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Last updated at May 29, 2018 by Teachoo

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 Therefore, the inequality 𝑎 ⋅ 𝑏 ≤ | 𝑎| | 𝑏| is satisfied trivially Now, 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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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.