Mathematics · Oct / Nov 2022 · Paper 41 · Question 3
YMA01
As printed in the paper
The question as text
3. R Q P Figure 1 Figure 1 shows a sketch of triangle PQR. Given that → = 2i – 3j + 4k • PQ → = 8i – 5j + 3k • PR →
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
3(a)[2]
y
−
2
Condone slips, for example in the signs and miscopies/misreads, but the overall
mechanics of the
attempt should be sound.
In E.g.II. an attempt must be made to make progress towards y = f (x), so allow for
example when
an intermediate line such as g( ) h( )
y
×
x
=
b
or g( )
y
=
h( )
x
where g( )
a linear expression in y
7
x
−
1
−+
1 7
x
A1: Correct equation
y
=
but allow
y
=
x
x
3
3
Alt Method:
t
ax
−
1
1
into
M1: Substitutes both
y
=
+
2
and
x
= −
y
=
and makes progress to a point in
t
t
bx
3
which sides can be compared
t
a
×
−
1
a
−
(
)
1
t
−
3
1
3
b
+
bt
=
a
−
+
2
=
=
+
or
2
1
t
t
b
bt
b
×−
t
3
dM1: Compares terms to set up and solve simultaneous equations in both a and b.
a
−
FYI 3
1
=
1,
=
2
b
=
3,
2
b
=
a
−
1
or
b
b
7
x
−
1
y
=
NOT just for correct values of a and b
A1:
3
x
3
x
A
B
+
A
B
(
)(
)
Values for
and
2 (a)
2
x
−
1
x
−
2
x
−
x
−
2
1
2
A = − or
B =
One correct value, either
1
2
−
1
2
Correct PF form
+
2
x
−
1
x
−
2
−
(
)
(
) ( )
1
2
1
+
d
x
= −
ln 2
x
−
1
+
2ln
x
−
2
+
c
x
−
x
−
2
1
2
2
25
25
3
x
1
(
)
(
)
x
= −
x
−
+
x
−
d
ln 2
1
2ln
2
(
)(
)
x
−
x
−
2
2
1
2
5
5
1
1
= −
+
−−
+
ln 49
2ln 23
ln9
2ln3
2
2
529
ln 21
(a)
M1: Attempts a correct PF form leading to values for A and B .
3
x
A
B
C
+
+
A
B
Condone
Values for at least
and
(
)(
)
2
x
−
1
x
−
2
x
−
x
−
2
1
2
A1: One correct value for A or B or one correct term
−
1
2
A1: Correct PF form
+
.
x
−
x
−
2
1
2
This is not just for correct values for A and B but it may be awarded from work seen in (b)
(b)
A
B
(
)
(
)
M1:
+
d
x
=
C
ln 2
x
−
1
+
D
ln
x
−
2
o.e. Condone missing brackets here
x
−
x
−
2
1
2
A
B
A
(
)
(
) ( )
A1ft: Correct ft on their values for A and B
+
x
=
x
−
+
B
x
−
+
d
ln 2
1
ln
2
2
x
−
1
x
−
2
2
Do not condone missing brackets but the brackets can be implied by subsequent working,
E.g. ln 49
dM1: Attempts to substitute the limits 5 and 25 (no misreads on the limits but condone slips, so
don't accept a consistent lower limit of 0 for instance) into an expression of the correct
form (with brackets which may be implied) and subtracts, either way around. Condone
attempts where the substitutions were made into incorrectly combined ln terms.
529
A1: cso
ln 21 Also allow in mixed number form
ln25
4
21
Attempts (
) (
)
3 (a)
i
−
j
+
k
−
i
−
j
+
k
2
3
4
8
5
3
RQ = −
6
i + j k
2
+
(b)
PQ RQ = ×−+ −×+ ×
Attempts
2
6
3 2
4 1
Full attempt to find cos PQR E.g.
2
×−+ −×+ ×=
6
3 2
4 1
29 41cos PQR
Angle PQR = 114°
(a)
Attempts to subtract vectors PQ
and PR
either way around. Look for PQ
PR
M1:
−
or
PR
−
PQ
.
If a method is not shown it can be implied by two correct components of
±
6
i
±
2
j
±
k Note that an attempt such as PR
−
QP
is M0
−
6
−
6
i
RQ = −
i + j k o.e. such as
A1:
6
2
+
2
Do not accept coordinates or indeed
2
j
1
k
(b)
Attempts scalar product of PQ
and their RQ
M1:
. Look for an attempt at multiplying
together the components and adding. There will be some confusion over direction so allow
for sight of
(
) (
) (
)
±×±
2 '' 6''
+ ±×±
3 '' 2''
+ ±×±
4 '' 1''
This cannot be scored if they attempt a scalar product of PQ
and PR
for instance.
dM1: Full attempt to find cos PQR using .
ab
=
a
b
θ
using vectors PQ
±
cos
and their
±
RQ
.
There must be an attempt at both moduli with at least one correct (which may be
unsimplified)
but you should ft on their RQ
. Don't be concerned whether the angle is acute or obtuse.
