Question 6 - A-Level Maths

Edexcel P4 2024 June — Question 6 10 marks

Exam Board	Edexcel
Module	P4 (Pure Mathematics 4)
Year	2024
Session	June
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Vectors: Cross Product & Distances
Type	Find parameter value for geometric condition
Difficulty	Standard +0.3 This is a structured multi-part vectors question with clear signposting. Part (a) involves substituting into the magnitude formula and algebraic manipulation (routine). Parts (b)(i-ii) require solving a quadratic and substituting back (standard). Part (c) uses the cross product formula for triangle area, which is a direct application of a known result. All steps are procedural with no novel insight required, making this slightly easier than average for Further Maths P4.
Spec	1.05c Area of triangle: using 1/2 ab sin(C)1.10a Vectors in 2D: i,j notation and column vectors 1.10b Vectors in 3D: i,j,k notation 1.10c Magnitude and direction: of vectors 1.10d Vector operations: addition and scalar multiplication 1.10e Position vectors: and displacement 1.10f Distance between points: using position vectors 4.04a Line equations: 2D and 3D, cartesian and vector forms

With respect to a fixed origin $O$, the line $l_1$ is given by the equation $$\mathbf{r} = \mathbf{i} + 2\mathbf{j} + 5\mathbf{k} + \lambda(8\mathbf{i} - \mathbf{j} + 4\mathbf{k})$$ where $\lambda$ is a scalar parameter. The point $A$ lies on $l_1$ Given that $|\overrightarrow{OA}| = 5\sqrt{10}$

show that at $A$ the parameter $\lambda$ satisfies $$81\lambda^2 + 52\lambda - 220 = 0$$ [3]

Hence

1. show that one possible position vector for $A$ is $-15\mathbf{i} + 4\mathbf{j} - 3\mathbf{k}$
2. find the other possible position vector for $A$. [3]

The line $l_2$ is parallel to $l_1$ and passes through $O$. Given that • $\overrightarrow{OA} = -15\mathbf{i} + 4\mathbf{j} - 3\mathbf{k}$ • point $B$ lies on $l_2$ where $|\overrightarrow{OB}| = 4\sqrt{10}$

find the area of triangle $OAB$, giving your answer to one decimal place. [4]

Show mark scheme Show mark scheme source

Question 6:

Answer	Marks
6(a)	2 ( ) 2 ( ) 2 ( ) 2
O A 1 8 2 5 4    = + + − + +	M1

( )2 ( )2 ( )2

Answer	Marks
OA =5 10  1+8 + 2− + 5+4 =250	M1

6 4 2 1 6 1 2 4 4 1 6 2 4 0 2 5 2 5 0        + + + − + + + + =

Answer	Marks
8 1 2 5 2 2 2 0 0 *    + − =	A1*

(3)

Answer	Marks
(b)	( ) ( ) 1 1 0  

8 1 2 5 2 2 2 0 0 8 1 1 1 0 2 0 . .. 2 ,      + − =  − + =  = −

8 1

1 8 " t h e i r "   +  

O A 2 " t h e i r "   = −

Answer	Marks
5 4 " t h e i r "  + 	M1

 1 1 0   9 6 1 

1 + 8 

1  8 −15 8 1 8 1

      1 1 0 5 2

E.g OA= 2 −2 −1 =  4* ; or O A = 2 − =

8 1 8 1

     

5 4 −3

      1 1 0 8 4 5

5 + 4 

Answer	Marks
8 1 8 1	A1*;

A1

(3)

Answer	Marks
(c)	1 5 8  −   

4 . 1 −

3 4 −

c o s . ..  =  =

Answer	Marks
1 5 2 4 2 3 2 8 2 1 2 4 2 + + + +	M1

=  − 1 3 6 ( =  0 .9 5 5 7 ... ) or e.g. = a w r t 0 .2 9 9 ( r a d ) / 1 7 .1  

Answer	Marks
4 5 1 0	A1

1 A r e a O A B 5 1 0 4 1 0 s i n ...  =  =

Answer	Marks
2	M1
= awrt 29.4	A1

(4)

(10 marks)

Alt

Answer	Marks
(c)	 − 1 2 7 

1 1

Midpoint of possible A’s is M = 1 8 8 and OM = 1272 +1882 +3012 =...

