Date | November 2018 | Marks available | 2 | Reference code | 18N.2.SL.TZ0.S_8 |
Level | Standard Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Find | Question number | S_8 | Adapted from | N/A |
Question
Consider the points A(−3, 4, 2) and B(8, −1, 5).
A line L has vector equation . The point C (5, , 1) lies on line L.
Find .
Find .
Find the value of .
Show that .
Find the angle between and .
Find the area of triangle ABC.
Markscheme
Note: There may be slight differences in answers, depending on which combination of unrounded values and previous correct 3 sf values the candidates carry through in subsequent parts. Accept answers that are consistent with their working.
valid approach (M1)
eg B − A, AO + OB,
A1 N2
[2 marks]
Note: There may be slight differences in answers, depending on which combination of unrounded values and previous correct 3 sf values the candidates carry through in subsequent parts. Accept answers that are consistent with their working.
correct substitution into formula (A1)
eg
12.4498
(exact), 12.4 A1 N2
[2 marks]
Note: There may be slight differences in answers, depending on which combination of unrounded values and previous correct 3 sf values the candidates carry through in subsequent parts. Accept answers that are consistent with their working.
valid approach to find (M1)
eg , ,
(seen anywhere) (A1)
attempt to substitute their parameter into the vector equation (M1)
eg ,
A1 N2
[3 marks]
Note: There may be slight differences in answers, depending on which combination of unrounded values and previous correct 3 sf values the candidates carry through in subsequent parts. Accept answers that are consistent with their working.
correct approach A1
eg , AO + OC,
AG N0
Note: Do not award A1 in part (ii) unless answer in part (i) is correct and does not result from working backwards.
[2 marks]
Note: There may be slight differences in answers, depending on which combination of unrounded values and previous correct 3 sf values the candidates carry through in subsequent parts. Accept answers that are consistent with their working.
finding scalar product and magnitude (A1)(A1)
scalar product = 11 × 8 + −5 × −10 + 3 × −1 (=135)
evidence of substitution into formula (M1)
eg
correct substitution (A1)
eg , ,
0.565795, 32.4177°
= 0.566, 32.4° A1 N3
[5 marks]
Note: There may be slight differences in answers, depending on which combination of unrounded values and previous correct 3 sf values the candidates carry through in subsequent parts. Accept answers that are consistent with their working.
correct substitution into area formula (A1)
eg ,
42.8660
area = 42.9 A1 N2
[2 marks]