OCR Further Pure Core 1 2021 November — Question 4 11 marks

Exam BoardOCR
ModuleFurther Pure Core 1 (Further Pure Core 1)
Year2021
SessionNovember
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors: Cross Product & Distances
TypeFind foot of perpendicular from point to line
DifficultyStandard +0.3 This is a standard Further Maths vectors/3D geometry question requiring routine techniques: finding a line equation from two points, using perpendicularity to find the closest point, and calculating triangle area using the cross product. While it involves multiple steps and Further Maths content, each part follows well-established procedures without requiring novel insight or particularly challenging problem-solving.
Spec4.04a Line equations: 2D and 3D, cartesian and vector forms4.04g Vector product: a x b perpendicular vector
Points \(A\), \(B\) and \(C\) have coordinates \((4, 2, 0)\), \((1, 5, 3)\) and \((1, 4, -2)\) respectively. The line \(l\) passes through \(A\) and \(B\).
  1. Find a cartesian equation for \(l\). [3]
\(M\) is the point on \(l\) that is closest to \(C\).
  1. Find the coordinates of \(M\). [4]
  2. Find the exact area of the triangle \(ABC\). [4]
Question 4:
AnswerMarks Guidance
4(a) −3
uuur
 
AB= 3 oe
 
 
 3 
x 4 −3
     
Equation of AB is y = 2 +λ 3 oe
     
     
z 0  3 
AnswerMarks
⇒4−x= y−2= zB1
M1
AnswerMarks
A11.1
1.1
AnswerMarks
1.1soi
−3 x
   
their 3 soi r = or y =is not required for M1
   
   
 3  z
Allow equivalent equations
e.g. ⇒1−x= y−5= z−3 from using B
[3]
AnswerMarks Guidance
4(b) −3 4 −3 4−3λ
uuur uuur
       
AB= 3 , OM is 2 +λ 3 = 2+3λ
       
       
 3  0  3   3λ 
4−3λ  1  3−3λ
uuur
     
CM = 2+3λ − 4 = 3λ−2
     
     
3λ −2 3λ+2
  
3−3λ −3
uuur
   
Perpendicular to AB⇒ 3λ−2 . 3 =0
   
   
3λ+2  3 
⇒−9+9λ+9λ−6+9λ+6=0
1
⇒27λ=9⇒λ= .
3
AnswerMarks
⇒Coordinates of M are (3, 3, 1)M1
A1
M1
AnswerMarks
A13.1a
1.1
1.1
AnswerMarks
1.1Attempt to find general point on AB to get vector CM.
Can use (1, 5, 3)
−1
 
Allow working throughout that uses e.g. 1
 
 
 1 
ft their vector from (i).
Use of dot product to solve
Do not accept a vector answer
AnswerMarks
Alternative method for last two marksExpress as a function of λ and minimise the quadratic
in λ
2
3−3λ
uuur
2  
Minimise CM = 3λ−2
 
 
3λ+2
=(3−3λ)2 +(3λ−2)2 +(3λ+2)2
1
⇒λ= ⇒Coordinates of M are (3, 3, 1)
3
M1
A1
[4]
Express as a function of λ and minimise the quadratic
in λ
AnswerMarks Guidance
4(c) CM2 =22 +12 +32 =14
AB2 =32 +32 +32 =27
uuur uuur
1
⇒Area = AB.CM
2
3
= 42
AnswerMarks
2B1
B1
M1
AnswerMarks
A11.1
1.1
3.1a
AnswerMarks
1.1B1 for each distance ft their M
ft their AB
Formula for area
ft their M
AnswerMarks
Alternative method 1M1
M1
A1
AnswerMarks
A1Formula for area
Cross product
 3  0 −12
uuur uuur
1 1     1  
Area = AB×BC = −3 × 1 = −15
     
