Question 3:
3 | 3
Question 8 Notes
8. (a)
(b)
(c) | M1
A1
B1ft
Note
M1
M1
A1
Note
Note
Note |  
Finding the difference (either way) between OB andOA.
If no “subtraction” seen, you can award M1 for 2 out of 3 correct components of the difference.
 1 1
 
i −j+ k or −1 or (1, −1,1) or benefit of the doubt −1
 
 
 1 1
−2  1 −1  1
         
{r}= 4 +λ −1 or {r}= 3 +λ −1 , with AB or BA correctly followed through from (a).
       
       
 7  1  8  1
r = is not needed.
 
An attempt to find either the vector PB or BP.
If no “subtraction” seen, you can award M1 for 2 out of 3 correct components of the difference.
   
( ) ( )
Applies dot product formula between their AB or BA and their PB or BP .
1
Obtains {cosθ}= by correct solution only.
3
 
AB •PB
If candidate starts by applying   correctly (without reference to cosθ=...)
AB .PB
they can gain both 2nd M1 and A1 mark.
1
Award the final A1 mark if candidate achieves {cosθ}= by either taking the dot product between
3
 1 −1 −1  1
       
(i) −1 and 1 or (ii) 1 and −1 . Ignore if any of these vectors are labelled incorrectly.
       
       
 1  5 −1 −5
Award final A0, cso for those candidates who take the dot product between
 1  1 −1 −1
       
(iii) −1 and −1 or (iv) 1 and 1
       
       
 1 −5 −1  5
1 1
They will usually find {cosθ}= − or may fudge{cosθ}= .
3 3
If these candidates give a convincing detailed explanation which must include reference to the direction
of their vectors then this can be given A1 cso
(c) | Alternative Method 1: The Cosine Rule
−1 0 −1  1
            Mark in the same way
PB = OB−OP=  3  −  2  =  1  or BP =  −1  M1
as the main scheme.
       
 8 3  5 −5
  
Note PB = 27, AB = 3 and PA = 24
( )2 ( )2 ( )2 ( )( ) Applies the cosine rule
24 = 27 + 3 − 2 27 3 cosθ M1 oe
the correct way round
27 + 3− 24 1
cosθ = = Correct proof A1 cso
18 3
[3]
8. (c) | Alternative Method 2: Right-Angled Trigonometry
−1 0 −1  1
            Mark in the same way
PB = OB−OP=  3  −  2  =  1  or BP =  −1  M1
as the main scheme.
       
 8 3  5 −5
( )2 ( )2 ( )2
Either 24 + 3 = 27
 1 − 2
      Confirms ∆PAB is right-angled M1
or AB•PA = −1 • 2 = −2 − 2 + 4 = 0
   
   
 1  4
 AB  3 1
So, cosθ = ⇒cosθ = = Correct proof A1 cso
 PB  27 3
[3]
(d)
(e)
(f) | M1
A1ft
Note
Note
M1
Note
A1ft
A1ft
Note | 0
   
Writing down a line in the form p +λd or p +µdwith either a = 2 or d = their AB d = their AB,
 
 
3

or a multiple of their ABfound in part (a).
0  1 0
       
Writing 2 +µ −1 or 2 +µd, where d = their AB or a multiple of their ABfound in part (a).
     
     
3  1 3
r = is not needed.
Using the same scalar parameter as in part (b) is fine for A1.
   
EitherOP + theirAB or OP − theirAB.
This can be implied at least two out of three correct components for either their C or their D.
At least one set of coordinates are correct. Ignore labelling of C, D
Both sets of coordinates are correct. Ignore labelling of C, D

You can follow through either or both accuracy marks in this part using their AB from part (a).
M1
Note | h
Way 1:  =sinθ
their PB
 
Way 2: Attempts PA or CB
1
Way 3: Attempts (their PB)(their AB)sinθ
2
Finding AD by itself is M0.
A1 | Either
 
• h = 27sin(70.5...) or PA = CB = 24 or equivalent. (See Way 1 and Way 2)
or
• the area of either triangle APB or APD or BDP = 1 3 ( 3 3 ) sin(70.5...) o.e. (See Way 3).
2
dM1
A1
Note | which is dependent on the 1st M1 mark.
A full method to find the area of trapezium ABCD. (See Way 1, Way 2 and Way 3).
9 2 from a correct solution only.
A decimal answer of 12.7279... (without a correct exact answer) is A0.
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