Question 6:
6 | (a) | 4 −2  −5   1 
       
4 × 3 = −10 =−5 2
       
       
3  1   20  −4 | B1 | 1.1 | Finding a normal to Π. | Any valid method; for example
1
 
using a and setting the dot
 
 
b
product with both vectors in Π to 0.
 1   2 
2 . 0 ( = 2 − 1 2 )
− 4 3 | M1 | 1.1 | Any clear attempt to find the
angle between the normals (can
be implied by dotting the two
normals together).
1 0 −
c o s  =
1 2 2 2 ( 4 ) 2 2 2 3 2 + + − + | M1 | 1.1 | Using the definition of dot
product to find cos in
unsimplified numerical form
Could see modulus signs. | This mark can be awarded after
M0
−10
cos𝜃 = 𝑠𝑜 𝜃 = 127.2
√273
so required angle is 52.8 (1 dp) | A1 | 1.1 | Or directly to answer.
Final answer | awrt 0.921 rads
[4]
(b) | −1  1 
   
2 . 2 =−1+4−4 (=−1)
   
   
 1  −4
 9   1   1 
     
 −7  +  2  .  2  =−1
     
20 −4 −4
9 – 14 – 80 + (1 + 4 + 16)ν = –1 => ν = 4
 9   1  13
     
r = −7 +4 2 = 1
F      
     
20 −4  4
soF is (13, 1, 4) | M1
M1
A1
A1 | 3.1a
1.1
1.1
1.1 | Dotting their normal and a point
on Π.
Forming the equation of the line
AF and intersecting with Π to
find the value of the parameter
for the PoI.
Condone presentation as
position vector of F. | Condone poor notation up to final
A mark as long as method clear
Or M1 for using formula to find
84
distance AF (= ), and dividing
√21
this by magnitude their n.
Alternative method | M1 | Forms equations
9 1
(−7)+𝜈( 2 )
20 −4
−1 4 −2
= ( 2 )+𝜆(4)+𝜇( 3 )
1 3 1
so 4𝜆−2𝜇−𝜈 = 10
4𝜆+3𝜇−2𝜈 = −9
3𝜆+𝜇+4𝜈 = 19
(𝜆 = 2 𝜇 = −3)𝜈 = 4 | M1A1 | BC | If not BC, then M1 for two
F is (13, 1, 4) | A1 | equations in two unknowns
M1
Forms equations
2ndAlternative method
F i s a p o i n t o n s o 
9 1 4 2  −   −     − 
A F A O O F 7 2 4 3   = + = + + +
2 0 1 3 1 −
1 0 4 2  −     − 
9 4 3   = + +
1 9 3 1 −
4  
A F 4 0 6 1 4 1 7 0   . =  − + + =
3
( 4 1 7 6 1 )   + =
2  − 
A F 3 0 2 8 7 1 4 0   . =  + + =
1
( 2 4 )   + = − | F i s a p o i n t o n s o 
9 1 4 2  −   −     − 
A F A O O F 7 2 4 3   = + = + + +
2 0 1 3 1 −
1 0 4 2  −     − 
9 4 3   = + +
1 9 3 1 − | Using F, a general point on Π,
equates A F b. or A F c. to 0
2 equations in λ and μ
4  
A F 4 0 6 1 4 1 7 0   . =  − + + =
3
( 4 1 7 6 1 )   + =
M1
2  − 
A F 3 0 2 8 7 1 4 0   . =  + + =
1
( 2 4 )   + = −
M1
2 , 3    = = −
1 4 2 1 3  −     −   
O F 2 2 4 3 3 1 = + − =
1 3 1 4
S o F i s (1 3 , 1 , 4 ) | A1 | Solves (BC)
A1
Using F, a general point on Π,
equates A F b. or A F c. to 0
2 equations in λ and μ
3ndAlternative method | M1
A1
M1
A1 | Finds perpendicular distance
Finds normal vector from A to
Π
Uses their normal vector to find
F
D , p e rp d is t fro m A to is g iv e n b y 
9 1    
   
7 2 ( 1 ) . − −     −
   
2 0 4 9 1 4 8 0 1 8 4 8 4 −     − − + −
D 2 1 4 2 1 = = = = =
2 1 2 2 ( 4 2 ) 1 4 1 6 2 1 2 1 + + − + +
1 1 4      
1
A F 4 ˆ 2 1 4 2 1 2 4 2 8  n = =  = =
2 1
4 4 1 6 − −  −
9 4 1 3      
O F O A  = + A F 7 8 1 = − + =
2 0 1 6 4 −
S o c o -o rd s o f F a re (1 3 , 1 , 4 )
1    
O r c o u ld s e e A F 2  =
4 −
2 2 ( 2 ) ( 4 2 ) 2 1 4 2 1 4       + + − = =  =
[4]
M1
A1
M1
A1
Finds perpendicular distance
Finds normal vector from A to
Π
Uses their normal vector to find
F