Question 11:
11 | (a) |  0   1 8 
r = 1 1 + . .. o r r = 2 + . ..
2   0 2    
7 1 6
or . .. 1  r = + − or ... 1 1 1  + −
1 7 1
0 2    
1 1 1  r = + −
7 1 | B1*
B1dep*
[2] | 3.3
3.3 | For two correct components – the
components are
• r = ... or r =... ONLY
• position vector,
• direction vector in lowest terms
with parameter.
For a correct equation with direction
vector in lowest terms. Condone lack
of (or incorrect) range of values for
parameter. Other common answers are:
0 2    − 
1 1 1 ,  r = +
7 1 −
1 8 2    − 
2 1 ,  r = +
1 6 1 −
1 8 2    
2 1 ,  r = + −
1 6 1
There are other possible answers, e.g.
6 2    
8 1 ,  r = + −
1 0 1
(b) | 6  2   8 
     
7  −1 = −4
     
     
1  1   −20
  0    0   8 
 
0 − 1 1 − 4
 
 
1 8 7   − 2 0
( D = )
 8 
− 4
− 2 0
2 2 3 0
= ( = 8 . 0 3 3 2 6 4 . . . )  5 so yes (the site) passes Test 1.
1 5 | M1
A1
[2] | 3.4
2.2a |  6 
Calculate vector product n with 7 and
1
the direction vector of their T from part (a)
with at least one component correct
(following through their vectors), and
setting up the equation
  0  
0 − a n
1 8
D = where a is a point on
n
their T from part (a) and their vector n.
Condone lack of modulus signs in
equation. Note that  these vectors are
valid as are scalar multiples of n.
Correct calculation (either exact e.g.
1 7 6
) or at least 2 significant
4 3 0
figures), including reference to 5 and
conclusion.
(c) | 0 6 0 6             
( P Q ) 0 7 1 1 7 1 1   =  + − =  −
1 8 1 7 1 1  +
( )
P Q ( 6 ) 2 ( 7 1 1 ) 2 ( 1 1 ) 2    = + − + +
(6)2 +(7−11)2 +(+11)2 192 (862 −132−119 0)
2 .1 7 ...   or 0 .6 3 7 ...  − but 0 2  so no, it’s not possible. | B1
B1
B1
[3] | 3.4
3.4
3.1b | No MR in this part.
For ( 0 , 1 1 , 7 ) to a general point on S.
Allow correct un-simplified. Can be
implied by a correct expression for the
distance (or distance squared)
2
Expression for PQ or PQ - allow un-
simplified.
Solving inequality/equation for 𝜆 and
concluding no + reason. Condone “no” as
conclusion. Critical values of correct to
at least 2 sf. Allow strict inequalities.
Alternative method
Distance between ( 0 , 1 1 , 7 ) and ( 0 , 0 , 1 8 ) is 112 +(7−18)2 | Distance between ( 0 , 1 1 , 7 ) and ( 0 , 0 , 1 8 ) is 112 +(7−18)2 | B1 | B1 | Find distance (or distance squared)
between (0, 11, 7) and (0, 0, 18) – allow
un-simplified. | Find distance (or distance squared)
between (0, 11, 7) and (0, 0, 18) – allow
Distance between ( 0 , 1 1 , 7 ) and (1 2 , 1 4 , 2 0 ) is 122 +(14−11)2 +(20−7)2 | B1 | Find distance (or distance squared)
between (0, 11, 7) and (12, 14, 20) – allow
un-simplified
1 1 2 = 1 5 . 5 5 6 3 . . . and 3 2 2 = 1 7 .9 4 4 3 ... which are both less than 19 and the
two point (0, 0, 18) and (12, 14, 20) are the points furthest from (0, 11, 7) so no,
it’s not possible. | B1 | Correct values given to at least 3 sf (or in a
form in which all three can be compared
e.g. 322, 242and 361) and some
indication that these two points are the
furthest from the camera and conclude no.
[3]
Find distance (or distance squared)
between (0, 11, 7) and (12, 14, 20) – allow
1 1 2 = 1 5 . 5 5 6 3 . . . and 3 2 2 = 1 7 .9 4 4 3 ... which are both less than 19 and the
two point (0, 0, 18) and (12, 14, 20) are the points furthest from (0, 11, 7) so no,
it’s not possible.