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

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