Check sibling questions

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

Example 20 - Chapter 10 Class 12 Vector Algebra - Part 2
Example 20 - Chapter 10 Class 12 Vector Algebra - Part 3


Transcript

Example 20 For any two vectors π‘Ž βƒ— and 𝑏 βƒ— , we always have |π‘Ž βƒ— + 𝑏 βƒ—| ≀ |π‘Ž βƒ—| + |𝑏 βƒ—| (triangle inequality).To Prove: |π‘Ž βƒ— + 𝑏 βƒ—| ≀ |π‘Ž βƒ—| + |𝑏 βƒ—| We first prove trivially, |π‘Ž βƒ— + 𝑏 βƒ—| = |0 βƒ—" + " 𝑏 βƒ— | = |𝑏 βƒ— | "|" π‘Ž βƒ—"| + |" 𝑏 βƒ—"|" = 0 + |𝑏 βƒ— | = |𝑏 βƒ— | |π‘Ž βƒ— + 𝑏 βƒ—| = |π‘Ž βƒ—+0 βƒ— | = |π‘Ž βƒ— | "|" π‘Ž βƒ—"| + |" 𝑏 βƒ—"|" = |π‘Ž βƒ— | + 0 = |π‘Ž βƒ— | Therefore, the inequality |π‘Ž βƒ— + 𝑏 βƒ—| ≀ |π‘Ž βƒ—| + |𝑏 βƒ—| is satisfied trivially. Let us assume 𝒂 βƒ— β‰  𝟎 βƒ— & 𝒃 βƒ— β‰  𝟎 βƒ— |π‘Ž βƒ— + 𝑏 βƒ—|2 = (π‘Ž βƒ— + 𝑏 βƒ—) . (π‘Ž βƒ— + 𝑏 βƒ—) = π‘Ž βƒ— . π‘Ž βƒ— + π‘Ž βƒ— . 𝑏 βƒ— + 𝒃 βƒ— . 𝒂 βƒ— + 𝑏 βƒ—. 𝑏 βƒ— = π‘Ž βƒ— . π‘Ž βƒ— + π‘Ž βƒ— . 𝑏 βƒ— + 𝒂 βƒ— . 𝒃 βƒ— + 𝑏 βƒ—. 𝑏 βƒ— = 𝒂 βƒ— . 𝒂 βƒ— + 2π‘Ž βƒ—. 𝑏 βƒ— + 𝒃 βƒ—. 𝒃 βƒ— = |𝒂 βƒ—|2 + 2π‘Ž βƒ—. 𝑏 βƒ— + |𝒃 βƒ—|2 = |π‘Ž βƒ—|2 + 2|𝒂 βƒ—||𝒃 βƒ—| cos ΞΈ + |𝑏 βƒ—|2 Thus, |π‘Ž βƒ— + 𝑏 βƒ—|2 = |π‘Ž βƒ—|2 + 2|π‘Ž βƒ—||𝑏 βƒ—| cos ΞΈ + |𝑏 βƒ—|2 (Using prop : π‘Ž βƒ—. π‘Ž βƒ— = |π‘Ž βƒ—|2) (Using prop : π‘Ž βƒ—. π‘Ž βƒ— = |π‘Ž βƒ—|2) We know that cos ΞΈ ≀ 1 Multiplying 2|π‘Ž βƒ—||𝑏 βƒ—| on both sides 2|𝒂 βƒ—||𝒃 βƒ—| cos ΞΈ ≀ 2|𝒂 βƒ—||𝒃 βƒ—| Adding |π‘Ž βƒ—|2 + |𝑏 βƒ—|2 on both sides, |𝒂 βƒ—|2 + |𝒃 βƒ—|2 + 2|𝒂 βƒ—||𝒃 βƒ—| cos ΞΈ ≀ |π‘Ž βƒ—|2 + |𝑏 βƒ—|2 + 2 |π‘Ž βƒ—| |𝑏 βƒ—| |𝒂 βƒ— + 𝒃 βƒ—|2 ≀ (|π‘Ž βƒ—| + |𝑏 βƒ—|) 2 Taking square root both sides |π‘Ž βƒ— + 𝑏 βƒ—| ≀ (|π‘Ž βƒ—| + |𝑏 βƒ—|) Hence proved.

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.