Mathematics · May / June 2025 · Paper 41 · Question 6

YMA01

As printed in the paper

The question as text

6. Relative to a fixed origin O, • the point A has position vector i + 2j – 3k • the point B has position vector 5i + 2j + k The line l passes through A and B.

Read automatically from the paper; the printed page above is exact.

Show the answer and mark scheme

6(a)[2]
±
(
+
+
) (
−
+
−
)
= ±
(
+
)
Attempts
5
i
2
j
k
i
2
j
3
k
4
i
4
k
= +
−
+
λ
+
=
+
+
+
λ
+
e.g.
r
i
2
j
3
k
(4
i
4 )
k or
r
5
i
2
j k
(4
i
4 )
k oe
6(b)[6]
AC
=
2
i
+
(
α
−
2)
j
+
8
k
AB =
4
2
+
4
2
or
AC
=
2
2
+
(
α
−
2)
2
+
8
2
AB =
32
and
=
+
α
−
2
AC
68
(
2)
8
+
32
°=
cos45
×
+
α
−
2
32
68
(
2)
40
cos45
°=
α
2
−
4
α
−
28
=
0
×
+
α
−
32
68
(
2)
2
α=
±
2
4 2
Allow vectors to be written in any acceptable form. E.g. column vectors are fine
(a)
M1:
Attempts to find the difference between the two vectors.
If no method is shown look for two correct components from (
)
±
4 ), (0
i
j )
,
±
(4
k
A1:
Any correct equation for the line in the form given.
Examples of correct equations are;
= +
−
+
λ
+
=
+
+
+
μ
+
r
i
2
j
3
k
(4
i
4 )
k ,
r
5
i
2
j
1
k
(
i
k
)
−
−
5
4
2
2
3
1
r
=
2
+
λ
0
,
r
=
2
+
k
0
and
r
=
2
+
0
1
−
4
−
2
2
−
1
−
1
Note that the LHS must be correct so
l = +
i
2
j
−
3
k
+
λ
(4
i
+
4 )
k is A0
Condone the vectors not being bold or underlined
−
5
i
Condone incorrect vector notation for the M but not the A mark. E.g.
=
+
λ
r
2
j
−
k
(b) There are many ways to attempt this question.
The following is via the scalar product of AB
and AC
. Note that it could be attempted via the
scalar product of (
)
±
+
1
i
1
k and AC
instead of AB
and AC
. Attempts using an incorrect vector
, can only score a maximum of 1 mark (1st M1)
pair, e.g AB
and BC
(
)
AC =
2 , (
i
α
−
2) , 8
j
k
Condone incorrect vector notation throughout (b). E.g.
B1:
A correct
=
+
α
−
+
(or
= −
−
α
−
−
)
AC
2
i
(
2)
j
8
k
CA
2
i
(
2)
j
8
k
It must be simplified but may be implied by further work.
Allow in coordinate form (
)
2,
α
−
2, 8
if used correctly
M1:
Attempts to find the magnitude of either their AB
or their AC
For example, for an attempt at the magnitude of AC
there should be attempt to subtract each component of OA
and OC
the three resulting values should then be squared and added
the answer to this is then square rooted
If the method is not apparent then look for two correct components
(
AB =
32
and
=
+
α
−
2
=
−
α
+
2
A1:
Correct magnitude of both
AC
68
(
2)
72
4
a
They must be simplified (as above) but can be scored following sign slips on either vector.
You can imply this mark from further work
dM1: Attempts the scalar product of '
±
AB
'
and '
±
AC
'
±
±
AB
and AC
must be attempted via a correct method. If the method is not directly seen it may
be implied by two correct components
There must be a similar correct attempt at the magnitudes of AB
±
and AC
±
All should be connected by the equation AB
±
±
AC
= ±
AB
±
AC
cos45
ddM1: Allowable method to find at least one value of αfrom a quadratic equation.
For example,
(
)
α
−
2
=
α
=
2
32
...
α
2
−
4
α
−
28
=
0
α
=
...
This can only be awarded if the equation has real roots.
It is dependent upon both previous M’s
A1:
α=
2
±
4 2
or exact equivalent
Note: Alternative method via cosine rule
=
+
α
−
+
= −
−
α
−
−
B1:
AC
2
i
(
2)
j
8
k
(or
CA
2
i
(
2)
j
8
k
) May be implied by further work.
M1:
Attempts to find the magnitude of either AB
or AC
(
)
AB =
=
+
α
−
2
=
−
α
+
2
A1:
32
and
AC
68
(
2)
72
4
a
2
2
2
dM1: Attempts
BC
=
AB
+
AC
−
2
AB AC
cos45
(
)
(
) (
)
(
) (
(
)
(
)
(
)
2
+ α−
2
+
2
=
2
+ α−
2
+
2
+
2
+
2
−
2
+ α−
2
+
2
FYI
2
2
4
2
2
8
4
4
2
2
2
4
ddM1: Rearranges and proceeds to at least one value for α. This can only be awarded if the equation is quadratic
and has real roots
α=
±
A1:
2
4 2
or exact equivalent

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