Mathematics · Oct / Nov 2021 · Paper 41 · Question 7
YMA01
As printed in the paper
The question as text
7. With respect to a fixed origin O, 4 − 4 • the line l has equation r = + λ where λ is a scalar constant 2 − 3 − 3 5 • the point A has position vector 9i − 3j + 2k Given that X is the point on l nearest to A,
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
7(a)[7]
Coordinates or position vector of a point on l =
2
−
3
λ
−+
3
5
λ
4
4
9
5
4
−
λ
−−
λ
2
3
3
5
3
AX
=
−
λ
−−
=
−
λ
−+
3
5
λ
2
−+
5
5
λ
−
4
2
Uses
AX
.
−
3
=
0
⇒
20 16
+
λ −
15
+ λ −
9
25
+
25
λ =
0
⇒λ =
5
5
4
−
4
λ
12 4
(i) Substitutes their λ into
2
−
3
λ
⇒
X
=
,
, 1
−
5
5
−+
3
5
λ
33
19
AX = −
i
+
j
−
3
k
(ii)
5
5
2
2
33
19
(
3
)
2
67
Shortest distance =
−
+
+ −
=
5
5
7(b)[2]
Uses AX
=
XB
or similar correct method
21
23
point B has position vector
−
i
+
j
−
4
k
5
5
4
−
4
λ
7 (a)
ALT way Coordinates or position vector of a point on l =
2
−
3
λ
−+
3
5
λ
4
−
4
λ
9
−−
5
4
λ
AX
=
2
−
3
λ
−−
3
=
5
−
3
λ
−+
3
5
λ
2
−+
5
5
λ
2
(
5
4
)
2
(
5
3
)
2
(
5
5
)
2
Uses
d = −−
λ
+
−
λ
+ −+
λ
which is smallest when its
2
(
)
(
)
(
)
⇒−
8
−−
5
4
λ
−
6 5
−
3
λ
+
10
−+
5
5
λ
=
0
⇒
λ
=
gradient =0
5
(Note that this could be done by completing the square)
4
−
4
λ
12 4
(i) Substitutes their λ into
2
−
3
λ
⇒
X
=
,
, 1
−
5
5
−+
3
5
λ
2
2
2
2
2
2
(ii) Shortest distance =
−−×
5
4
+
5
−×
3
+ −+ ×
5
5
⇒
d
=
67
5
5
5
(a)
4
−
4
λ
4
−
4
λ
9
M1: States or uses a general point on l =
2
−
3
λ
and attempt
AX
=
2
−
3
λ
−−
−+
3
5
λ
−+
3
5
λ
2
either way around with their general point. Condone slips
dM1:
Uses a correct method to find a value for λ
This could be
−
4
via scalar products
AX
.
−
3
=
0
5
via differentiation using minimum distance of
(
)
2
(
)
2
(
)
2
−−
5
4
λ
+
5
−
3
λ
+ −+
5
5
λ
in this method condone slips on the
differentiation
2
λ =
A1:
5
(i)
dM1:
Uses their λ to find the coordinates (or position vector) of X This is dependent upon the previous M
12 4
A1:
Finds coordinates for
X
=
,
, 1
−
Condone use of position vector
5
5
(ii)
M1: Uses a correct method to find distance AX or distance AX 2 using A and their X .
Award if a correct method is seen for two of the three coordinates
Usually look for an attempt at (
)
2
(
)
2
(
)
2
−−
5
4
λ
+
5
−
3
λ
+ −+
5
5
λ
with their λ .
A1:
67 Note that d =
67 so d = 67 is quite common. This is fine as d is the constant given in the question
(b)
M1: Uses a correct method to find B.
The method can be implied by two correct coordinates for their coordinates for X and A.
21
23
A1: Correct position vector for B.
−
i
+
j
−
4
k . Condone use of coordinates
5
5
Note that it is possible (though
unlikely) to attempt this questions
via scalar product using CA and CX.
CACX
.
cos
FYI
CX
=
CA
θ =
CX
2 2
2
λ =
=
Hence
50
5
∫
∫
1
1
1
x
2
ln
x x
d
=
x
3
ln
x
−
x
3
×
d
x
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