Mathematics · Oct / Nov 2024 · Paper 41 · Question 8
YMA01
As printed in the paper
The question as text
8. Relative to a fixed origin O • the point A has coordinates (–10, 5, –4) • the point B has coordinates (–6, 4, –1) The straight line l1 passes through A and B.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
8(a)[2]
(
(
)
)
±−
6
i
+
4
j
−
k
−−
10
i
+
5
j
−
4
k
(
)
r
= −
10
i
+
5
j
−
4
k
±
λ
4
i
−+
j
3
k
e.g.
(
)
r
= −
6
i
+
4
j k
−
±
λ
4
i
−+
j
3
k
e.g.
8(b)[3]
3
+
3
μ
= −
6
μ
= −
3
p
+
12
=
4
p
=
...
q
−
3
= −
1
q
=
...
or
p
= −
8,
q
=
2
(
) (
)
8(c)[3]
4
i
j
3
k
3
i
4
j k
12
4
3
−+
−
+
=
+ +
(
) (
)
4
i
−+
j
3
k
3
i
−
4
j
+
k
=
26 26 cos
θ
=
19
cos
θ
=
...
) 19
(
cos
θ=
26
8(d)[2]
2
19
AC
=
AB
sin
θ
=
26 1
−
26
3 910
o.e.
26
(a)
M1: Attempts the direction of 1l . The expression is sufficient condoning one sign slip or may be
implied by 2 correct components of (
)
±
4
i
−+
j
3
k
A1: Any correct equation including ‘r =’. Allow any parameter to be used including μ.
−
10
4
−
6
−
(
)
r
=
5
±
λ
−
1
r
=
4
±
λ
Allow
r
= −
10
i
+
5
j
−
4
k
±
λ
4
i
−+
j
3
k ,
,
−
4
3
−
1
−
Do not accept i, j, k notation in either bracket using column vector notation
10
4
i
−
r
=
5
±
λ
−
1
j
e.g.
−
4
3
k
(b)
M1: Equates the x component of 2l to 6
− and proceeds to find a value for μ.
dM1: Uses the value of μ to find p or q. It is dependent on the previous method mark.
Alternatively,
forms a correct equation for the x component using their 1l and 2lproceeds to find λuses this value of λ with the value of μ correctly to find p or q by forming anequation for the y or z component.
Do not be concerned by the mechanics of the rearrangement in solving.
A1: Correct values
Alt (b)
M1: Forms three correct equations 3
+
3
μ
= −
6,
p
−
4
μ
=
4,
q
+
μ
= −
1
dM1: Solves simultaneously to find a value for p or q. Do not be concerned by the mechanics of the
rearrangement.
A1: Correct values
(c)
M1: Attempts the scalar product using their direction vectors or may find another direction vector
which should be a multiple of the one for 1l or 2l (condoning slips e.g. if they restart using
another point on 2l to find a direction vector). Look for at least two correct products using
their 1l . May be implied by “±19”. May find the obtuse angle which can still score this mark.
M1: Completes the scalar product method using their direction vectors to find cos θ (acute or
obtuse). Attempts the magnitude of both of their direction vectors and uses these values in the
correct positions in the equation for cos θ
19
A1:
26 or exact equivalent. Do not accept fractions where the numerator or denominator is not
an integer (square roots must be evaluated). Isw if they proceed to find θ
Alt (c) There will be many alternatives such as the one below. Send to review if unsure.
M1: The first mark should be for finding the required components to use in the cosine rule e.g.
attempts to find another point P on 2l and attempts to find the lengths of AB, BP and PA
M1: The second mark should be using all the components required in the cosine rule. E.g.uses
their lengths for AB, BP and PA in the correct positions in the cosine rule.
A1: As above in the main scheme notes.
(d)
M1: Fully correct method to find the exact value for the length of AC.
May alternatively attempt e.g.
47
26
3
13
+
3
μ
2
2
2
97
25
47
25
59
−
4
−13
−
4
μ
=
0
μ
= −
AC
=
AC
=
+
+
26
13
26
13
26
1
6
+
μ
59
26
213 90
45
Note the coordinates for C are
−
,
,
−
and they may use their coordinates for C
26
13
26
with point A to find the distance AC. Condone one sign slip in the distance formula. i.e. there
must be sufficient evidence that they were attempting the differences between appropriate
coordinates.
Look out for other fully correct methods to find the exact value for the length of AC.
e.g. attempts to find BC and uses either trigonometry or Pythagoras’ Theorem.
315
35
3 26
A1: Correct value. Allow
26 or
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