8 1 81

Answer	Marks
3 0 1	M1
= 4.6534…	A1

1 Area =  4 1 0  4 . 6 5 3 4 . .. ;= 29.43…

Answer	Marks
2	M1; A1

(4)

Answer	Marks	Guidance
Question	Scheme	Marks

Question 6:
--- 6(a) ---
6(a) | 2 ( ) 2 ( ) 2 ( ) 2
O A 1 8 2 5 4    = + + − + + | M1
( )2 ( )2 ( )2
OA =5 10  1+8 + 2− + 5+4 =250 | M1
6 4 2 1 6 1 2 4 4 1 6 2 4 0 2 5 2 5 0        + + + − + + + + =
8 1 2 5 2 2 2 0 0 *    + − = | A1*
(3)
(b) | ( ) ( ) 1 1 0  
8 1 2 5 2 2 2 0 0 8 1 1 1 0 2 0 . .. 2 ,      + − =  − + =  = −
8 1
1 8 " t h e i r "   +  
O A 2 " t h e i r "   = −
5 4 " t h e i r "  +  | M1
 1 1 0   9 6 1 
1 + 8 
1  8 −15 8 1 8 1
      1 1 0 5 2
E.g OA= 2 −2 −1 =  4* ; or O A = 2 − =
8 1 8 1
     
5 4 −3
      1 1 0 8 4 5
5 + 4 
8 1 8 1 | A1*;
A1
(3)
(c) | 1 5 8  −   
4 . 1 −
3 4 −
c o s . ..  =  =
1 5 2 4 2 3 2 8 2 1 2 4 2 + + + + | M1
=  − 1 3 6 ( =  0 .9 5 5 7 ... ) or e.g. = a w r t 0 .2 9 9 ( r a d ) / 1 7 .1  
4 5 1 0 | A1
1
A r e a O A B 5 1 0 4 1 0 s i n ...  =  =
2 | M1
= awrt 29.4 | A1
(4)
(10 marks)
Alt
(c) |  − 1 2 7 
1 1
Midpoint of possible A’s is M = 1 8 8 and OM = 1272 +1882 +3012 =...
8 1 81
3 0 1 | M1
= 4.6534… | A1
1
Area =  4 1 0  4 . 6 5 3 4 . .. ;= 29.43…
2 | M1; A1
(4)
Question | Scheme | Marks

Show LaTeX source

With respect to a fixed origin $O$, the line $l_1$ is given by the equation

$$\mathbf{r} = \mathbf{i} + 2\mathbf{j} + 5\mathbf{k} + \lambda(8\mathbf{i} - \mathbf{j} + 4\mathbf{k})$$

where $\lambda$ is a scalar parameter.

The point $A$ lies on $l_1$

Given that $|\overrightarrow{OA}| = 5\sqrt{10}$

\begin{enumerate}[label=(\alph*)]
\item show that at $A$ the parameter $\lambda$ satisfies

$$81\lambda^2 + 52\lambda - 220 = 0$$
[3]
\end{enumerate}

Hence

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item \begin{enumerate}[label=(\roman*)]
\item show that one possible position vector for $A$ is $-15\mathbf{i} + 4\mathbf{j} - 3\mathbf{k}$

\item find the other possible position vector for $A$.
[3]
\end{enumerate}
\end{enumerate}

The line $l_2$ is parallel to $l_1$ and passes through $O$.

Given that
• $\overrightarrow{OA} = -15\mathbf{i} + 4\mathbf{j} - 3\mathbf{k}$
• point $B$ lies on $l_2$ where $|\overrightarrow{OB}| = 4\sqrt{10}$

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item find the area of triangle $OAB$, giving your answer to one decimal place.
[4]
\end{enumerate}

\hfill \mbox{\textit{Edexcel P4 2024 Q6 [10]}}

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