Question 8 - A-Level Maths

OCR Further Pure Core 1 2023 June — Question 8 15 marks

Exam Board	OCR
Module	Further Pure Core 1 (Further Pure Core 1)
Year	2023
Session	June
Marks	15
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Vectors: Cross Product & Distances
Type	Volume of tetrahedron using scalar triple product
Difficulty	Challenging +1.2 This is a multi-part Further Maths question requiring standard vector techniques (cross product for angle/plane, distance formula, rotation matrix) but each part follows routine procedures. Part (a) uses standard dot/cross product formulas, (b) finds plane equation from normal vector, (c) applies distance formula and volume formula as given, and (d) uses booklet formula for rotation matrix. While lengthy (4 parts) and requiring careful calculation, it demands no novel insight—just systematic application of FM1 techniques.
Spec	4.04c Scalar product: calculate and use for angles 4.04g Vector product: a x b perpendicular vector

8 The points \(P , Q\) and \(R\) have coordinates \(( 0,2,3 ) , ( 2,0,1 )\) and \(( 1,3,0 )\) respectively.
The acute angle between the line segments \(P Q\) and \(P R\) is \(\theta\).

Show that \(\sin \theta = \frac { 2 } { 11 } \sqrt { 22 }\). The triangle \(P Q R\) lies in the plane \(\Pi\).
Determine an equation for \(\Pi\), giving your answer in the form \(\mathrm { ax } + \mathrm { by } + \mathrm { cz } = \mathrm { d }\), where \(a , b , c\) and \(d\) are integers. The point \(S\) has coordinates \(( 5,3 , - 1 )\).
By finding the shortest distance between \(S\) and the plane \(\Pi\), show that the volume of the tetrahedron \(P Q R S\) is \(\frac { 14 } { 3 }\).
[0pt] [The volume of a tetrahedron is \(\frac { 1 } { 3 } \times\) area of base × perpendicular height] The tetrahedron \(P Q R S\) is transformed to the tetrahedron \(\mathrm { P } ^ { \prime } \mathrm { Q } ^ { \prime } \mathrm { R } ^ { \prime } \mathrm { S } ^ { \prime }\) by a rotation about the \(y\)-axis.
The \(x\)-coordinate of \(S ^ { \prime }\) is \(2 \sqrt { 2 }\).
By using the matrix for a rotation by angle \(\theta\) about the \(y\)-axis, as given in the Formulae Booklet, determine in exact form the possible coordinates of \(R ^ { \prime }\).

Show mark scheme Show mark scheme source

Question 8:

Answer	Marks	Guidance
8	(a)	 2   1 

P Q = − 2 , P R = 1

− 2 − 3

2 1    

2 . 1 −

2 3 − −

c o s  =

( 2 ) 2 ( 2 2 ) ( 2 ) 2 ( 1 ) 2 ( 1 ) 2 ( 3 2 ) + − + −  + + −

6 =

1 2 1 1

2  

6 8 2 2 2  

s i n 1 2 2 2   = − = = =

Answer	Marks
1 2 1 1 1 1 1 2 1 1 1	B1

M1

Answer	Marks
A1	Both correct or other way round

Correct use of scalar product including correct method for magnitudes

and dot product

AG All working must be using exact forms

Answer	Marks	Guidance
Alternative for M1 A1	M1
A1	Correct use of vector product to find	sinθ

2 1

for magnitudes, and correct method for (−2)×( 1 ) soi. Condone

−2 −3

Answer	Marks
sinθ instead of	sinθ

AG.

 2   1 

   

−2  1

   

−2 −3

 

sin=

( 2 )2 +(−2 )2 +(−2 )2  ( 1 )2 +( 1 )2 +(−3 )2

2

 

4 1

 

1 4 6 4 2

= = = = 22

12 11 132 22 11

[3]

M1

A1

Answer	Marks	Guidance
Correct use of vector product to find	sinθ	including correct method

2 1

for magnitudes, and correct method for (−2)×( 1 ) soi. Condone

−2 −3

Answer	Marks
sinθ instead of	sinθ

AG.

Answer	Marks	Guidance
8	(a)	Alternative method:

M1

A1

B1

M1

Answer	Marks
A1	Sides of triangle

Cosine rule

Sides of triangle

Recognises an isosceles triangle so uses a median line

Q R 1 1 , R P 1 1 , P Q 1 2 = = =

1 2 1 1 1 1 3 + −

C o s i n e r u l e c o s =   =

2 1 1 1 2 1 1  

3 8 2

s i n 1 2 2   = − = =

1 1 1 1 1 1

Or:

Answer	Marks	Guidance
QR	= 11,	RP

Drop perpendicular from R to QP at M

2 1 

RM = 11−  12 = 8

2 

8 2

sin= = 22

11 11

[3]

B1

M1

A1

B1

M1

A1

Sides of triangle

Cosine rule

Sides of triangle

Recognises an isosceles triangle so uses a median line

Answer	Marks	Guidance
8	(b)	 2   1   2 

P Q  P R = − 2  1 = 4 1

− 2 − 3 1

2x+ y+z =d

Answer	Marks
Sub for a pointd =5 2x+ y+z =5	B1

M1

Answer	Marks
A1	BC or any other relevant vector product. Can be awarded if seen in (a)

Attempt to use r.n = d to form linear equation.

ft their vector product.

