Edexcel CP1 2021 June — Question 7 8 marks

Exam BoardEdexcel
ModuleCP1 (Core Pure 1)
Year2021
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Lines & Planes
TypePerpendicular distance point to plane
DifficultyStandard +0.8 This is a multi-part Further Maths question requiring cross products to find a normal vector, conversion between vector and Cartesian plane equations, and using perpendicular distance formula with parameter solving. Part (c) requires solving a quadratic equation from the distance formula, making it more demanding than standard A-level questions but still following established techniques without requiring novel insight.
Spec4.04b Plane equations: cartesian and vector forms4.04j Shortest distance: between a point and a plane
The plane \(\Pi\) has equation $$\mathbf{r} = \begin{pmatrix} 3 \\ 3 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}$$ where \(\lambda\) and \(\mu\) are scalar parameters.
  1. Show that vector \(\mathbf{2i + 3j - 4k}\) is perpendicular to \(\Pi\). [2]
  2. Hence find a Cartesian equation of \(\Pi\). [2]
The line \(l\) has equation $$\mathbf{r} = \begin{pmatrix} 4 \\ -5 \\ 2 \end{pmatrix} + t \begin{pmatrix} 1 \\ 6 \\ -3 \end{pmatrix}$$ where \(t\) is a scalar parameter. The point \(A\) lies on \(l\). Given that the shortest distance between \(A\) and \(\Pi\) is \(2\sqrt{29}\)
  1. determine the possible coordinates of \(A\). [4]
Question 7:
AnswerMarks
7(a) − 1   2   2   2 
2 . 3 = − 2 + 6 − 4 = 0 and 0 . 3 = 4 + 0 − 4 = 0
1 − 4 1 − 4
−1 2  21−10 
     
Alt: 2  0 = −(−11−12) =...
     
     
AnswerMarks Guidance
 1 1  −10−22 M1 1.1b
As 2i + 3j – 4k is perpendicular to both direction vectors (two non-
AnswerMarks Guidance
parallel vectors) of Π then it must be perpendicular to ΠA1 2.2a
(2)
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(b) x   2   3   2 
y . 3 = 3 . 3  . . .
AnswerMarks Guidance
z − 4 2 − 4M1 1.1a
2 x + 3 y − 4 z = 7A1 2.2a
(2)
AnswerMarks
(c)2 ( 4 + t ) + 3 ( − 5 + 6 t ) − 4 ( 2 − 3 t ) − 7
= 2 2 9  t = . . .
AnswerMarks Guidance
2 2 + 3 2 + ( − 4 2 )M1 3.1a
9 5
t = − and t =
AnswerMarks Guidance
8 2A1 1.1b
 4   1   4   1 
9 5
r = − 5 − 6 = ... or r = − 5 + 6 = ...
8 2
AnswerMarks Guidance
2 − 3 2 − 3M1 1.1b
23 47 43  1 3 1 1 
 ,− ,  and , 1 0 , −
AnswerMarks Guidance
 8 4 8  2 2A1 2.2a
(4)
(8 marks)
Notes:
(a)
M1: Attempts the scalar product of each direction vector and the vector 2i + 3j – 4k. Some
numerical calculation is required, just “= 0” is insufficient. Alternatively, attempts the cross product
(allow sign slips) with the two direction vectors.
A1: Shows that both scalar products = 0 (minimum −2+6−4=0and 4−4=0) and makes a
minimal conclusion with no erroneous statements. If using cross product, the calculation must be
correct, and a minimal conclusion given with no erroneous statements.
(b)
x  2 3  2
       
M1: Applies  y  .  3  =  3  .  3  ...
       
 z  −4 2  −4
A1: 2x+3y−4z =7
(c)
M1: A fully correct method for finding a value of t. Other methods are possible, but must be valid
and lead to a value of t. Examples of other methods:
 
2 ( 4 + t ) + 3 − ( 5 + 6 t ) − 4 ( 2 − 3 t ) 7
• 2 2 9 =  − using plane parallel to through origin
2 2 + 3 2 + ( − 4 2 ) 2 9
and shortest distance from plane to origin.
• 2 ( 4 + t ) + 3 ( − 5 + 6 t ) − 4 ( 2 − 3 t ) = 7  t = t (t at intersection of line and plane) and
i
( 2 , 3 , 4 ) T . (1 , 6 , 3 ) T − −
s i n  = (sine of angle between line and plane) followed by
2 9 4 6
2 2 9
s i n k .. . t t k  =  =  = 
i
k 4 6
A1: Correct values for t. Both are required.
M1: Uses a value of t to find a set of coordinates for A.
A1: Both correct sets of coordinates for A.
AnswerMarks Guidance
QuestionScheme Marks 
Question 7:
--- 7(a) ---
7(a) |  − 1   2   2   2 
2 . 3 = − 2 + 6 − 4 = 0 and 0 . 3 = 4 + 0 − 4 = 0
1 − 4 1 − 4
−1 2  21−10 
     
Alt: 2  0 = −(−11−12) =...
     
