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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+s602. Point P(15,9,c)P(15,9,c) lies on L1L1.

Find cc.

[4]
a.

A second line, L2L2, is parallel to L1L1 and passes through (1, 2, 3).

Write down a vector equation for L2L2.

[2]
b.

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=153+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]

a.

r=(123)+t(602)r=123+t602 (=(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+t602, A1 for L2=(123)+t(602)L2=123+t602A0 for r=(602)+t(123)r=602+t123.

[2 marks]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 3— Geometry and trigonometry » AHL 3.14—Vector equation of line
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Topic 3— Geometry and trigonometry

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