Mathematics · May / June 2024 · Paper 41 · Question 2
YMA01
As printed in the paper
The question as text
2. With respect to a fixed origin, O, the point A has position vector 7 → = OA 2 5 Given that 2 → = AB 4 3
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
2(a)[2]
7
−
2
OB
=
OA
+
AB
=
2
+
4
=
...
−
5
3
5
k or (
)
=
6
, 5
+
6
−
2
5,6, 2
−
i
j
−
2
2(b)[4]
a
5
BC
=
OC
−
''
OB
''
=
5
−
''
6 ''
=
...
−
1
−
2
a
a
−
5
(
)
OC BC
.
''
''
=
0
5 .''
−
1 ''
=
0
a a
−
5
−−=
5 1
0
−
1
1
(
)(
)
a
2
−
5
a
−
6
=
0
a
−
6
a
+
1
=
0
a
=
...
a = −
1
, 6
(a)
M1: Attempts to add OA
and AB
, which may be implied by two correct coordinates out of three if there
is no direct statement as in the main scheme.
A1: Correct coordinates, in either vector or Cartesian form. (See main scheme).
Withhold the mark for hybrid versions such as (
)
5 ,6 , 2
i
j
−
k but isw after you have seen a correct answer
(b)
M1: Attempts BC
by subtracting their OB
(look at their answer to (a)) and OC
, subtracted either way
round, which may be implied by two correct coordinates out of three following through on their (a).
dM1: Applies the scalar product with OC
and their BC
, sets equal to 0 and forms a quadratic equation
in a. Dependent upon the previous M.
An alternative method is via Pythagoras’ theorem
OC
2
BC
2
OB
2
+
=
(
) (
)
a
2
+
5
2
+
1
2
+
(
a
−
5)
2
+
1
2
+
1
2
=
5
2
+
6
2
+ −
( 2)
2
2
a
2
−
10
a
−
12
=
0
ddM1: Depends on both previous Ms. Solves their 3 term quadratic equation (which must have real
roots) by any valid means including a calculator (check roots if done this way).
A1: Correct values for a. If the candidate goes on to reject a value it is A0.
8
3
3
2
2
9
x
−
y
+
xy
=
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