Check sibling questions

If aΒ Μ‚ and bΒ Μ‚ are unit vectors, then prove that

|aΒ Μ‚+bΒ Μ‚ |= 2π‘π‘œπ‘  ΞΈ/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

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.