Uses formula for shortest distance.
5
--- 6(ii) ---
6(ii) | i j k −1
 
n= 1 −1 1 =t 4
 
 
3 2 −1 5
  | M1 A1 | Finds normal to plane.
( )+4 ( )+5 ( )=0
− 0 0 0 | M1 | Uses point on plane.
−x+4y+5z =0 | A1 | AEF
4
Question | Answer | Marks | Guidance
With $O$ as the origin, the points $A$, $B$, $C$ have position vectors
$$\mathbf{i} - \mathbf{j}, \quad 2\mathbf{i} + \mathbf{j} + 7\mathbf{k}, \quad \mathbf{i} - \mathbf{j} + \mathbf{k}$$
respectively.

\begin{enumerate}[label=(\roman*)]
\item Find the shortest distance between the lines $OC$ and $AB$. [5]

\item Find the cartesian equation of the plane containing the line $OC$ and the common perpendicular of the lines $OC$ and $AB$. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE FP1 2019 Q6 [9]}}
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