     
 1 1  −10−22  | M1 | 1.1b
As 2i + 3j – 4k is perpendicular to both direction vectors (two non-
parallel vectors) of Π then it must be perpendicular to Π | A1 | 2.2a
(2)
(b) |  x   2   3   2 
y . 3 = 3 . 3  . . .
z − 4 2 − 4 | M1 | 1.1a
2 x + 3 y − 4 z = 7 | A1 | 2.2a
(2)
(c) | 2 ( 4 + t ) + 3 ( − 5 + 6 t ) − 4 ( 2 − 3 t ) − 7
= 2 2 9  t = . . .
2 2 + 3 2 + ( − 4 2 ) | M1 | 3.1a
9 5
t = − and t =
8 2 | A1 | 1.1b
 4   1   4   1 
9 5
r = − 5 − 6 = ... or r = − 5 + 6 = ...
8 2
2 − 3 2 − 3 | M1 | 1.1b
23 47 43  1 3 1 1 
 ,− ,  and , 1 0 , −
 8 4 8  2 2 | A1 | 2.2a
(4)
(8 marks)
Notes:
(a)
M1: Attempts the scalar product of each direction vector and the vector 2i + 3j – 4k. Some
numerical calculation is required, just “= 0” is insufficient. Alternatively, attempts the cross product
(allow sign slips) with the two direction vectors.
A1: Shows that both scalar products = 0 (minimum −2+6−4=0and 4−4=0) and makes a
minimal conclusion with no erroneous statements. If using cross product, the calculation must be
correct, and a minimal conclusion given with no erroneous statements.
(b)
x  2 3  2
       
M1: Applies  y  .  3  =  3  .  3  ...
       
 z  −4 2  −4
A1: 2x+3y−4z =7
(c)
M1: A fully correct method for finding a value of t. Other methods are possible, but must be valid
and lead to a value of t. Examples of other methods:
 
2 ( 4 + t ) + 3 − ( 5 + 6 t ) − 4 ( 2 − 3 t ) 7
• 2 2 9 =  − using plane parallel to through origin
2 2 + 3 2 + ( − 4 2 ) 2 9
and shortest distance from plane to origin.
• 2 ( 4 + t ) + 3 ( − 5 + 6 t ) − 4 ( 2 − 3 t ) = 7  t = t (t at intersection of line and plane) and
i
( 2 , 3 , 4 ) T . (1 , 6 , 3 ) T − −
s i n  = (sine of angle between line and plane) followed by
2 9 4 6
2 2 9
s i n k .. . t t k  =  =  = 
i
k 4 6
A1: Correct values for t. Both are required.
M1: Uses a value of t to find a set of coordinates for A.
A1: Both correct sets of coordinates for A.
Question | Scheme | Marks | AOs
The plane $\Pi$ has equation

$$\mathbf{r} = \begin{pmatrix} 3 \\ 3 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ 0 \\ 1 \end{pmatrix}$$

where $\lambda$ and $\mu$ are scalar parameters.

\begin{enumerate}[label=(\alph*)]
\item Show that vector $\mathbf{2i + 3j - 4k}$ is perpendicular to $\Pi$.
[2]

\item Hence find a Cartesian equation of $\Pi$.
[2]
\end{enumerate}

The line $l$ has equation

$$\mathbf{r} = \begin{pmatrix} 4 \\ -5 \\ 2 \end{pmatrix} + t \begin{pmatrix} 1 \\ 6 \\ -3 \end{pmatrix}$$

where $t$ is a scalar parameter.

The point $A$ lies on $l$.

Given that the shortest distance between $A$ and $\Pi$ is $2\sqrt{29}$

\begin{enumerate}[label=(\alph*)]
\setcounter{enumii}{2}
\item determine the possible coordinates of $A$.
[4]
\end{enumerate}

\hfill \mbox{\textit{Edexcel CP1 2021 Q7 [8]}}
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