Date | May 2017 | Marks available | 2 | Reference code | 17M.2.AHL.TZ2.H_9 |
Level | Additional Higher Level | Paper | Paper 2 | Time zone | Time zone 2 |
Command term | Show that | Question number | H_9 | Adapted from | N/A |
Question
The points A, B and C have the following position vectors with respect to an origin O.
i + j – 2k
i – j + 2k
i + 3j + 3k
The plane Π contains the points O, A and B and the plane Π contains the points O, A and C.
Find the vector equation of the line (BC).
Determine whether or not the lines (OA) and (BC) intersect.
Find the Cartesian equation of the plane Π, which passes through C and is perpendicular to .
Show that the line (BC) lies in the plane Π.
Verify that 2j + k is perpendicular to the plane Π.
Find a vector perpendicular to the plane Π.
Find the acute angle between the planes Π and Π.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
= (i + 3j + 3k) (2i j + 2k) = i + 4j + k (A1)
r = (2i j + 2k) + (i + 4j + k)
(or r = (i + 3j + 3k) + (i + 4j + k) (M1)A1
Note: Do not award A1 unless r = or equivalent correct notation seen.
[3 marks]
attempt to write in parametric form using two different parameters AND equate M1
A1
attempt to solve first pair of simultaneous equations for two parameters M1
solving first two equations gives (A1)
substitution of these two values in third equation (M1)
since the values do not fit, the lines do not intersect R1
Note: Candidates may note that adding the first and third equations immediately leads to a contradiction and hence they can immediately deduce that the lines do not intersect.
[6 marks]
METHOD 1
plane is of the form r (2i + j 2k) = d (A1)
d = (i + 3j + 3k) (2i + j 2k) = 1 (M1)
hence Cartesian form of plane is A1
METHOD 2
plane is of the form (A1)
substituting (to find gives ) (M1)
hence Cartesian form of plane is A1
[3 marks]
METHOD 1
attempt scalar product of direction vector BC with normal to plane M1
(i + 4j + k) (2i + j 2k)
A1
hence BC lies in Π AG
METHOD 2
substitute eqn of line into plane M1
A1
hence BC lies in Π AG
Note: Candidates may also just substitute into the plane since they are told C lies on .
Note: Do not award A1FT.
[2 marks]
METHOD 1
applying scalar product to and M1
(2j + k) (2i + j 2k) = 0 A1
(2j + k) (2i j + 2k) =0 A1
METHOD 2
attempt to find cross product of and M1
plane Π has normal = 8j 4k A1
since 8j 4k = 4(2j + k), 2j + k is perpendicular to the plane Π R1
[3 marks]
plane Π has normal = 9i 8j + 5k A1
[1 mark]
attempt to use dot product of normal vectors (M1)
(M1)
(A1)
Note: Accept . acute angle between planes A1
[4 marks]