Mathematics · Oct / Nov 2020 · Paper 41 · Question 8
YMA01
As printed in the paper
The question as text
8. Relative to a fixed origin O, the lines l1 and l2 are given by the equations 4 3 l1: r = + λ where λ is a scalar parameter 3 2 2 1 2 2 l2: r = + μ where μ is a scalar parameter 0 1 9 3 Given that l1 and l2 meet at the point X,
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
8(a)[5]
−
3
+
λ
−
2
=
0
+
µ
−
1
⇒−−
3
2
λ
=
0 1
−
µ
any two of these
2
−
1
−
9
−
3
2 1
−
λ
= −−
9
3
µ
Full method to find either λ or µ eg (1) +3(3)
⇒
µ
=
...
Either
λ = − or
4
µ = −
5
4
3
2
2
Position vector of intersection is
−
3
−
4
−
2
=
OR
0
−
5
−
1
=
2
−
1
−
9
−
3
−
8
=
5
6
2
+
2
µ
8(b)[5]
Coordinates or position vector of point Q =
0 1
−
µ
−−
9
3
µ
2
+
2
µ
10
2
µ
−
8
PQ
=
0 1
−
µ
−−
7
=
7 1
−
µ
−−
9
3
µ
0
−−
9
3
µ
2
2
Uses
PQ
.
−
1
=
0
⇒
4
µ
−
16
−
7
+
µ
+
27
+
9
µ
=
0
⇒
µ
= −
7
3
−
2
2
+
µ
10 2
57
Substitutes their µ into
0 1
Q
,
,
−
µ
⇒
=
−
7
7
7
−−
9
3
µ
M1: For writing down any two equations that give the coordinates of the point of intersection.
Accept two of 4
+
3
λ
= +
2
2
µ
, 3
−−
2
λ
=
0 1
−
µ
, 2 1
−
λ
= −−
9
3
µ
There must be an attempt to set the coordinates equal but condone one slip.
Setting the vectors equal to each other is insufficient for this but correct calculations will imply this.
M1: A full method to find either λ or µ . There are a few answers with limited or no working.
In such cases this M mark can be implied only if either λ or µ are correct for their equations.
A1: Either value correct
µ = −or
5
λ = −. Correct value(s) following correct equations implies M1 A1
4
dM1: Substitutes their value of λ into l1 to find the coordinates or position vector of the point of intersection.
It is dependent upon having scored second method mark. Alternatively substitutes their value of µ into
l2 to find the coordinates or position vector of the point of intersection. It can be implied in cases
where there is limited working by two correct coordinates for their λ or µ .
−
8
A1: Either
5
or 8
i
5
j
6
k but not the coordinate (
8, 5, 6
)
−
+
+
−
6
However ISW after sight of the correct vector**
(b)
2
2
2
+
2
µ
10
+
µ
M1: States or uses a general point on 2l =
0 1
and attempt
PQ
=
0 1
−
µ
−−
7
=
−
µ
9
3
−−
9
3
µ
0
−−
µ
way around with their general point. Condone slips
2
dM1: Uses the scalar product with fact that PQ
and
−
1
are perpendicular vectors to find µ
−
3
See scheme but again condone slips.
Alternatively uses the scalar product with fact that PQ
and XQ
are perpendicular vectors to find µ
(
)(
) (
)(
) (
)(
)
PQ XQ
.
=
0
⇒
2
µ
−
8
2
µ
+
10
+
7
−
µ
−
µ
−
5
+ −−
9
3
µ
−
15
−
3
µ
=
0
⇒
µ
= −
2
A1:
µ = −
7
ddM1: Uses their µ to find the coordinates of Q.
If no method is seen imply by two correct coordinates for their µ
10 2
57
but not (**) 10
2
57
,
,
i +
j
k o.e
A1:
Q
=
−
−
7
7
7
7
7
7
** Only penalise incorrect notation once, the first time that it occurs. **
2
+
2
µ
(b)
Coordinates or position vector of point Q =
0 1
−
µ
−−
9
3
µ
2
+
2
µ
10
2
µ
−
8
PQ
=
0 1
−
µ
−−
7
=
7 1
−
µ
−−
9
3
µ
0
−−
9
3
µ
Uses Pythagoras with
2
2
2
2
2
PQ
+
QX
=
PX
⇒
28
µ
+
148
µ
+
544
=
504
⇒
µ
= −
5
,
−
7
2
+
2
µ
10 2
57
Substitutes their µ into
0 1
−
µ
⇒
Q
=
,
,
−
7
7
7
−−
9
3
µ
Alt using minimum distance and differentiation
2
+
2
µ
(b)
Coordinates or position vector of point Q =
0 1
−
µ
−−
9
3
µ
2
+
2
µ
10
2
µ
−
8
PQ
=
0 1
−
µ
−−
7
=
7 1
−
µ
−−
9
3
µ
0
−−
9
3
µ
Uses Pythagoras with
2
(
)
2
(
)
2
(
)
2
PQ
=
2
µ
−
8
+
7
−
µ
+ −−
9
3
µ
d
2
2
(
)
(
)
(
)
PQ
=
0
⇒
4 2
µ
−
8
−
2 7
−
µ
−
6
−−
9
3
µ
=
0
⇒
µ
= −
d
µ
7
2
+
2
µ
10 2
57
Substitutes their µ into
0 1
−
µ
⇒
Q
=
,
,
−
7
7
7
−−
9
3
µ
3
d
A
A
∫
d
A
∫
d
t
2
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