Mark scheme Core Pure (AS/Year 1) Unit Test 7: Vectors
Pearson
Progression Step
Q Scheme Marks AOs
and Progress
Descriptor
1 States or implies x = 3 − y = −5 + 2 and z = 3 − 3 M1 1.1b 5th
Solve coordinate
Writes M1 2.2a geometry
( )
2
(3 − − (−2))2 + (−5 + 2 − (−2))2 + (3 − 3 − 2)2 = 7 3 problems
involving lines in
three dimensions
Simplfies to obtain 14 2 − 28 + 35 = 147 M1 1.1b
Reduces and factorises: 2 − 2 − 8 = 0 ( − 4)( + 2) = 0 M1 1.1b
Finds = −2 and states A (5, − 9, 9) A1 1.1b
Finds = 4 and states B (−1, 3, − 9) A1 1.1b
(6 marks)
Notes
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Mark scheme Core Pure (AS/Year 1) Unit Test 7: Vectors
Pearson
Progression Step
Q Scheme Marks AOs
and Progress
Descriptor
2 Demonstrates an understanding of perpendicular vectors M1 3.1a 5th
For example, Solve coordinate
geometry
x x problems
states a y = 0 or b y = 0 , or states scalar product is zero involving lines in
z z three dimensions
Writes two vector equations, 3x + y − z = 0 and M1 1.1b
− x + 5 y + 3z = 0
Makes an attempt to solve these equtions by setting either x, y or M1 2.2a
z equal to a constant, most likely 1
Solves to find x, y or z and states the solution: (1, − 1, 2) or a A1 1.1b
multiple thereof
(4 marks)
Notes
Some may use cross or vector product to do this question.
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Mark scheme Core Pure (AS/Year 1) Unit Test 7: Vectors
Pearson
Progression Step
Q Scheme Marks AOs
and Progress
Descriptor
3 Finds any two vectors M1 3.1a 5th
−4 5 Use the scalar
product to solve
For example, AB = 11 and AC = 4
−4 −11 coordinate
geometry
problems in three
Finds the scalar product of these two vectors, M1 1.1b dimensions
−4 5
AB AC = 11 4 = −20 + 44 + 44 = 68
−4 −11
Finds the magnitude of each vector, M1 1.1b
AB = (−4)2 + (11)2 + (−4)2 = 153 and
AC = (5)2 + (4)2 + (−11)2 = 162
AB AC 68
Uses cos (BAC ) = = to find A1 2.2a
AB AC 153 162
cos (BAC ) = 0.4319 and therefore BAC = 64.410
1
States area = AB AC sin (BAC ) M1 2.2a
2
1
or writes, area = 153 162 sin (64.41...)
2
States correct answer = 71.0, accept awrt 71.0 A1 1.1b
(6 marks)
Notes
Some may use vector product
1
The triangle area is AB AC
2
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Mark scheme Core Pure (AS/Year 1) Unit Test 7: Vectors
Pearson
Progression Step
Q Scheme Marks AOs
and Progress
Descriptor
4a Demonstrates an understanding of perpendicular vectors by M1 3.1a 5th
writing,
Solve coordinate
3 2p geometry
problems
−4 1 =0 involving lines in
−2 p
three dimensions
Solves 6 p − 4 − 2 p = 0 writing p = 1 A1 1.1b
(2)
4b Writes any two of the following by equating x, y and z M1 3.1a 5th
compnents,
Solve coordinate
1 + 3 = −16 + 2 , −3 − 4 = 5 + or 1 − 2 = 3 + geometry
problems
Solves any pair of simultaneous equations to find, M1 1.1b involving lines in
three dimensions
= −3 and = 4
Checks that = −3 and = 4 satisfies the third equation M1 2.3
States the correct point of intersection, (−8, 9, 7) A1 1.1b
(4)
(6 marks)
Notes
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Mark scheme Core Pure (AS/Year 1) Unit Test 7: Vectors
Pearson
Progression Step
Q Scheme Marks AOs
and Progress
Descriptor
5 6 + 4 M1 1.1b 6th
States that a general point on l1 has position vector 2 + 5 Find the point of
−2 − intersection of a
line and a plane
6 + 4 M1 3.1a
Makes an attempt to substitute 2 + 5 into 2 x − y + 4 z = 4
−2 −
For example, 2(6 + 4 ) − (2 + 5 ) + 4(−2 − ) = 4 is seen
Solves the equation to find = −2 and concludes that the point M1 1.1b
of intersection is (−2, −8,0)
States that a vector equation of the line through (6, 2, −2) and M1 3.1a
