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 π‘Žο·― and 𝑏﷯ , we always have | π‘Žο·― + 𝑏﷯| ≀ | π‘Žο·―| + | 𝑏﷯| (triangle inequality). To Prove: | π‘Žο·― + 𝑏﷯| ≀ | π‘Žο·―| + | 𝑏﷯| We first prove trivially, Therefore, the inequality | π‘Žο·― + 𝑏﷯| ≀ | π‘Žο·―| + | 𝑏﷯| is satisfied trivially. Now, let us assume π‘Žο·― β‰  0ο·― & 𝑏﷯ β‰  0ο·― | π‘Žο·― + 𝑏﷯|2 = ( π‘Žο·― + 𝑏﷯) . ( π‘Žο·― + 𝑏﷯) = π‘Žο·― . π‘Žο·― + π‘Žο·― . 𝑏﷯ + 𝒃﷯ . 𝒂﷯ + 𝑏﷯. 𝑏﷯ = π‘Žο·― . π‘Žο·― + π‘Žο·― . 𝑏﷯ + 𝒂﷯ . 𝒃﷯ + 𝑏﷯. 𝑏﷯ = 𝒂﷯ . 𝒂﷯ + 2 π‘Žο·―. 𝑏﷯ + 𝒃﷯. 𝒃﷯ = | 𝒂﷯|2 + 2 π‘Žο·―. 𝑏﷯ + | 𝒃﷯|2 = | π‘Žο·―|2 + 2| 𝒂﷯|| 𝒃﷯| cos ΞΈ + | 𝑏﷯|2 ∴ | π‘Žο·― + 𝑏﷯|2 = | π‘Žο·―|2 + 2| π‘Žο·―|| 𝑏﷯| cos ΞΈ + | 𝑏﷯|2 We know that cos ΞΈ ≀ 1 Multiplying 2| π‘Žο·―|| 𝑏﷯| on both sides 2| π‘Žο·―|| 𝑏﷯| cos ΞΈ ≀ 2| π‘Žο·―|| 𝑏﷯| Adding | π‘Žο·―|2 + | 𝑏﷯|2 on both sides, | π‘Žο·―|2 + | 𝑏﷯|2 + 2| π‘Žο·―|| 𝑏﷯| cos ΞΈ ≀ | π‘Žο·―|2 + | 𝑏﷯|2 + 2 | π‘Žο·―| | 𝑏﷯| | π‘Žο·― + 𝑏﷯|2 ≀ (| π‘Žο·―| + | 𝑏﷯|) 2 Taking square root both sides | π‘Žο·― + 𝑏﷯| ≀ (| π‘Žο·―| + | 𝑏﷯|) Hence proved.

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