Question 15:
--- 15(a) ---
15(a) | Obtains two vectors in
the plane | 1.1a | B1 | 0 −4
𝐴𝐴� ���𝐵𝐵�⃗ = �−6� �𝐵𝐵���𝐶𝐶�⃗ = �7.2�
−8 9.6
Normal vector or
𝐢𝐢 0 −5 0
𝐧𝐧 = �𝐣𝐣 3 9 � = �−20�
𝐤𝐤 4 12 15
0
� −4�
Eq3uation of plane :
Π
−4𝑦𝑦+3𝑧𝑧 = 16
Selects a Πmethod to find
a vector normal to the
plane either by taking
the vector product of
their twΠo vectors in or
by taking the scalar
product of a generaΠl
vector with their two
vectors in | 3.1a | M1
Obtains a Πcorrect
Cartesian equation of
the plane | 1.1b | A1
--- 15(b)(i) ---
15(b)(i) | Selects a Πmethod to
show that lies in
either by 𝐿𝐿1 Π
substituting a general
point on into their
equation of ,
𝐿𝐿1
or by
Π
substituting a point on
in their equation of
and using a scalar
𝐿𝐿p 1 roduct or a vector Π
product to show that the
direction vector of is
perpendicular to their
𝐿𝐿1
normal to ` | 3.1a | M1 | For any point on ,
𝐿𝐿1
and
𝑦𝑦 = −0.4+3𝜇𝜇
𝑧𝑧 = 4.8+4𝜇𝜇
∴ −4𝑦𝑦+3𝑧𝑧 = −4(−0.4+3𝜇𝜇 )+3(4.8+ 4𝜇𝜇 )
So the =po1in.6t −lie1s2 i𝜇𝜇n +the1 4p.4la+ne1 2 𝜇𝜇an=d1 6
therefore lies in the plane .
Π
𝐿𝐿1 Π
CompletesΠ a rigorous
argument to show that
lies in | 2.1 | R1
--- 15(b)(ii) ---
15(b)(ii) | Selects a method to
show that every point
on is equidistant
from and by finding
the𝐿𝐿 g 1 eneral vector from
a poin𝐵𝐵t on 𝐶𝐶 to or
or by finding the
midpoint (5𝐿𝐿, 1 -0.4𝐵𝐵, 4.8)𝐶𝐶 of | 3.1a | M1 | Midpoint of is (5, -0.4, 4.8), which lies on
Consider the direction vectors of and :
𝐵𝐵𝐶𝐶 𝐿𝐿1
𝐿𝐿1 𝐵𝐵𝐶𝐶
15 −5
� 3 �⋅� 9 � = −75+27+48 = 0
4 12
is perpendicular to
Si n c1e the midpoint of a lso lies on then
∴ 𝐿𝐿 𝐵𝐵 𝐶𝐶
is the perpendicular bisector of and
𝐵𝐵𝐶𝐶 𝐿𝐿1
hence every point on is equidistant from
a𝐿𝐿1 𝐵𝐵𝐶𝐶
n d
𝐿𝐿1 𝐵𝐵
𝐶𝐶
𝐵𝐵𝐶𝐶
Finds the vector or
forms an expression for
ge𝐵𝐵�� ��𝐶𝐶�⃗
the length of a n eral
vector from a point on
to or | 1.1a | M1
𝐿𝐿T 1 akes𝐵𝐵 the 𝐶𝐶scalar
product of and the
direction vector of or
forms an e𝐵𝐵q ���� u𝐶𝐶�⃗ ation for
the distances from𝐿𝐿 a 1
point on to and
being equal.
1 𝐵𝐵 𝐶𝐶 | 1.1a | M1
𝐿𝐿
Completes a reasoned
argument to show that
every point on is
equidistant from and
𝐿𝐿1
𝐵𝐵 | 2.1 | R1
--- 15(c) ---
15(c) | Deduces that the
direction vector of is
perpendicular to AB or to
𝐿𝐿2
their normal to | 2.2a | M1 | is the perpendicular bisector of AB in
the plane .
𝐿𝐿 2
Midpoint oΠf AB is M (7, -1, 4)
AB
Direction vector for is perpendicular
to both AB and .
𝐬𝐬 𝐿𝐿2
𝐧𝐧
so let
𝐢𝐢 0 0 25 1
�𝐣𝐣 3 −4� = � 0 � 𝐬𝐬 = �0�
𝐤𝐤 4 3 0 0
:
7 1
𝐿𝐿2 𝐫𝐫 = �−1�+𝜆𝜆�0�
4 0
Deduces that Π passes
through the midpoint of
𝐿𝐿2 | 2.2a | M1
S𝐴𝐴𝐵𝐵elects a method to find
the direction vector of
by taking the vector
𝐿𝐿2
product of and their
normal to
�𝐴𝐴���𝐵𝐵�⃗ | 3.1a | M1
CompletesΠ a reasoned
mathematical argument
to obtain a correct
equation of | 2.1 | R1
--- 15(d) ---
15(d) | Deduces tha𝐿𝐿t 2 is at the
intersection of and
𝐷𝐷 | 2.2a | B1 | is the point of intersection of and
.
𝐷𝐷 𝐿𝐿1 𝐿𝐿2
5+15𝜇𝜇 = 𝜆𝜆+ 7
0.4+3𝜇𝜇 = −1
4.8+4𝜇𝜇 = 4
𝜇𝜇 = −0.2
2
𝐫𝐫 = � −1�
D 4
Equates equati𝐿𝐿o 1 ns of 𝐿𝐿2
and their
𝐿𝐿1 | 1.1a | M1
Obtains th𝐿𝐿e 2 correct
coordinates of
FT their equation of
𝐷𝐷
Condone position ve c2tor
𝐿𝐿
of
𝐷𝐷 | 1.1b | A1
Total | 16 | (2,−1,4)