User interface language: English | Español

Date May 2018 Marks available 2 Reference code 18M.1.SL.TZ2.S_1
Level Standard Level Paper Paper 1 Time zone Time zone 2
Command term Find Question number S_1 Adapted from N/A

Question

Let  OA = ( 2 1 3 ) and AB = ( 1 3 1 ) , where O is the origin. L1 is the line that passes through A and B.

Find a vector equation for L1.

[2]
a.

The vector ( 2 p 0 ) is perpendicular to AB . Find the value of p.

[3]
b.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

any correct equation in the form r = a + tb (accept any parameter for t)

where a is  ( 2 1 3 ) , and b is a scalar multiple of  ( 1 3 1 )      A2 N2

eg r =  ( 2 1 3 ) = t ( 1 3 1 ) r = 2i + j + 3k + s(i + 3j + k)

Note: Award A1 for the form a + tb, A1 for the form L = a + tb, A0 for the form r = b + ta.

[2 marks]

a.

METHOD 1

correct scalar product     (A1)

eg  (1 × 2) + (3 × p) + (1 × 0), 2 + 3p

evidence of equating their scalar product to zero     (M1)

eg  a•b = 0, 2 + 3p = 0, 3p = −2

p = 2 3        A1 N3

 

METHOD 2

valid attempt to find angle between vectors      (M1)

correct substitution into numerator and/or angle       (A1)

eg  cos θ = ( 1 × 2 ) + ( 3 × p ) + ( 1 × 0 ) | a | | b | , cos θ = 0

p = 2 3        A1 N3

[3 marks]

b.

Examiners report

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

Syllabus sections

Topic 3—Geometry and trigonometry » AHL 3.11—Vector equation of a line in 2d and 3d
Show 51 related questions
Topic 3—Geometry and trigonometry » AHL 3.13—Scalar and vector products
Topic 3—Geometry and trigonometry

View options