Edexcel P4 2024 June — Question 2 6 marks

Exam BoardEdexcel
ModuleP4 (Pure Mathematics 4)
Year2024
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors 3D & Lines
TypePerpendicularity conditions
DifficultyModerate -0.8 Part (a) is a straightforward vector addition requiring only OB = OA + AB. Part (b) requires finding BC, setting up a dot product equation BC·OC = 0, and solving a quadratic, but these are all standard techniques with no novel insight needed. This is easier than average for A-level, being mostly routine manipulation.
Spec1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement4.04c Scalar product: calculate and use for angles
With respect to a fixed origin, \(O\), the point \(A\) has position vector $$\overrightarrow{OA} = \begin{pmatrix} 7 \\ 2 \\ -5 \end{pmatrix}$$ Given that $$\overrightarrow{AB} = \begin{pmatrix} -2 \\ 4 \\ 3 \end{pmatrix}$$
  1. find the coordinates of the point \(B\). [2]
The point \(C\) has position vector $$\overrightarrow{OC} = \begin{pmatrix} a \\ 5 \\ -1 \end{pmatrix}$$ where \(a\) is a constant. Given that \(\overrightarrow{OC}\) is perpendicular to \(\overrightarrow{BC}\)
  1. find the possible values of \(a\). [4]
Question 2:
AnswerMarks
2 7   − 2   a 
O A = 2 A B = 4 O C = 5
− 5 3 − 1
AnswerMarks
(a) 7 −2
   
OB=OA+ AB=  2 +  4 =...
   
−5 3
AnswerMarks
   M1
 5 
( )
= 6 , 5 i + 6 j − 2 k or 5 , 6 , − 2
AnswerMarks
− 2A1
(2)
AnswerMarks
(b) a   5 
B C = O C − ''O B '' = 5 − '' 6 '' = ...
AnswerMarks
− 1 − 2M1
 a   a − 5 
( )
O C . '' B C '' = 0  5 . '' − 1 '' = 0  a a − 5 − 5 − 1 = 0
AnswerMarks
− 1 1dM1
( ) ( )
AnswerMarks
a 2 − 5 a − 6 = 0  a − 6 a + 1 = 0  a = . . .ddM1
a = − 1 , 6A1
(4)
(6 marks)
AnswerMarks Guidance
QuestionScheme Marks 
Question 2:
2 |  7   − 2   a 
O A = 2 A B = 4 O C = 5
− 5 3 − 1
(a) |  7 −2
   
OB=OA+ AB=  2 +  4 =...
   
−5 3
    | M1
 5 
( )
= 6 , 5 i + 6 j − 2 k or 5 , 6 , − 2
− 2 | A1
(2)
(b) |  a   5 
B C = O C − ''O B '' = 5 − '' 6 '' = ...
− 1 − 2 | M1
 a   a − 5 
( )
O C . '' B C '' = 0  5 . '' − 1 '' = 0  a a − 5 − 5 − 1 = 0
− 1 1 | dM1
( ) ( )
a 2 − 5 a − 6 = 0  a − 6 a + 1 = 0  a = . . . | ddM1
a = − 1 , 6 | A1
(4)
(6 marks)
Question | Scheme | Marks
With respect to a fixed origin, $O$, the point $A$ has position vector

$$\overrightarrow{OA} = \begin{pmatrix} 7 \\ 2 \\ -5 \end{pmatrix}$$

Given that

$$\overrightarrow{AB} = \begin{pmatrix} -2 \\ 4 \\ 3 \end{pmatrix}$$

\begin{enumerate}[label=(\alph*)]
\item find the coordinates of the point $B$.
[2]
\end{enumerate}

The point $C$ has position vector

$$\overrightarrow{OC} = \begin{pmatrix} a \\ 5 \\ -1 \end{pmatrix}$$

where $a$ is a constant.

Given that $\overrightarrow{OC}$ is perpendicular to $\overrightarrow{BC}$

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find the possible values of $a$.
[4]
\end{enumerate}

\hfill \mbox{\textit{Edexcel P4 2024 Q2 [6]}}
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