Date | May 2019 | Marks available | 2 | Reference code | 19M.1.SL.TZ1.S_2 |
Level | Standard Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 1 |
Command term | Write down | Question number | S_2 | Adapted from | N/A |
Question
A line, L1L1, has equation r=(−3910)+s(602)r=⎛⎜⎝−3910⎞⎟⎠+s⎛⎜⎝602⎞⎟⎠. Point P(15,9,c)P(15,9,c) lies on L1L1.
Find cc.
A second line, L2L2, is parallel to L1L1 and passes through (1, 2, 3).
Write down a vector equation for L2L2.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
correct equation (A1)
eg −3+6s=15−3+6s=15, 6s=186s=18
s=3s=3 (A1)
substitute their ss value into zz component (M1)
eg 10+3(2)10+3(2), 10+610+6
c=16c=16 A1 N3
[4 marks]
r=(123)+t(602)r=⎛⎜⎝123⎞⎟⎠+t⎛⎜⎝602⎞⎟⎠ (=(i + 2j + 3k) + tt(6i + 2k)) A2 N2
Note: Accept any scalar multiple of (602)⎛⎜⎝602⎞⎟⎠ for the direction vector.
Award A1 for (123)+t(602)⎛⎜⎝123⎞⎟⎠+t⎛⎜⎝602⎞⎟⎠, A1 for L2=(123)+t(602)L2=⎛⎜⎝123⎞⎟⎠+t⎛⎜⎝602⎞⎟⎠, A0 for r=(602)+t(123)r=⎛⎜⎝602⎞⎟⎠+t⎛⎜⎝123⎞⎟⎠.
[2 marks]