Substitutes coordinates of P, Q or R to find d. (multiples accepted)

[3]

Answer	Marks
(c)	 5   2 

3 . 1 − 5



− 1 1 7

D = =

 2  6

1 1 1  

V P Q P R s i n D  =  

3 2

1 1 2 7

2 3 1 1 2 2 =     

3 2 1 1 6

1 4

=

Answer	Marks
3	M1

A1

M1

Answer	Marks
A1	Uses the formula given in formula book, or any other complete method for the

shortest distance, ft their 𝑃⃗⃗⃗⃗𝑄⃗ ×𝑃⃗⃗⃗⃗𝑅⃗ .

Correct shortest distance.

2 1

Answer	Marks	Guidance
Uses 1	𝑃⃗⃗⃗⃗𝑄⃗

2 2

−2 −3

2 1

1

Answer	Marks	Guidance
multiplied by their D/3, but must indicate that	(−2)×( 1 )	is the area of

2 −2 −3

the base.

AG.

[4]

Answer	Marks	Guidance
8	(d)	c o s 0 s i n 5 2 2        

0 1 0 3 . . . =

s i n 0 c o s 1 . . .   − −

5 c o s s i n 2 2    − =

( ) 2

s i n 2 5 c o s 2 2    = −

1 c o s 2 2 5 c o s 2 2 0 2 c o s 8     − = − +

2 6 c o s 2 2 0 2 c o s 7 0    − + =

2 7 2

c o s ,   =

2 2 6

2 1 7 2

s i n 5 c o s 2 2 ,   = − = −

2 2 6

 cos 0 sin1  cos

    

0 1 0 3 = 3

    

 −sin 0 cos 0  −sin 

 2   7 2 

   

 2   26 

 3  ,  3 

   

 2  17 2 

−   

 2   26 

 2 2  7 2 17 2 

i.e. R'= ,3,−  or ,3, 

   

Answer	Marks
 2 2   26 26 	M1

M1

A1

M1

Answer	Marks
A1	𝑂⃗⃗⃗⃗𝑅⃗ 𝑂⃗⃗⃗⃗⃗𝑆

For rotation matrix multiplied by or .

For a correct step to form quadratic equation in sin𝜙 or cos𝜙 only.

For reference: 26sin2+4 2sin−17=0

Solves quadratic equation in sin𝜙 or cos𝜙. (Exact answers required)

Uses their sin𝜙 or cos𝜙 with 5cos𝜙−sin𝜙 = 2√2to find cos𝜙 or sin𝜙

respectively, (even if only one root)

or fromsin𝜙 = √1−cos2𝜙 or cos𝜙 = √1−sin2𝜙 (condone inclusion of ±),

or repeats previous method

and multiplies out the matrices.

For both, and no others. Accept given as 𝑂⃗⃗⃗⃗𝑅⃗⃗⃗ ′.

If M0M0A0M0A0, SCB1 for any R′with y-coordinate = 3, and no other y-

coordinates.

[5]

Answer	Marks	Guidance
8	(d)	Alternative method for first 3 marks:

M1

Answer	Marks
A1	𝑂⃗⃗⃗⃗𝑅⃗ 𝑂⃗⃗⃗⃗⃗𝑆

For rotation matrix multiplied by or .

Expressing in the form 𝑅cos(𝜙+𝛼) or 𝑅sin(𝜙+𝛼) oe. Note

that 5cos𝜙−sin𝜙

1 = √26sin(𝜙+arctan(− )+𝜋)

5 √2 17√2

Solves for sin𝜙 or cos𝜙. For reference, ≈ 0.707,− ≈

2 26

7√2

−0.925 and ≈ 0.381.

26 5 c o s s i n 2 2   − =

5 c o s s i n 2 6 c o s ( ) 2 2      − = + =

1 1 5

w h e r e t a n s i n , c o s    =  =

5 2 6 2 6

( 2 )

2 6 s i n ( ) 2 6 2 2 3 2   + =  − = 

c o s c o s ( ( ) ) c o s ( ) c o s s i n ( ) s i n           = + − = + + +

2 2 5 3 2 1

=   

2 6 2 6 2 6 2 6

2 7 2

c o s ,   =

Answer	Marks	Guidance
2 2 6	M1	𝑂⃗⃗⃗⃗𝑅⃗ 𝑂⃗⃗⃗⃗⃗𝑆

For rotation matrix multiplied by or .