2 2 2
     
 −3  5   3 
1 1 3
= 122 +152 +32 = 378 = 42
2 2 2
AnswerMarks
Alternative method 2M1
A1
M1
AnswerMarks
A1For use of dot product, formula for area
Pythagoras to find sinθ
uuur uuur uuur uuur uuur uuur
1
Area = AB BC sinθ where AB.BC = AB BC cosθ
2
−12 −4
⇒cosθ= =
27 26 78
8 31
⇒sinθ= 1− =
39 39
1 31 3
⇒Area = 27 26 = 42
2 39 2
[4]
M1
M1
A1
A1
Formula for area
Cross product
M1
A1
M1
A1
For use of dot product, formula for area
Pythagoras to find sinθ
Take z =cosθ+isinθ⇒ z−1 =cosθ−isinθ
1
⇒ z− =2isinθ
z
5
 1 10 5 1
z− =32isin5θ= z5 −5z3 +10z− + −
 z z z3 z5
 1   1   1
⇒32isin5θ= z5 − −5z3 − +10z−
  
 z5   z3   z
=2isin5θ−10isin3θ+20isinθ
5 5 1
⇒sin5θ= sinθ− sin3θ+ sin5θ.
8 16 16
5 5 1
i.e. A= , B=− , C =
AnswerMarks
8 16 16M1
A1
M1
AnswerMarks
A12.1
1.1
1.1
AnswerMarks
2.2aUse of z and de Moivre Sight of i is necessary
Both sides
Attempt conversion into sin soi
All three stated
Alternative method 1
eiθ−e−iθ 5
(sinθ)5 = 
 2i 
1 ( e5iθ−5e3iθ+10eiθ−10e−iθ+5e−3iθ−e−5iθ)
=
(2i)5
1 (( e5iθ−e−5iθ) ( e3iθ−e−3iθ) ( eiθ−e−iθ))
= −5 +10
32i
1
= (sin5θ−5sin3θ+10sinθ)
16
5 5 1
⇒sin5θ= sinθ− sin3θ+ sin5θ
8 16 16
5 5 1
⇒ A= , B=− , C=
AnswerMarks
8 16 16M1
A1
M1
AnswerMarks
A1Sight of i in the denominator is necessary
Collection to convert back
All three stated
Alternative method 1
eiθ−e−iθ 5
(sinθ)5 = 
 2i 
1 ( e5iθ−5e3iθ+10eiθ−10e−iθ+5e−3iθ−e−5iθ)
=
(2i)5
1 (( e5iθ−e−5iθ) ( e3iθ−e−3iθ) ( eiθ−e−iθ))
= −5 +10
32i
1
= (sin5θ−5sin3θ+10sinθ)
16
5 5 1
⇒sin5θ= sinθ− sin3θ+ sin5θ
8 16 16
5 5 1
⇒ A= , B=− , C=
8 16 16
M1
A1
M1
A1
Alternative method 2
z=cosθ+isinθ
⇒z5=cos5θ+isin5θ
=(cosθ+isinθ)5
=cos5θ+5icos4θsinθ−10cos3θsin2θ−10icos2θsin3θ+5cosθsin4θ+isin5θ
⇒sin5θ=5cos4θsinθ−10cos2θsin3θ+sin5θ
=5 ( 1−sin2θ)2 sinθ−10 ( 1−sin2θ) sin3θ+sin5θ
=5sinθ−10sin3θ+5sin5θ−10sin3θ+10sin5θ+sin5θ
=5sinθ−20sin3θ+16sin5θ
z3=cos3θ+isin3θ
=(cosθ+isinθ)5=cos3θ+3icos2θsinθ−3cosθsin2θ−isin3θ
⇒sin3θ=3cos2θsinθ−sin3θ
=3sinθ−4sin3θ
⇒sin5θ−5sin3θ=−10sinθ+16sin5θ
⇒16sin5θ=10sinθ−5sin3θ+sin5θ
10 5 5 1
⇒A= = ,B=− ,C=
AnswerMarks
16 8 8 16M1
A1
M1
A1
De Moivre
for both
Eliminate sin3θ
All three stated
[4]
Alternative method 2
z=cosθ+isinθ
⇒z5=cos5θ+isin5θ
=(cosθ+isinθ)5
=cos5θ+5icos4θsinθ−10cos3θsin2θ−10icos2θsin3θ+5cosθsin4θ+isin5θ
⇒sin5θ=5cos4θsinθ−10cos2θsin3θ+sin5θ
=5 ( 1−sin2θ)2 sinθ−10 ( 1−sin2θ) sin3θ+sin5θ
=5sinθ−10sin3θ+5sin5θ−10sin3θ+10sin5θ+sin5θ
=5sinθ−20sin3θ+16sin5θ
z3=cos3θ+isin3θ
=(cosθ+isinθ)5=cos3θ+3icos2θsinθ−3cosθsin2θ−isin3θ
⇒sin3θ=3cos2θsinθ−sin3θ
=3sinθ−4sin3θ
⇒sin5θ−5sin3θ=−10sinθ+16sin5θ
⇒16sin5θ=10sinθ−5sin3θ+sin5θ
10 5 5 1
⇒A= = ,B=− ,C=
16 8 8 16
M1
A1
M1
A1
Question 4:
4 | (a) | −3
uuur
 
