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Example 20 - Show |a + b| < |a| + |b| (triangle inequality) - Examples

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  1. Chapter 10 Class 12 Vector Algebra
  2. Serial order wise
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Example 20 For any two vectors 𝑎﷯ and 𝑏﷯ , we always have | 𝑎﷯ + 𝑏﷯| ≤ | 𝑎﷯| + | 𝑏﷯| (triangle inequality). To Prove: | 𝑎﷯ + 𝑏﷯| ≤ | 𝑎﷯| + | 𝑏﷯| We first prove trivially, Therefore, the inequality | 𝑎﷯ + 𝑏﷯| ≤ | 𝑎﷯| + | 𝑏﷯| is satisfied trivially. Now, let us assume 𝑎﷯ ≠ 0﷯ & 𝑏﷯ ≠ 0﷯ | 𝑎﷯ + 𝑏﷯|2 = ( 𝑎﷯ + 𝑏﷯) . ( 𝑎﷯ + 𝑏﷯) = 𝑎﷯ . 𝑎﷯ + 𝑎﷯ . 𝑏﷯ + 𝒃﷯ . 𝒂﷯ + 𝑏﷯. 𝑏﷯ = 𝑎﷯ . 𝑎﷯ + 𝑎﷯ . 𝑏﷯ + 𝒂﷯ . 𝒃﷯ + 𝑏﷯. 𝑏﷯ = 𝒂﷯ . 𝒂﷯ + 2 𝑎﷯. 𝑏﷯ + 𝒃﷯. 𝒃﷯ = | 𝒂﷯|2 + 2 𝑎﷯. 𝑏﷯ + | 𝒃﷯|2 = | 𝑎﷯|2 + 2| 𝒂﷯|| 𝒃﷯| cos θ + | 𝑏﷯|2 ∴ | 𝑎﷯ + 𝑏﷯|2 = | 𝑎﷯|2 + 2| 𝑎﷯|| 𝑏﷯| cos θ + | 𝑏﷯|2 We know that cos θ ≤ 1 Multiplying 2| 𝑎﷯|| 𝑏﷯| on both sides 2| 𝑎﷯|| 𝑏﷯| cos θ ≤ 2| 𝑎﷯|| 𝑏﷯| Adding | 𝑎﷯|2 + | 𝑏﷯|2 on both sides, | 𝑎﷯|2 + | 𝑏﷯|2 + 2| 𝑎﷯|| 𝑏﷯| cos θ ≤ | 𝑎﷯|2 + | 𝑏﷯|2 + 2 | 𝑎﷯| | 𝑏﷯| | 𝑎﷯ + 𝑏﷯|2 ≤ (| 𝑎﷯| + | 𝑏﷯|) 2 Taking square root both sides | 𝑎﷯ + 𝑏﷯| ≤ (| 𝑎﷯| + | 𝑏﷯|) Hence proved.

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