Answer	Marks
M1	Expressing in the form 𝑅cos(𝜙+𝛼) or 𝑅sin(𝜙+𝛼) oe. Note

that 5cos𝜙−sin𝜙

1 = √26sin(𝜙+arctan(− )+𝜋)

5

Answer	Marks
A1	√2 17√2

Solves for sin𝜙 or cos𝜙. For reference, ≈ 0.707,− ≈

2 26

7√2

−0.925 and ≈ 0.381.

26 [3]

Question 8:
8 | (a) |  2   1 
P Q = − 2 , P R = 1
− 2 − 3
2 1    
2 . 1 −
2 3 − −
c o s  =
( 2 ) 2 ( 2 2 ) ( 2 ) 2 ( 1 ) 2 ( 1 ) 2 ( 3 2 ) + − + −  + + −
6
=
1 2 1 1
2  
6 8 2 2 2  
s i n 1 2 2 2   = − = = =
1 2 1 1 1 1 1 2 1 1 1 | B1
M1
A1 | Both correct or other way round
Correct use of scalar product including correct method for magnitudes
and dot product
AG All working must be using exact forms
Alternative for M1 A1 | M1
A1 | Correct use of vector product to find |sinθ| including correct method
2 1
for magnitudes, and correct method for (−2)×( 1 ) soi. Condone
−2 −3
sinθ instead of |sinθ|
AG.
 2   1 
   
−2  1
   
   
−2 −3
 
sin=
( 2 )2 +(−2 )2 +(−2 )2  ( 1 )2 +( 1 )2 +(−3 )2
2
 
4 1
 
 
1 4 6 4 2
= = = = 22
12 11 132 22 11
[3]
M1
A1
Correct use of vector product to find |sinθ| including correct method
2 1
for magnitudes, and correct method for (−2)×( 1 ) soi. Condone
−2 −3
sinθ instead of |sinθ|
AG.
8 | (a) | Alternative method: | B1
M1
A1
B1
M1
A1 | Sides of triangle
Cosine rule
Sides of triangle
Recognises an isosceles triangle so uses a median line
Q R 1 1 , R P 1 1 , P Q 1 2 = = =
1 2 1 1 1 1 3 + −
C o s i n e r u l e c o s =   =
2 1 1 1 2 1 1  
3 8 2
s i n 1 2 2   = − = =
1 1 1 1 1 1
Or:
QR | = 11, | RP | = 11, | PQ | = 12
Drop perpendicular from R to QP at M
2
1 
RM = 11−  12 = 8
2 
8 2
sin= = 22
11 11
[3]
B1
M1
A1
B1
M1
A1
Sides of triangle
Cosine rule
Sides of triangle
Recognises an isosceles triangle so uses a median line
8 | (b) |  2   1   2 
P Q  P R = − 2  1 = 4 1
− 2 − 3 1
2x+ y+z =d
Sub for a pointd =5 2x+ y+z =5 | B1
M1
A1 | BC or any other relevant vector product. Can be awarded if seen in (a)
Attempt to use r.n = d to form linear equation.
ft their vector product.
Substitutes coordinates of P, Q or R to find d. (multiples accepted)
[3]
(c) |  5   2 
3 . 1 − 5

− 1 1 7
D = =
 2  6
1
1
1 1  
V P Q P R s i n D  =  
3 2
1 1 2 7
2 3 1 1 2 2 =     
3 2 1 1 6
1 4
=
3 | M1
A1
M1
A1 | Uses the formula given in formula book, or any other complete method for the
shortest distance, ft their 𝑃⃗⃗⃗⃗𝑄⃗ ×𝑃⃗⃗⃗⃗𝑅⃗ .
Correct shortest distance.
2 1
Uses 1 |𝑃⃗⃗⃗⃗𝑄⃗ ||𝑃⃗⃗⃗⃗𝑅⃗ |sin𝜃oe, multiplied by their D/3, or 1 |(−2)×( 1 )|
2 2
−2 −3
2 1
1
multiplied by their D/3, but must indicate that |(−2)×( 1 )| is the area of
2
−2 −3
the base.
AG.
[4]
8 | (d) | c o s 0 s i n 5 2 2        
0 1 0 3 . . . =
s i n 0 c o s 1 . . .   − −
5 c o s s i n 2 2    − =
( ) 2
s i n 2 5 c o s 2 2    = −
1 c o s 2 2 5 c o s 2 2 0 2 c o s 8     − = − +
2 6 c o s 2 2 0 2 c o s 7 0    − + =
2 7 2
c o s ,   =
2 2 6
2 1 7 2
s i n 5 c o s 2 2 ,   = − = −
2 2 6
 cos 0 sin1  cos
    