AB= 3 oe
 
 
 3 
x 4 −3
     
Equation of AB is y = 2 +λ 3 oe
     
     
z 0  3 
⇒4−x= y−2= z | B1
M1
A1 | 1.1
1.1
1.1 | soi
−3 x
   
their 3 soi r = or y =is not required for M1
   
   
 3  z
Allow equivalent equations
e.g. ⇒1−x= y−5= z−3 from using B
[3]
4 | (b) | −3 4 −3 4−3λ
uuur uuur
       
AB= 3 , OM is 2 +λ 3 = 2+3λ
       
       
 3  0  3   3λ 
4−3λ  1  3−3λ
uuur
     
CM = 2+3λ − 4 = 3λ−2
     
     
3λ −2 3λ+2
  
3−3λ −3
uuur
   
Perpendicular to AB⇒ 3λ−2 . 3 =0
   
   
3λ+2  3 
⇒−9+9λ+9λ−6+9λ+6=0
1
⇒27λ=9⇒λ= .
3
⇒Coordinates of M are (3, 3, 1) | M1
A1
M1
A1 | 3.1a
1.1
1.1
1.1 | Attempt to find general point on AB to get vector CM.
Can use (1, 5, 3)
−1
 
Allow working throughout that uses e.g. 1
 
 
 1 
ft their vector from (i).
Use of dot product to solve
Do not accept a vector answer
Alternative method for last two marks | Express as a function of λ and minimise the quadratic
in λ
2
3−3λ
uuur
2  
Minimise CM = 3λ−2
 
 
3λ+2
=(3−3λ)2 +(3λ−2)2 +(3λ+2)2
1
⇒λ= ⇒Coordinates of M are (3, 3, 1)
3
M1
A1
[4]
Express as a function of λ and minimise the quadratic
in λ
4 | (c) | CM2 =22 +12 +32 =14
AB2 =32 +32 +32 =27
uuur uuur
1
⇒Area = AB.CM
2
3
= 42
2 | B1
B1
M1
A1 | 1.1
1.1
3.1a
1.1 | B1 for each distance ft their M
ft their AB
Formula for area
ft their M
Alternative method 1 | M1
M1
A1
A1 | Formula for area
Cross product
 3  0 −12
uuur uuur
1 1     1  
Area = AB×BC = −3 × 1 = −15
     
2 2 2
     
 −3  5   3 
1 1 3
= 122 +152 +32 = 378 = 42
2 2 2
Alternative method 2 | M1
A1
M1
A1 | For use of dot product, formula for area
Pythagoras to find sinθ
uuur uuur uuur uuur uuur uuur
1
Area = AB BC sinθ where AB.BC = AB BC cosθ
2
−12 −4
⇒cosθ= =
27 26 78
8 31
⇒sinθ= 1− =
39 39
1 31 3
⇒Area = 27 26 = 42
2 39 2
[4]
M1
M1
A1
A1
Formula for area
Cross product
M1
A1
M1
A1
For use of dot product, formula for area
Pythagoras to find sinθ
Take z =cosθ+isinθ⇒ z−1 =cosθ−isinθ
1
⇒ z− =2isinθ
z
5
 1 10 5 1
z− =32isin5θ= z5 −5z3 +10z− + −

