| Date | November 2016 | Marks available | 2 | Reference code | 16N.1.hl.TZ0.4 |
| Level | HL only | Paper | 1 | Time zone | TZ0 |
| Command term | Find | Question number | 4 | Adapted from | N/A |
Question
Consider the vectors a == i − 3− 3j − 2− 2k, b =− 3=− 3j + 2+ 2k.
Find a ×× b.
[2]
a.
Hence find the Cartesian equation of the plane containing the vectors a and b, and passing through the point (1, 0, −1)(1, 0, −1).
[3]
b.
Markscheme
a ×× b =−12=−12i − 2− 2j − 3− 3k (M1)A1
[2 marks]
a.
METHOD 1
−12x−2y−3z=d−12x−2y−3z=d M1
−12×1−2×0−3(−1)=d−12×1−2×0−3(−1)=d (M1)
⇒d=−9⇒d=−9 A1
−12x−2y−3z=−9 (or 12x+2y+3z=9)−12x−2y−3z=−9 (or 12x+2y+3z=9)
METHOD 2
(xyz)∙(−12−2−3)=(10−1)∙(−12−2−3)⎛⎜⎝xyz⎞⎟⎠∙⎛⎜⎝−12−2−3⎞⎟⎠=⎛⎜⎝10−1⎞⎟⎠∙⎛⎜⎝−12−2−3⎞⎟⎠ M1A1
−12x−2y−3z=−9 (or 12x+2y+3z=9)−12x−2y−3z=−9 (or 12x+2y+3z=9) A1
[3 marks]
b.
Examiners report
[N/A]
a.
[N/A]
b.