0 1 0 3 = 3
    
    
 −sin 0 cos 0  −sin 
 2   7 2 
   
 2   26 
 3  ,  3 
   
 2  17 2 
−   
 2   26 
 2 2  7 2 17 2 
i.e. R'= ,3,−  or ,3, 
   
 2 2   26 26  | M1
M1
A1
M1
A1 | 𝑂⃗⃗⃗⃗𝑅⃗ 𝑂⃗⃗⃗⃗⃗𝑆
For rotation matrix multiplied by or .
For a correct step to form quadratic equation in sin𝜙 or cos𝜙 only.
For reference: 26sin2+4 2sin−17=0
Solves quadratic equation in sin𝜙 or cos𝜙. (Exact answers required)
Uses their sin𝜙 or cos𝜙 with 5cos𝜙−sin𝜙 = 2√2to find cos𝜙 or sin𝜙
respectively, (even if only one root)
or fromsin𝜙 = √1−cos2𝜙 or cos𝜙 = √1−sin2𝜙 (condone inclusion of ±),
or repeats previous method
and multiplies out the matrices.
For both, and no others. Accept given as 𝑂⃗⃗⃗⃗𝑅⃗⃗⃗ ′.
If M0M0A0M0A0, SCB1 for any R′with y-coordinate = 3, and no other y-
coordinates.
[5]
8 | (d) | Alternative method for first 3 marks: | M1
M1
A1 | 𝑂⃗⃗⃗⃗𝑅⃗ 𝑂⃗⃗⃗⃗⃗𝑆
For rotation matrix multiplied by or .
Expressing in the form 𝑅cos(𝜙+𝛼) or 𝑅sin(𝜙+𝛼) oe. Note
that 5cos𝜙−sin𝜙
1
= √26sin(𝜙+arctan(− )+𝜋)
5
√2 17√2
Solves for sin𝜙 or cos𝜙. For reference, ≈ 0.707,− ≈
2 26
7√2
−0.925 and ≈ 0.381.
26
5 c o s s i n 2 2   − =
5 c o s s i n 2 6 c o s ( ) 2 2      − = + =
1 1 5
w h e r e t a n s i n , c o s    =  =
5 2 6 2 6
( 2 )
2 6 s i n ( ) 2 6 2 2 3 2   + =  − = 
c o s c o s ( ( ) ) c o s ( ) c o s s i n ( ) s i n           = + − = + + +
2 2 5 3 2 1
=   
2 6 2 6 2 6 2 6
2 7 2
c o s ,   =
2 2 6 | M1 | 𝑂⃗⃗⃗⃗𝑅⃗ 𝑂⃗⃗⃗⃗⃗𝑆
For rotation matrix multiplied by or .
M1 | Expressing in the form 𝑅cos(𝜙+𝛼) or 𝑅sin(𝜙+𝛼) oe. Note
that 5cos𝜙−sin𝜙
1
= √26sin(𝜙+arctan(− )+𝜋)
5
A1 | √2 17√2
Solves for sin𝜙 or cos𝜙. For reference, ≈ 0.707,− ≈
2 26
7√2
−0.925 and ≈ 0.381.
26
[3]

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8 The points $P , Q$ and $R$ have coordinates $( 0,2,3 ) , ( 2,0,1 )$ and $( 1,3,0 )$ respectively.\\
The acute angle between the line segments $P Q$ and $P R$ is $\theta$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\sin \theta = \frac { 2 } { 11 } \sqrt { 22 }$.

The triangle $P Q R$ lies in the plane $\Pi$.
\item Determine an equation for $\Pi$, giving your answer in the form $\mathrm { ax } + \mathrm { by } + \mathrm { cz } = \mathrm { d }$, where $a , b , c$ and $d$ are integers.

The point $S$ has coordinates $( 5,3 , - 1 )$.
\item By finding the shortest distance between $S$ and the plane $\Pi$, show that the volume of the tetrahedron $P Q R S$ is $\frac { 14 } { 3 }$.\\[0pt]
[The volume of a tetrahedron is $\frac { 1 } { 3 } \times$ area of base × perpendicular height]

The tetrahedron $P Q R S$ is transformed to the tetrahedron $\mathrm { P } ^ { \prime } \mathrm { Q } ^ { \prime } \mathrm { R } ^ { \prime } \mathrm { S } ^ { \prime }$ by a rotation about the $y$-axis.\\
The $x$-coordinate of $S ^ { \prime }$ is $2 \sqrt { 2 }$.
\item By using the matrix for a rotation by angle $\theta$ about the $y$-axis, as given in the Formulae Booklet, determine in exact form the possible coordinates of $R ^ { \prime }$.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 1 2023 Q8 [15]}}

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Q1 3 Q2 6 Q3 5 Q4 11 Q5 6 Q6 4 Q7 11 Q8 15 Q9 14