Mathematics · May / June 2021 · Paper 41 · Question 7
YMA01
As printed in the paper
The question as text
7. Relative to a fixed origin O, the line l has equation 1 4 r = + λ where λ is a scalar parameter − 10 4 − 9 2 → is a unit vector parallel to l, Given that OA →
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
7(a)[2]
Finds
+
+
=
and attempts
''6''
2
2
23
23
OA
=
oe
13
1
+
4
λ
7(b)[5]
Coordinates or position vector of point X =
−
10
+
4
λ
−+
9
2
λ
4
=
OX
. 4
0
4 1
(
4
)
4
(
10
4
)
2
(
9
2
)
0
⇒
+
λ
+
−
+
λ
+
−+
λ
=
2
36
λ
=
54
⇒
λ
=
1.5
(
)
X =
7, 4, 6
−
−
2
2
2
7(c)[3]
7
( 4)
( 6)
101
Finds
OX =
+ −
+ −
=
and OA = 1
Area OXA = 1
101
× ×
1
101
=
2
2
Handy diagram
(a)
and use of r
M1: Correct attempt at the unit vector. Look for an attempt at
4
2
+
4
2
+
2
2
r where
(
)
or equivalent vector form such as
r
=
4
i
+
4
j
+
2
k . Condone an attempt to find the coordinates of A
or equivalent such as
OA
4
OA =
4
2
A1:
=
or
+
+
23
OA
=
6
6
2
13
not in coordinate form.
(b)
M1: For an attempt at the coordinates or position vector of point X = (
1
+
4 , 10
λ
−
+
4 , 9
λ
−+
2
λ
)
or
4
=
dM1: For using
OX
. 4
0
to set up an equation in λ
2
2
Alternatively finds
OX
(
1
4
)
2
(
10
4
)
2
(
9
2
)
2
and attempts to differentiate and set = 0
=
+
λ
+ −
+
λ
+ −+
λ
2
2
2
May use
OX
+
OA
=
AX
to set up an equation in λ
ddM1: Solves for λ This is dependent upon having scored both previous M's
A1: Correct value for
λ
=
1.5
(
)
A1: Correct coordinates for
X =
7, 4, 6
−
−
. Condone position vector form 7
i
−
4
j
−
6
k o.e.
(c)
M1: Finds all elements required to calculate the area.
In the main scheme this would be the distance OX or OX 2 AND distance OA or OA2 (which is 1)
dM1: Correct method of finding the area of OXA = 1
×
OX
×
1
2
101
A1: Area =
o.e
2
.............................................................................................................................................................................
There are various alternatives for part (c). Amongst others are;
Alt I via vector product.
i
j
k
1
1
8
19
22
1
8
2
19
2
22
2
101
×
7
−
4
−
6
=
×
i
−
j
+
i
=
×
+ −
+
=
2
2
3
3
3
2
3
3
3
2
2
2
1
3
3
3
i
j
k
M1: For an attempt at 1
1
8
19
22
i
×
7
−
4
−
6
=
×
i
−
j
+
2
2
3
3
3
2
2
1
3
3
3
dM1: Followed by an attempt at finding the modulus of the resulting vector multiplied by 1/2
Alt II via scalar products
M1: Attempts to find all three components required to find the area triangle OXA
E.g. Angle OXA with length of side OX and XA
Alternatively angle OAX with length of side OA and XA
For this to be scored
appropriate gradient vectors need to be attempted by subtractinga correct attempt at using .a b
a b
cos
θ to find cosθ or θ
=
1
(
)
dM1: Full attempt at area of triangle using
2 OX XA
sin
OXA or equivalent
∫
1
∫
−
−
2
x
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