| Date | May 2017 | Marks available | 4 | Reference code | 17M.1.hl.TZ1.5 |
| Level | HL only | Paper | 1 | Time zone | TZ1 |
| Command term | Determine | Question number | 5 | Adapted from | N/A |
Question
ABCD is a parallelogram, where →AB−−→AB = –i + 2j + 3k and →AD−−→AD = 4i – j – 2k.
Find the area of the parallelogram ABCD.
[3]
a.
By using a suitable scalar product of two vectors, determine whether AˆBCA^BC is acute or obtuse.
[4]
b.
Markscheme
→AB×→AD=−−−→AB×−−→AD=−i +10+10j – 7k M1A1
area=|→AB×→AD| = √12+102+72area=∣∣∣−−→AB×−−→AD∣∣∣ = √12+102+72
=5√6(√150)=5√6(√150) A1
[3 marks]
a.
METHOD 1
→AB∙→AD=−4−2−6−−→AB∙−−→AD=−4−2−6 M1A1
=−12=−12
considering the sign of the answer
→AB∙→AD<0−−→AB∙−−→AD<0, therefore angle DˆABD^AB is obtuse M1
(as it is a parallelogram), AˆBCA^BC is acute A1
[4 marks]
METHOD 2
→BA∙→BC=+4+2+6−−→BA∙−−→BC=+4+2+6 M1A1
=12=12 considering the sign of the answer M1
→BA∙→BC>0⇒AˆBC−−→BA∙−−→BC>0⇒A^BC is acute A1
[4 marks]
b.
Examiners report
[N/A]
a.
[N/A]
b.
