Date | May 2018 | Marks available | 2 | Reference code | 18M.2.sl.TZ2.8 |
Level | SL only | Paper | 2 | Time zone | TZ2 |
Command term | Find | Question number | 8 | Adapted from | N/A |
Question
Two points P and Q have coordinates (3, 2, 5) and (7, 4, 9) respectively.
Let →PR = 6i − j + 3k.
Find →PQ.
Find |→PQ|.
Find the angle between PQ and PR.
Find the area of triangle PQR.
Hence or otherwise find the shortest distance from R to the line through P and Q.
Markscheme
valid approach (M1)
eg (7, 4, 9) − (3, 2, 5) A − B
→PQ= 4i + 2j + 4k (=(424)) A1 N2
[2 marks]
correct substitution into magnitude formula (A1)
eg √42+22+42
|→PQ|=6 A1 N2
[2 marks]
finding scalar product and magnitudes (A1)(A1)
scalar product = (4 × 6) + (2 × (−1) + (4 × 3) (= 34)
magnitude of PR = √36+1+9=(6.782)
correct substitution of their values to find cos Q∧PR M1
eg cos Q∧PR = 24−2+12(6)×(√46),0.8355
0.581746
Q∧PR = 0.582 radians or Q∧PR = 33.3° A1 N3
[4 marks]
correct substitution (A1)
eg 12×|→PQ|×|→PR|×sinP,12×6×√46×sin0.582
area is 11.2 (sq. units) A1 N2
[2 marks]
recognizing shortest distance is perpendicular distance from R to line through P and Q (M1)
eg sketch, height of triangle with base [PQ],12×6×h,sin33.3∘=h√46
correct working (A1)
eg 12×6×d=11.2,|→PR|×sinP,√46×sin33.3∘
3.72677
distance = 3.73 (units) A1 N2
[3 marks]