Mathematics · Oct / Nov 2023 · Paper 41 · Question 6
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As printed in the paper
The question as text
6. The line l1 has equation r where λ is a scalar parameter. 3 2 7 2 2 4 r The line l2 has equation 3 1 where μ is a scalar parameter. 7 8 Given that l1 and l2 meet at the point P
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6(a)[1]
2, 3,
−
7
6(b)[3]
1
4
( )
Attempts
2
−1
= ×+ ×−+ ×=
1 4
2
1 2 8
18
2
8
( )
2
2
2
2
2
2
18
=
1
+
2
+
2
×
4
+ −
1
+
8 cos
θ
Attempts .
a b
=
a b
cosθ
:
2
cos
θ=
3
(b) Alt
4
1
3
1
2
3
e.g.
±
−
−
= ±−
8
2
6
( )
( )
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
3
+
3
+
6
=
1
+
2
+
2
+
4
+ −
1
+
8
−
2 1
+
2
+
2
×
4
+ −
1
+
8 cosθ
2
cos
θ=
3
6(c)[3]
Uses λ = 6 to find length PQ E.g.
( )
2
2
2
PQ
=
6
i
+
12
j
+
12
k
PQ
=
6
+
12
+
12
=
18
( )
2
2
2
Or
PQ = ×
6
1
+
2
+
2
=
18
2
1
1
2
2
sin
18
1
54 5
Area QPR =
ab
C
=
×
×
−
=
2
2
3
6(d)[3]
Attempts a correct method of finding at least one value for μ e.g.
(
)
2
(
)
2
2
2
2
2
4
μ
+
μ
+
8
μ
=
6
+
12
+
12
''18''
( )
μ
=
= ±
2
(
)
2
2
2
4
+ −
1
+
8
2
4
Attempts one correct position
r
=
3
±
2
−
1
−
7
8
Possible coordinates (
10,1, 9 and (
)
)
−
6, 5,
−
23
B1: Correct coordinates or position vector e.g. as shown or x = 2, y = 3, z = −7 or 2i + 3j – 7k or
Condone (
)
2 , 3 ,
i
j
−
7
k
(b)
1
4
M1: Attempts
2
−1
condoning slips. Note that any nonzero multiples of these vectors can be used.
2
8
If the method is not explicit then this mark may be implied by 2 correct components.
dM1: Full attempt at .
a b
=
a b
cosθ
where a and b are the direction vectors or multiples of them.
Depends on the first method mark.
4
−+
2 16
Note that
cos
θ=
would imply both method marks.
2
2
2
2
2
2
1
+
2
+
2
4
+
1
+
8
2
A1:
cos
θ=
3
(b) Alternative using the cosine rule:
4
1
M1: Attempts
±
α
−
1
−
β
2
8
2
dM1: Full attempt at the cosine rule using the appropriate lengths.
Depends on the first method mark.
2
A1:
cos
θ=
3
(c)
M1: Uses λ = 6 to find the length of PQ. E.g. uses Pythagoras to calculate 6
i
+
12
j
+
12
k or 6
If the method is not explicit then the attempt at PQ
may be implied by 2 correct components.
M1: Full attempt at 1
C where a = b = PQ
and C is their θ which may be attempted in decimals.
2 ab
sin
Their θ or sin θ must follow an attempt to use their cos θ.
A1: 54 5 or exact equivalent e.g. 162 5
3
M1: Attempts a correct method of finding at least one value for μ .
Pythagoras must be used correctly and both sides of their equation must be consistent e.g.
(
)
(
)
and (
)
(
)
2
2
2
2
2
2
2
4
μ
−
μ
+
8
μ
=
18
4
μ
+
μ
+
8
μ
=
18
both score M0
Must be correct work here so must be using the length of PQ not OQ.
2
4
r
=
3
±
2
−
1
dM1: Attempts one correct position
Depends on the first method mark.
7
8
−
A1: Gives both possible coordinates (
10, 1, 9 and (
)
)
6, 5,
23
−
−
Allow as coordinates or vectors or as x = …, y = …, z = …
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