6 2
perpendicular to is r = 2 + −1
−2 4
Attempts to finds a point on this line that is also on by M1 3.1a
substituting,
x = 6 + 2 , y = 2 − and z = −2 + 4 into 2 x − y + 4 z = 4
2 A1 1.1b
Solves 2(6 + 2 ) − (2 − ) + 4(−2 + 4 ) = 4 to obtain =
21
Concludes that the point, M1 3.2a
6 2
2
2 + 21 −1
−2 4
is halfway between (6, 2, −2) and a point on l2 ; therefore, a
point on l2 has position vector,
2 21
134
6
4
2 +
21 − 1 = 21
24
−2 4 12
− 21
(continued)
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Mark scheme Core Pure (AS/Year 1) Unit Test 7: Vectors
Attempts to find the equation of l2 using the points M1 1.1b
−2 134
2421
−
8 and 21
0 12
− 21
−2 44 A1 1.1b
Finds a correct vector equation of l2 : r = 8 + 48
0 −3
134 −2 44
2421
Accept 21 instead of −8 and any multiple of 48
12 0 −3
− 21
(9 marks)
Notes
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Mark scheme Core Pure (AS/Year 1) Unit Test 7: Vectors
Pearson
Progression Step
Q Scheme Marks AOs
and Progress
Descriptor
6a Finds two vectors in the plane, for example PQ = (−5,1, −5) and M1 3.1a 5th
PR = (3,3, −6) Use the scalar
product to find
Uses the dot product to find two vector equations, M1 1.1b normal vectors
−5 x + y − 5 z = 0 and 3x + 3 y − 6 z = 0
Makes an attempt to solve these equations by setting either x, y M1 2.2a
or z equal to a constant, most likely 1
Solves to find x, y or z and states the solution for a vector normal A1 2.2a
to the plane,
1 5
− i + j + k or (−i + 5j + 2k ) or a multiple thereof
2 2
Finds a unit vector normal to the plane using Pythagoras’ A1 1.1b
Theorem in three dimensions,
1
(−i + 5 j + 2k )
30
(5)
6b Uses the fact that a general vector on the plane M1 3.1a 5th
( x − 2, y − 1, z − 5) will be perpendicular to the normal vector
Find the Cartesian
x − 2 −1 equation of a
plane in three
(−1, 5,1) by writing y − 1 5 =0 dimensions
z −5 1
Solves to find: − x + 5 y + z = 8 A1 1.1b
(2)
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Mark scheme Core Pure (AS/Year 1) Unit Test 7: Vectors
6c States n1 = (−1, 5, 2) and n2 = (3, 2, − 8) and finds the scalar M1 1.1b 6th
product of these two vectors, Find angles
between lines and
−1 3
planes in three
n1 n2 = 5 2 = −3 + 10 − 16 = −9 dimensions
2 −8
Finds the magnitude of each vector, M1 1.1b
n1 = (−1) 2 + (5) 2 + (2) 2 = 30 and
n2 = (3) 2 + (2)2 + (−8)2 = 77
Uses M1 2.2a
n1 n2 −9
cos = = to find
n1 n1 30 77
cos = −0.18725 and finds = 100.79.27
States the acute angle between the planes is 79.2 A1 1.1b
(4)
(11 marks)
Notes
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Mark scheme Core Pure (AS/Year 1) Unit Test 7: Vectors
Pearson
Progression Step
Q Scheme Marks AOs
and Progress
Descriptor
7a Finds PQ = (6, − 13,14) M1 3.1a 7th
Find the shortest
6 10 M1 1.1b distance between
a point and a line
States that a general point, S, on PQ is S = −13 + 4
14 −2
6 + 10 10 M1 1.1b
At the shortest (perpendicular) distance −13 + 4 4 =0
14 − 2 −2
1 A1 1.1b
Solves 60 + 100 − 52 + 16 − 28 + 4 = 0 to obtain =
6
23 M1 3.1a
337
Substitutes to find the coordinates of S: S − 3 and
41
3
2 2 2
23 37 41
attempts to find the distance OS = + − +
3 3 3
1193 A1 1.1b
Finds minimum distance = 19.9 km
3
Concludes that as this is greater than 15 km, so submarine A can A1 ft 3.2a
move undetected.
(7)
7b Possible answers, B1 3.5b 7th
Sub A will not move in an exact straight line Find the shortest
distance between
Sub A might purposely deviate to avoid detection
a point and a line
Sub A might not be able to move in a straight line due to rocks
Sub B will not be stationary
(1)
(8 marks)
Notes
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