Example 19 - Show |a.b| <= |a| |b| (Cauchy-Schwartz inequality)

Example 19 - Chapter 10 Class 12 Vector Algebra - Part 2
Example 19 - Chapter 10 Class 12 Vector Algebra - Part 3

  1. Chapter 10 Class 12 Vector Algebra (Term 2)
  2. Serial order wise

Transcript

Example 19 For any two vectors π‘Ž βƒ— and 𝑏 βƒ— , we always have |π‘Ž βƒ— ⋅𝑏 βƒ—| ≀ |π‘Ž βƒ—||𝑏 βƒ—| (Cauchy - Schwartz inequality). To prove : |π‘Ž βƒ— ⋅𝑏 βƒ—| ≀ |π‘Ž βƒ—||𝑏 βƒ—| We check trivially first |π‘Ž βƒ— ⋅𝑏 βƒ—| = |"|" π‘Ž βƒ—"||" 𝑏 βƒ—"| cos ΞΈ" | = 0 Γ— |𝑏 βƒ—| cos ΞΈ = 0 "|" π‘Ž βƒ—"||" 𝑏 βƒ—"|" = 0 Γ— |𝑏 βƒ—| = 0 |π‘Ž βƒ— ⋅𝑏 βƒ—| = |"|" π‘Ž βƒ—"||" 𝑏 βƒ—"| cos ΞΈ" | = |π‘Ž βƒ—| Γ— 0 Γ— cos ΞΈ = 0 |π‘Ž βƒ— ⋅𝑏 βƒ—| = |"|" π‘Ž βƒ—"||" 𝑏 βƒ—"| cos ΞΈ" | = |π‘Ž βƒ—| Γ— 0 Γ— cos ΞΈ = 0 "|" π‘Ž βƒ—"||" 𝑏 βƒ—"|" = |π‘Ž βƒ—| Γ— 0 = 0 Therefore, the inequality |π‘Ž βƒ—" β‹…" 𝑏 βƒ— | ≀ |π‘Ž βƒ—| |𝑏 βƒ—| is satisfied trivially Let us assume 𝒂 βƒ— β‰  𝟎 βƒ— & 𝒃 βƒ— β‰  𝟎 βƒ— π‘Ž βƒ— ⋅𝑏 βƒ— = "|" π‘Ž βƒ—"||" 𝑏 βƒ—"| cos ΞΈ" Taking modulus on both sides, |𝒂 βƒ— ⋅𝒃 βƒ—| = |"|" 𝒂 βƒ—"||" 𝒃 βƒ—"| cos ΞΈ" | |π‘Ž βƒ— ⋅𝑏 βƒ—| = "|" π‘Ž βƒ—"||" 𝑏 βƒ—"|" |"cos ΞΈ" | |"cos ΞΈ" | = ("|" π‘Ž βƒ—" β‹…" 𝑏 βƒ—"|" )/("|" π‘Ž βƒ—"||" 𝑏 βƒ—"|" ) We know that βˆ’1 ≀ cos ΞΈ ≀ 1 0 ≀ "|cos ΞΈ|" ≀ 1 ∴ ("|" π‘Ž βƒ—" β‹…" 𝑏 βƒ—"|" )/("|" π‘Ž βƒ—"||" 𝑏 βƒ—"|" ) ≀ 1 So, |𝒂 ⃗⋅𝒃 βƒ—| ≀ "|" 𝒂 βƒ—"||" 𝒃 βƒ—"|" 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 10 years. He provides courses for Maths and Science at Teachoo.