If a Μ‚ and b Μ‚ are unit vectors, then prove that

|a Μ‚+b Μ‚ |= 2π‘π‘œπ‘  θ/2, where πœƒ is the angle between them.

This question is similar to MIsc 17 (MCQ) - Chapter 10 Class 12 - Vector Algebra

Slide9.JPG

Slide10.JPG
Slide11.JPG

Go Ad-free

Transcript

Question 3 If π‘Ž Μ‚ and 𝑏 Μ‚ are unit vectors, then prove that |π‘Ž Μ‚+𝑏 Μ‚ |= 2π‘π‘œπ‘  πœƒ/2, where πœƒ is the angle between them. Given π‘Ž Μ‚ & 𝑏 Μ‚ are unit vectors, So, |𝒂 Μ‚ | = 1 & |𝒃 Μ‚ | = 1 We need to find |𝒂 Μ‚+𝒃 Μ‚ | Let’s find |𝒂 Μ‚+𝒃 Μ‚ |^𝟐 instead Now, |𝒂 Μ‚+𝒃 Μ‚ |^𝟐=(𝒂 Μ‚+𝒃 Μ‚ ).(𝒂 Μ‚+𝒃 Μ‚ ) |π‘Ž Μ‚+𝑏 Μ‚ |^2=π‘Ž Μ‚.(π‘Ž Μ‚+𝑏 Μ‚ )+𝑏 Μ‚.(π‘Ž Μ‚+𝑏 Μ‚ ) (As |π‘Ž βƒ— |^2 = π‘Ž βƒ—.π‘Ž βƒ—) |π‘Ž Μ‚+𝑏 Μ‚ |^2=𝒂 Μ‚.𝒂 Μ‚+π‘Ž Μ‚.𝑏 Μ‚+𝑏 Μ‚.π‘Ž Μ‚+𝒃 Μ‚.𝒃 Μ‚ |π‘Ž Μ‚+𝑏 Μ‚ |^2=|𝒂 Μ‚ |^𝟐+π‘Ž Μ‚.𝑏 Μ‚+𝑏 Μ‚.π‘Ž Μ‚+|𝒃 Μ‚ |^𝟐 |π‘Ž Μ‚+𝑏 Μ‚ |^2=𝟏^𝟐+π‘Ž Μ‚.𝑏 Μ‚+𝑏 Μ‚.π‘Ž Μ‚+𝟏^𝟐 |π‘Ž Μ‚+𝑏 Μ‚ |^2=𝟐+π‘Ž Μ‚.𝑏 Μ‚+𝒃 Μ‚.𝒂 Μ‚ |π‘Ž Μ‚+𝑏 Μ‚ |^2=2+π‘Ž Μ‚.𝑏 Μ‚+𝒂 Μ‚.𝒃 Μ‚ |π‘Ž Μ‚+𝑏 Μ‚ |^2=2+2𝒂 Μ‚.𝒃 Μ‚ |π‘Ž Μ‚+𝑏 Μ‚ |^2=2+2|𝒂 Μ‚|.|𝒃 Μ‚ | 𝐜𝐨𝐬⁑𝜽 |π‘Ž Μ‚+𝑏 Μ‚ |^2=2+2 Γ— 1 Γ— 1 π‘π‘œπ‘ β‘πœƒ |π‘Ž Μ‚+𝑏 Μ‚ |^2=2+2 π‘π‘œπ‘ β‘πœƒ |π‘Ž Μ‚+𝑏 Μ‚ |^2=2(𝟏+π’„π’π’”β‘πœ½) Using cos 2ΞΈ = 2cos2 ΞΈ βˆ’ (As |π‘Ž βƒ— |^2 = π‘Ž βƒ—.π‘Ž βƒ—) (As |π‘Ž Μ‚ |=1 & |𝑏 Μ‚ |=1) (As |π‘Ž Μ‚ |=1 & |𝑏 Μ‚ |=1) |π‘Ž Μ‚+𝑏 Μ‚ |^2=2 Γ— πŸπ’„π’π’”^𝟐 𝜽/𝟐 |π‘Ž Μ‚+𝑏 Μ‚ |^2=4π‘π‘œπ‘ ^2 πœƒ/2 |π‘Ž Μ‚+𝑏 Μ‚ |^2=(2π‘π‘œπ‘  πœƒ/2)^2 Taking square root both sides |𝒂 Μ‚+𝒃 Μ‚ |=πŸπ’„π’π’” 𝜽/𝟐 Hence proved

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo