Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

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Example 26 (Supplementary NCERT) Deleted for CBSE Board 2024 Exams

Example 27 (Supplementary NCERT) Deleted for CBSE Board 2024 Exams

Example 28 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams

Example 29 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams

Example 30 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams

Example 31 (Supplementary NCERT) Important Deleted for CBSE Board 2024 Exams

Last updated at May 29, 2023 by Teachoo

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, |𝑎 ⃗ + 𝑏 ⃗| = |0 ⃗" + " 𝑏 ⃗ | = |𝑏 ⃗ | "|" 𝑎 ⃗"| + |" 𝑏 ⃗"|" = 0 + |𝑏 ⃗ | = |𝑏 ⃗ | |𝑎 ⃗ + 𝑏 ⃗| = |𝑎 ⃗+0 ⃗ | = |𝑎 ⃗ | "|" 𝑎 ⃗"| + |" 𝑏 ⃗"|" = |𝑎 ⃗ | + 0 = |𝑎 ⃗ | Therefore, the inequality |𝑎 ⃗ + 𝑏 ⃗| ≤ |𝑎 ⃗| + |𝑏 ⃗| is satisfied trivially. 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 ⃗ + 2𝑎 ⃗. 𝑏 ⃗ + 𝒃 ⃗. 𝒃 ⃗ = |𝒂 ⃗|2 + 2𝑎 ⃗. 𝑏 ⃗ + |𝒃 ⃗|2 = |𝑎 ⃗|2 + 2|𝒂 ⃗||𝒃 ⃗| cos θ + |𝑏 ⃗|2 Thus, |𝑎 ⃗ + 𝑏 ⃗|2 = |𝑎 ⃗|2 + 2|𝑎 ⃗||𝑏 ⃗| cos θ + |𝑏 ⃗|2 (Using prop : 𝑎 ⃗. 𝑎 ⃗ = |𝑎 ⃗|2) (Using prop : 𝑎 ⃗. 𝑎 ⃗ = |𝑎 ⃗|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.