A1: Angle PQR = awrt 114° ISW after sight of this, e.g followed by 66°
Alt (b)
M1: Attempts all three lengths or all three lengths 2 using Pythagoras' Theorem. Look for an
attempt to
square and add with at least one modulus or modulus2 correct.
dM1: Attempts to use the cosine rule with the lengths in the correct positions in order to find
angle PQR
2
2
2
PQ
+
QR
−
PR
PQR
=
Look for
cos
2
×
PQ QR
×
There are more round about methods including finding other angles first and then using
the sine rule
but this method mark can only be awarded when an attempt is made at angle PQR
A1: Angle PQR = awrt 114°
It must be found using ''correct'' vectors, e.g . (
k and (
)
)
±−
i + j
+
±
i
−
j
+
k
6
2
2
3
4
If, for example RQ
i
−
2 +
j k
, A0 will be awarded even if 114°is
is incorrect, e.g. 6
stated
1
−
1
(
)
1
2
−
2
3(b)[3]
+
d
x
= −
ln 2
x
−
1
+
2ln
x
−
2
+
c
x
−
x
−
2
1
2
2
25
25
3
x
1
(
)
(
)
x
= −
x
−
+
x
−
d
ln 2
1
2ln
2
(
)(
)
x
−
x
−
2
2
1
2
5
5
1
1
= −
+
−−
+
ln 49
2ln 23
ln9
2ln3
2
2
529
ln 21
(a)
M1: Attempts a correct PF form leading to values for A and B .
3
x
A
B
C
+
+
A
B
Condone
Values for at least
and
(
)(
)
2
x
−
1
x
−
2
x
−
x
−
2
1
2
A1: One correct value for A or B or one correct term
−
1
2
A1: Correct PF form
+
.
x
−
x
−
2
1
2
This is not just for correct values for A and B but it may be awarded from work seen in (b)
(b)
A
B
(
)
(
)
M1:
+
d
x
=
C
ln 2
x
−
1
+
D
ln
x
−
2
o.e. Condone missing brackets here
x
−
x
−
2
1
2
A
B
A
(
)
(
) ( )
A1ft: Correct ft on their values for A and B
+
x
=
x
−
+
B
x
−
+
d
ln 2
1
ln
2
2
x
−
1
x
−
2
2
Do not condone missing brackets but the brackets can be implied by subsequent working,
E.g. ln 49
dM1: Attempts to substitute the limits 5 and 25 (no misreads on the limits but condone slips, so
don't accept a consistent lower limit of 0 for instance) into an expression of the correct
form (with brackets which may be implied) and subtracts, either way around. Condone
attempts where the substitutions were made into incorrectly combined ln terms.
529
A1: cso
ln 21 Also allow in mixed number form
ln25
4
21
Attempts (
) (
)
3 (a)
i
−
j
+
k
−
i
−
j
+
k
2
3
4
8
5
3
RQ = −
6
i + j k
2
+
(b)
PQ RQ = ×−+ −×+ ×
Attempts
2
6
3 2
4 1
Full attempt to find cos PQR E.g.
2
×−+ −×+ ×=
6
3 2
4 1
29 41cos PQR
Angle PQR = 114°
(a)
Attempts to subtract vectors PQ
and PR
either way around. Look for PQ
PR
M1:
−
or
PR
−
PQ
.
If a method is not shown it can be implied by two correct components of
±
6
i
±
2
j
±
k Note that an attempt such as PR
−
QP
is M0
−
6
−
6
i
RQ = −
i + j k o.e. such as
A1:
6
2
+
2
Do not accept coordinates or indeed
2
j
1
k
(b)
Attempts scalar product of PQ
and their RQ
M1:
. Look for an attempt at multiplying
together the components and adding. There will be some confusion over direction so allow
for sight of
(
) (
) (
)
±×±
2 '' 6''
+ ±×±
3 '' 2''
+ ±×±
4 '' 1''
This cannot be scored if they attempt a scalar product of PQ
and PR
for instance.
dM1: Full attempt to find cos PQR using .
ab
=
a
b
θ
using vectors PQ
±
cos
and their
±
RQ
.
There must be an attempt at both moduli with at least one correct (which may be
unsimplified)
but you should ft on their RQ
. Don't be concerned whether the angle is acute or obtuse.
A1: Angle PQR = awrt 114° ISW after sight of this, e.g followed by 66°
Alt (b)
M1: Attempts all three lengths or all three lengths 2 using Pythagoras' Theorem. Look for an
attempt to
square and add with at least one modulus or modulus2 correct.
dM1: Attempts to use the cosine rule with the lengths in the correct positions in order to find
angle PQR
2
2
2
PQ
+
QR
−
PR
PQR
=
Look for
cos
2
×
PQ QR
×
There are more round about methods including finding other angles first and then using
the sine rule
but this method mark can only be awarded when an attempt is made at angle PQR
A1: Angle PQR = awrt 114°
It must be found using ''correct'' vectors, e.g . (
k and (
)
)
±−
i + j
+
±
i
−
j
+
k
6
2
2
3
4
If, for example RQ
i
−
2 +
j k
, A0 will be awarded even if 114°is
is incorrect, e.g. 6
stated
1
−
1
(
)
1
2
−
2
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