Loading [MathJax]/jax/element/mml/optable/Latin1Supplement.js

User interface language: English | Español

Date May 2021 Marks available 4 Reference code 21M.1.AHL.TZ2.5
Level Additional Higher Level Paper Paper 1 (without calculator) Time zone Time zone 2
Command term Show that Question number 5 Adapted from N/A

Question

Given any two non-zero vectors, a and b, show that |a×b|2=|a|2|b|2-(a·b)2.

Markscheme

METHOD 1

use of |a×b|=|a||b|sinθ on the LHS            (M1)

|a×b|2=|a|2|b|2sin2θ                   A1

=|a|2|b|2(1-cos2θ)            M1

=|a|2|b|2-|a|2|b|2cos2θ  OR  =|a|2|b|2-(|a||b|cosθ)2                   A1

=|a|2|b|2-(a·b)2                   AG

 

METHOD 2

use of a·b=|a||b|cosθ on the RHS            (M1)

=|a|2|b|2-|a|2|b|2cos2θ                   A1

=|a|2|b|2(1-cos2θ)            M1

=|a|2|b|2sin2θ  OR  =(|a||b|sinθ)2                   A1

=|a×b|2                   AG

 

Note: If candidates attempt this question using cartesian vectors, e.g

a=(a1a2a3)  and  b=(b1b2b3),

award full marks if fully developed solutions are seen.
Otherwise award no marks.

 

[4 marks]

Examiners report

[N/A]

Syllabus sections

Topic 3— Geometry and trigonometry » AHL 3.13—Scalar (dot) product
Show 77 related questions
Topic 3— Geometry and trigonometry » AHL 3.16—Vector product
Topic 3— Geometry and trigonometry

View options