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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

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.