 z z z3 z5
 1   1   1
⇒32isin5θ= z5 − −5z3 − +10z−
  
 z5   z3   z
=2isin5θ−10isin3θ+20isinθ
5 5 1
⇒sin5θ= sinθ− sin3θ+ sin5θ.
8 16 16
5 5 1
i.e. A= , B=− , C =
8 16 16 | M1
A1
M1
A1 | 2.1
1.1
1.1
2.2a | Use of z and de Moivre Sight of i is necessary
Both sides
Attempt conversion into sin soi
All three stated
Alternative method 1
eiθ−e−iθ 5
(sinθ)5 = 
 2i 
1 ( e5iθ−5e3iθ+10eiθ−10e−iθ+5e−3iθ−e−5iθ)
=
(2i)5
1 (( e5iθ−e−5iθ) ( e3iθ−e−3iθ) ( eiθ−e−iθ))
= −5 +10
32i
1
= (sin5θ−5sin3θ+10sinθ)
16
5 5 1
⇒sin5θ= sinθ− sin3θ+ sin5θ
8 16 16
5 5 1
⇒ A= , B=− , C=
8 16 16 | M1
A1
M1
A1 | Sight of i in the denominator is necessary
Collection to convert back
All three stated
Alternative method 1
eiθ−e−iθ 5
(sinθ)5 = 
 2i 
1 ( e5iθ−5e3iθ+10eiθ−10e−iθ+5e−3iθ−e−5iθ)
=
(2i)5
1 (( e5iθ−e−5iθ) ( e3iθ−e−3iθ) ( eiθ−e−iθ))
= −5 +10
32i
1
= (sin5θ−5sin3θ+10sinθ)
16
5 5 1
⇒sin5θ= sinθ− sin3θ+ sin5θ
8 16 16
5 5 1
⇒ A= , B=− , C=
8 16 16
M1
A1
M1
A1
Alternative method 2
z=cosθ+isinθ
⇒z5=cos5θ+isin5θ
=(cosθ+isinθ)5
=cos5θ+5icos4θsinθ−10cos3θsin2θ−10icos2θsin3θ+5cosθsin4θ+isin5θ
⇒sin5θ=5cos4θsinθ−10cos2θsin3θ+sin5θ
=5 ( 1−sin2θ)2 sinθ−10 ( 1−sin2θ) sin3θ+sin5θ
=5sinθ−10sin3θ+5sin5θ−10sin3θ+10sin5θ+sin5θ
=5sinθ−20sin3θ+16sin5θ
z3=cos3θ+isin3θ
=(cosθ+isinθ)5=cos3θ+3icos2θsinθ−3cosθsin2θ−isin3θ
⇒sin3θ=3cos2θsinθ−sin3θ
=3sinθ−4sin3θ
⇒sin5θ−5sin3θ=−10sinθ+16sin5θ
⇒16sin5θ=10sinθ−5sin3θ+sin5θ
10 5 5 1
⇒A= = ,B=− ,C=
16 8 8 16 | M1
A1
M1
A1
De Moivre
for both
Eliminate sin3θ
All three stated
[4]
Alternative method 2
z=cosθ+isinθ
⇒z5=cos5θ+isin5θ
=(cosθ+isinθ)5
=cos5θ+5icos4θsinθ−10cos3θsin2θ−10icos2θsin3θ+5cosθsin4θ+isin5θ
⇒sin5θ=5cos4θsinθ−10cos2θsin3θ+sin5θ
=5 ( 1−sin2θ)2 sinθ−10 ( 1−sin2θ) sin3θ+sin5θ
=5sinθ−10sin3θ+5sin5θ−10sin3θ+10sin5θ+sin5θ
=5sinθ−20sin3θ+16sin5θ
z3=cos3θ+isin3θ
=(cosθ+isinθ)5=cos3θ+3icos2θsinθ−3cosθsin2θ−isin3θ
⇒sin3θ=3cos2θsinθ−sin3θ
=3sinθ−4sin3θ
⇒sin5θ−5sin3θ=−10sinθ+16sin5θ
⇒16sin5θ=10sinθ−5sin3θ+sin5θ
10 5 5 1
⇒A= = ,B=− ,C=
16 8 8 16
M1
A1
M1
A1
Points $A$, $B$ and $C$ have coordinates $(4, 2, 0)$, $(1, 5, 3)$ and $(1, 4, -2)$ respectively. The line $l$ passes through $A$ and $B$.
\begin{enumerate}[label=(\alph*)]
\item Find a cartesian equation for $l$. [3]
\end{enumerate}
$M$ is the point on $l$ that is closest to $C$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the coordinates of $M$. [4]
\item Find the exact area of the triangle $ABC$. [4]
\end{enumerate}

\hfill \mbox{\textit{OCR Further Pure Core 1 2021 Q4 [11]}}
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