Question 12:
--- 12(a) ---
12(a) | Forms the product of
the matrix and the
direction/position
vector of L
1 | 2.2a | M1 |  2 1 0   1   5 

3 4 6 3 = 9 = d ir e c t io n v e c t o r o f L
 2
– – 5 2 3  – 1 4
    1 5
   
• = 3 9 2 8
   
    – 1 4    
2 8
θ = c o s
1 1 1 2 2
θ = 0 . 7 0 1
( 3 d p )
Obtains the correct
direction vector of L
2
possibly embedded | 1.1b | A1
Use the scalar
product of the
direction vectors of
L and their L to
1 2
obtain their correct
28. | 1.1a | M1
Completes a
reasoned argument
to obtain 0 . 7 0 1
Must see
28
cosθ =
11 122
or
cosθ =0.7643
or
θ = 0 . 7 0 0 7 . . . | 2.1 | R1
Subtotal | 4
Q | Marking instructions | AO | Marks | Typical solution
--- 12(b) ---
12(b) | Forms the product of the matrix
and a point on L
1 | 2.2a | M1 |  2 1 04  10 
    
3 4 6 2 = 26
    
 –5 2 –3   1   –19 
Vector connecting L to L
1 2
 10  4  6 
     
AB = 26 – 2 = 24
     
 –19  1  –20
  
Vector perpendicular to both lines
5  1 –21
     
n= 9 × 3 = 9
     
 4   –1    6  
n•AB = –30
n•AB 10 5 62
Distance= = =
n 62 31
Obtains a correct point on
 1 0   5 
2 6 + μ 9
– 1 9 4 | 1.1b | A1
Obtains a vector connecting a
point on each line.
or
Obtains the equations of two
correct parallel planes PI | 1.1b | B1
Obtains a vector perpendicular
to both lines.
or
Calculates the scalar product of
the general vector between the
lines with both direction vectors
to obtain a pair of simultaneous
equations. | 3.1a | M1
Uses a correct method to obtain
the shortest distance between
the lines.
Condone a negative distance
or
Obtains solutions to their pair of
simultaneous equations and
obtains the vector between the
two closest points | 3.1a | M1
5 6 2
Obtains OE
3 1 | 1.1b | A1
Subtotal | 6
Question total | 10
Alternative Typical Solution 1 | Alternative Typical Solution 2
 2 1 04 10 
    
3 4 6 2 = 26
    
 –5 2 –3   1   –19 
10  5
   
Equationof L :r= 26 +μ 9
2    
 –19  4
 
General vector
 10+5μ   4+  6+5μ− 
     
AB= 26+9μ – 2+3 = 24+9μ−3
     
 –19+4μ   1−    –20+4μ+ 
Scalar product
1  6+5μ− 
   
0= 3 • 24+9μ−3 =98+28μ−11
   
 −1  –20+4μ+
 
11−28μ=98
5  6+5μ− 
   
0= 9 • 24+9μ−3 =166+122μ−28
   
 4  –20+4μ+
 
28−122μ=166
406 51
= , μ=
31 31
 35 31 
 
AB= –15 31
 
 –10 31 
5 62
Distance=
31 |  2 1 0    4  1 0 
  
3 4 6 2 = 2 6
  
– – 5 2 3   1   – 1 9 
    5 1  – 2 1 
  
× = 9 3 9
  
   – 4 1    6
P a r a l l e l p l a n e s
 − x 2 1 9 y − 6 z  = 2 1 4 − 9  2 − 6  1 = 6 0
1
 − x 2 1 9 y − 6 z  = 2 1 1 0 − 9  2 6 − 6  − 1 9 = 9 0
2
9 0 − 6 0 5 6 2
d = is t a n c e =
2 2 + 1 9 2 2 + 6 3 1
Q | Marking
instructions | AO | Marks | Typical solution