Mathematics · May / June 2024 · Paper 41 · Question 6
YMA01
As printed in the paper
The question as text
6. With respect to a fixed origin O, the line l1 is given by the equation r = i + 2j + 5k + λ(8i – j + 4k) where λ is a scalar parameter. The point A lies on l1 → = 5 10 Given that OA
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
6(a)[3]
1 8
2
5
4
OA
=
+
λ
+
−
λ
+
+
λ
(
)
(
)
(
)
2
2
2
OA
=
5 10
1 8
+
λ
+
2
−
λ
+
5
+
4
λ
=
250
64
λ
2
+
16
λ
+ +
1
λ
2
−
4
λ
+
4 16
+
λ
2
+
4
0
λ
+
25
=
250
81
λ
2
+
5
2
λ
−
2
20
=
0
*
6(b)[3]
110
(
)(
)
81
λ
2
+
5
2
λ
−
220
=
0
81
λ
−
110
λ
+
2
=
0
λ
=
...
−
2, 81
1 8
"their "
+
×
λ
OA
2
"thei
r "
=
−
λ
5
4
"their "
+
×
λ
110
961
1 8
+
×
81
81
1
8
−
1
5
110
52
E.g
OA
=
2
−
2
−
1
=
4
* ; or
OA
=
2
−
=
81
8
1
5
4
−
3
1
10
84
5
5
+
4
×
8
1
81
6(c)[4]
−
15
8
4 .
−
1
−
3
4
c
os
θ
=
±
=
..
.
1
5
2
+
4
2
+
3
2
8
2
+
1
2
+
4
2
−
136
(
)
or e.g.
θ
=
awrt 0.299(
rad
) /17.1
°
= ±
=
±
0.9557...
45 1
0
1
Area
OAB
=
5 10
×
4 10 sin
θ
=
...
2
= awrt 29.4
(a)
M1: Attempts to find the distance, or distance squared, from O to a general point on the line, or to A.
Condone slips. It is acceptable to use a different variable rather than .
λ
M1: Attempts to use
OA =
5 10
to set up a quadratic equation in λ, need not be expanded for this mark.
Condone slips on the 250 for
OA on this mark. Allow it to appear as 5 10
2
×
=
50
A1*: Expands and reaches the correct equation from correct work.
An intermediate line equivalent to
64
λ
2
+
16
λ
+ +
1
λ
2
−
4
λ
+
4 16
+
λ
2
+
40
λ
+
25
=
250
written out before the final given answer is seen.
(b)
M1: Correct attempt at solving the quadratic equation, (allow a calculator), and substitutes at least one of
their values for λ into the equation for the line to find a possible position vector. The (
given so they cannot just write λ = −2, A = (
)
−
15, 4,
−
3
without some evidence. However, sight of a
9
61
52
8
4
5
correct second coordinate
,
,
would imply M1
81
81
8
1
A1*: Correct value λ = −2 found, and substituted to reach the given position vector.
1 8
+
×−
2
−
15
Scored for sight of
OA
=
2
−−
(
2
)
=
4
o.e.
5
4
2
3
+
×
−
−
Also allow the coordinate (
15, 4,
3
)
but not an incorrect hybrid version. E.g.(
15 , 4 ,
−
−
−
i
9
61
52
8
4
A1: Correct second possible position vector but condone in coordinate form.
,
,
81
81
8
1
exact. Do not allow hybrid versions but if this form has been penalised already, allow here.
(c)
(
)
M1: Attempts the scalar product of
k and the direction vector of the line
±
OA
= ±−
15
i
+
4
j
−
3
(
)
8
1
4
±
i
−
j
+
k to find cosθ for the angle between OA
and l2. Condone slips but the intention must be
clear. If the method of finding modulus is not shown award if one of the two is correct. For example,
4 10 is a common error/slip on the modulus of (
)
±
8
−
1
+
4
i
j
k as this length is given in the question.
with (
)
±
8
μ
−
1
μ
+
4
μ
Note that it is acceptable to find the scalar product of OA
±
i
j
k as long as µ is used
consistently within the calculation on the numerator. Note any variable may be used here.
A1: Correct value for cosθ (either sign acceptable). Implied by use of
θ
=
awrt 0.299
rads or awrt 17.1°
M1: Uses formula for area of triangle in a complete method to find the area required. It is dependent upon
having found cosθ via a scalar product of the two correct vectors.
A1: Correct area, awrt 29.4
Alt: There are alternative methods most of which involve finding the perpendicular distance
between the two lines. An example of such a method is shown below.
Alt
−
127
1
1
(c)
Midpoint of possible A’s is M =
188
and
OM =
127
2
+
188
2
+
301
2
=
...
81
81
301
= 4.6534…
Area = 1
2 ×
4 10
×
4.6534...
;= 29.43…
M1: Finds the midpoint for the two possible positions for A and proceeds to find the distance from O to
this point.
A1: Correct distance between lines.
M1: Uses formula for area of triangle in a complete method to find the area required.
A1: Correct area, awrt 29.4
There may be longer alternatives, such as finding point B first and using a different angle, but the scheme
will apply in like manner.
d
x
1
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