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

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Transcript

Example 19 For any two vectors 𝑎 ⃗ and 𝑏 ⃗ , we always have |𝑎 ⃗ ⋅𝑏 ⃗| ≤ |𝑎 ⃗||𝑏 ⃗| (Cauchy - Schwartz inequality). To prove : |𝑎 ⃗ ⋅𝑏 ⃗| ≤ |𝑎 ⃗||𝑏 ⃗| We check trivially first |𝑎 ⃗ ⋅𝑏 ⃗| = |"|" 𝑎 ⃗"||" 𝑏 ⃗"| cos θ" | = 0 × |𝑏 ⃗| cos θ = 0 "|" 𝑎 ⃗"||" 𝑏 ⃗"|" = 0 × |𝑏 ⃗| = 0 |𝑎 ⃗ ⋅𝑏 ⃗| = |"|" 𝑎 ⃗"||" 𝑏 ⃗"| cos θ" | = |𝑎 ⃗| × 0 × cos θ = 0 |𝑎 ⃗ ⋅𝑏 ⃗| = |"|" 𝑎 ⃗"||" 𝑏 ⃗"| cos θ" | = |𝑎 ⃗| × 0 × cos θ = 0 "|" 𝑎 ⃗"||" 𝑏 ⃗"|" = |𝑎 ⃗| × 0 = 0 Therefore, the inequality |𝑎 ⃗" ⋅" 𝑏 ⃗ | ≤ |𝑎 ⃗| |𝑏 ⃗| is satisfied trivially Let us assume 𝒂 ⃗ ≠ 𝟎 ⃗ & 𝒃 ⃗ ≠ 𝟎 ⃗ 𝑎 ⃗ ⋅𝑏 ⃗ = "|" 𝑎 ⃗"||" 𝑏 ⃗"| cos θ" Taking modulus on both sides, |𝒂 ⃗ ⋅𝒃 ⃗| = |"|" 𝒂 ⃗"||" 𝒃 ⃗"| cos θ" | |𝑎 ⃗ ⋅𝑏 ⃗| = "|" 𝑎 ⃗"||" 𝑏 ⃗"|" |"cos θ" | |"cos θ" | = ("|" 𝑎 ⃗" ⋅" 𝑏 ⃗"|" )/("|" 𝑎 ⃗"||" 𝑏 ⃗"|" ) We know that −1 ≤ cos θ ≤ 1 0 ≤ "|cos θ|" ≤ 1 ∴ ("|" 𝑎 ⃗" ⋅" 𝑏 ⃗"|" )/("|" 𝑎 ⃗"||" 𝑏 ⃗"|" ) ≤ 1 So, |𝒂 ⃗⋅𝒃 ⃗| ≤ "|" 𝒂 ⃗"||" 𝒃 ⃗"|" 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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.