Date | May 2018 | Marks available | 5 | Reference code | 18M.1.AHL.TZ2.H_9 |
Level | Additional Higher Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 2 |
Command term | Show that | Question number | H_9 | Adapted from | N/A |
Question
The points A, B, C and D have position vectors a, b, c and d, relative to the origin O.
It is given that .
The position vectors , , and are given by
a = i + 2j − 3k
b = 3i − j + pk
c = qi + j + 2k
d = −i + rj − 2k
where p , q and r are constants.
The point where the diagonals of ABCD intersect is denoted by M.
The plane cuts the x, y and z axes at X , Y and Z respectively.
Explain why ABCD is a parallelogram.
Using vector algebra, show that .
Show that p = 1, q = 1 and r = 4.
Find the area of the parallelogram ABCD.
Find the vector equation of the straight line passing through M and normal to the plane containing ABCD.
Find the Cartesian equation of .
Find the coordinates of X, Y and Z.
Find YZ.
Markscheme
a pair of opposite sides have equal length and are parallel R1
hence ABCD is a parallelogram AG
[1 mark]
attempt to rewrite the given information in vector form M1
b − a = c − d A1
rearranging d − a = c − b M1
hence AG
Note: Candidates may correctly answer part i) by answering part ii) correctly and then deducing there
are two pairs of parallel sides.
[3 marks]
EITHER
use of (M1)
A1A1
OR
use of (M1)
A1A1
THEN
attempt to compare coefficients of i, j, and k in their equation or statement to that effect M1
clear demonstration that the given values satisfy their equation A1
p = 1, q = 1, r = 4 AG
[5 marks]
attempt at computing (or equivalent) M1
A1
area (M1)
= 15 A1
[4 marks]
valid attempt to find (M1)
A1
the equation is
r = or equivalent M1A1
Note: Award maximum M1A0 if 'r = …' (or equivalent) is not seen.
[4 marks]
attempt to obtain the equation of the plane in the form ax + by + cz = d M1
11x + 10y + 2z = 25 A1A1
Note: A1 for right hand side, A1 for left hand side.
[3 marks]
putting two coordinates equal to zero (M1)
A1
[2 marks]
M1
A1
[4 marks]