OCR Further Pure Core 2 2020 November — Question 4 9 marks

Exam BoardOCR
ModuleFurther Pure Core 2 (Further Pure Core 2)
Year2020
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypeLine intersection with plane
DifficultyStandard +0.3 This is a straightforward Further Maths vectors question requiring standard techniques: substituting a line equation into a plane equation to find intersection points, then calculating distance. Part (a) is verification (routine substitution), and part (b) requires finding Q similarly then using the distance formula. While it's Further Maths content, the methods are mechanical and well-practiced, making it slightly easier than average overall.
Spec4.04a Line equations: 2D and 3D, cartesian and vector forms4.04b Plane equations: cartesian and vector forms4.04f Line-plane intersection: find point
4 The equations of two intersecting lines \(l _ { 1 }\) and \(l _ { 2 }\) are \(l _ { 1 } : \mathbf { r } = \left( \begin{array} { l } 1 \\ 0 \\ a \end{array} \right) + \lambda \left( \begin{array} { r } 2 \\ 1 \\ - 3 \end{array} \right) \quad l _ { 2 } : \mathbf { r } = \left( \begin{array} { r } 7 \\ 9 \\ - 2 \end{array} \right) + \mu \left( \begin{array} { r } - 1 \\ 1 \\ 2 \end{array} \right)\) where \(a\) is a constant.
The equation of the plane \(\Pi\) is
r. \(\left( \begin{array} { l } 1 \\ 5 \\ 3 \end{array} \right) = - 14\). \(l _ { 1 }\) and \(\Pi\) intersect at \(Q\). \(l _ { 2 }\) and \(\Pi\) intersect at \(R\).
  1. Verify that the coordinates of \(R\) are (13, 3, -14).
  2. Determine the exact value of the length of \(Q R\).
Question 4:
AnswerMarks Guidance
4(a)  13  1
   
3 . 5 =13+15−42=−14 (so R is on Π)
   
   
−14 3
eg 7 – µ = 13 => µ = –6 =>
 7  −1  13 
     
r=  9  −6  1  =  3  (so R is also on l 2 )
     
AnswerMarks
−2  2  −14B1
B11.1
1.1AG. Intermediate working must
be seen
AG. Or 9 + µ = 3 or
– 2 + 2µ = –14 but must be
checked in other two equations.
AnswerMarks Guidance
Alternate methodM1 AG. Substituting in expression
of the point into the equation of
the plane to find a value for µ
1  7  −1
     
 5  . 9  +µ  1  =46+10µ=−14⇒µ=−6
  3      −2    2    
AnswerMarks Guidance
µ = –6 =>A1 AG.
acceptable.
 7  −1  13 
     
r= 9 −6 1 = 3 so R is (13, 3, –14)
     
     
−2  2  −14
[2]
M1
AG. Substituting in expression
of the point into the equation of
the plane to find a value for µ
A1
Answer in vector form is
acceptable.
(31−13)2 +(15−3)2 +(−40−−14)2
AnswerMarks
= 182 +122 +262 = 1144dep*M
1
A1
AnswerMarks
[7]2.1
3.2aMethod fully shown or at least 2
of 3 squared terms correct
AnswerMarks
2√286Depends on correct method
shown to find Q
(3+3a)2 +(4.5+1.5a)2 +
(−8.5−3.5a)2
Question 4:
4 | (a) |  13  1
   
3 . 5 =13+15−42=−14 (so R is on Π)
   
   
−14 3
eg 7 – µ = 13 => µ = –6 =>
 7  −1  13 
     
r=  9  −6  1  =  3  (so R is also on l 2 )
     
−2  2  −14 | B1
B1 | 1.1
1.1 | AG. Intermediate working must
be seen
AG. Or 9 + µ = 3 or
– 2 + 2µ = –14 but must be
checked in other two equations.
Alternate method | M1 | AG. Substituting in expression
of the point into the equation of
the plane to find a value for µ
1  7  −1
     
 5  . 9  +µ  1  =46+10µ=−14⇒µ=−6
  3      −2    2    
µ = –6 => | A1 | AG. | Answer in vector form is
acceptable.
 7  −1  13 
     
r= 9 −6 1 = 3 so R is (13, 3, –14)
     
     
−2  2  −14
[2]
M1
AG. Substituting in expression
of the point into the equation of
the plane to find a value for µ
A1
Answer in vector form is
acceptable.
(31−13)2 +(15−3)2 +(−40−−14)2
= 182 +122 +262 = 1144 | dep*M
1
A1
[7] | 2.1
3.2a | Method fully shown or at least 2
of 3 squared terms correct
2√286 | Depends on correct method
shown to find Q
(3+3a)2 +(4.5+1.5a)2 +
(−8.5−3.5a)2
4 The equations of two intersecting lines $l _ { 1 }$ and $l _ { 2 }$ are\\
$l _ { 1 } : \mathbf { r } = \left( \begin{array} { l } 1 \\ 0 \\ a \end{array} \right) + \lambda \left( \begin{array} { r } 2 \\ 1 \\ - 3 \end{array} \right) \quad l _ { 2 } : \mathbf { r } = \left( \begin{array} { r } 7 \\ 9 \\ - 2 \end{array} \right) + \mu \left( \begin{array} { r } - 1 \\ 1 \\ 2 \end{array} \right)$\\
where $a$ is a constant.\\
The equation of the plane $\Pi$ is\\
r. $\left( \begin{array} { l } 1 \\ 5 \\ 3 \end{array} \right) = - 14$.\\
$l _ { 1 }$ and $\Pi$ intersect at $Q$.\\
$l _ { 2 }$ and $\Pi$ intersect at $R$.
\begin{enumerate}[label=(\alph*)]
\item Verify that the coordinates of $R$ are (13, 3, -14).
\item Determine the exact value of the length of $Q R$.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 2 2020 Q4 [9]}}
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