Question 10:
--- 10(a) ---
10(a) | Carry out correct process for evaluating the scalar product of
direction vectors | *M1 | 3 1 3  1 
       
4 . 2 or 4 . 2
       
   a     2    a    2    
3(–1) + 4(2) + 2a or –3 + 8 + 2a or 5 + 2a.
Allow one slip in unsimplified form.
Using the correct process for the moduli, divide the scalar product
2
by the product of the moduli and equate to  ,
2
2
or equate the scalar product to the product of the moduli and 
2 | *M1 | *M1 marks independent of each other, so *M0 *M1 for failure to
use both direction vectors, but must be using scalar product and
same 2 vectors throughout.
2 2
Allow or − throughout question.
2 2
52a 2
State a correct equation in any form, e.g. 
3 25a2 2
Allow unsimplified as in guidance | A1 | 52a 2
 OE
916a2 144 2
2
E.g. 5 + 2a =  916a2 144
2
If moduli initially correct but later has errors, award A1 when using
2 2 2
or  or − .
2 2 2
Form a quadratic equation in a with 3 or more terms all on one side
and solve for a.
DM1 depends on BOTH *M1 | DM1 | Must square (5 + 2a) to get 3 terms and must remove square roots
from both terms on other side.
9
25 + 20a + 4a2 = (25 + a2)
2
a2 − 40a + 175 = 0 hence (a – 5)(a – 35) = 0.
10(a) | Obtain a = 5 and a = 35 | A2 | A1 for each, working not needed if quadratic correct.
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Question | Answer | Marks | Guidance
--- 10(b) ---
10(b) | Express general point of at least one line correctly in component
form,
 1  3   –3 –µ 
   
i.e. 1  4 or –1  2µ
   
   
2a  a  4  2µ  | B1 | Often the third point on the line occurs after M1 A1 is gained.
Equate at least two pairs of corresponding components and solve
for λ or µ or a | M1 | If solve for a first, they must have a complete method to eliminate
both λ and µ.
If using a to solve for λ or for µ, a must have been found from a
valid method.
Obtain λ = –1 or µ = –1 | A1
Obtain a = 2 | A1
Obtain position vector of the point of intersection is –2i – 3j + 2k
Two different answers for point of intersection scores A0 even if
one is correct | A1 | Accept coordinates, row or column, but not (–2i,– 3j,+ 2k) or
2i
 
–3j but ISW after correct form seen.
 
 
 2k
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