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

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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

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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, Social Science, Physics, Chemistry, Computer Science at Teachoo.