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Example 26 (Supplementary NCERT) Deleted for CBSE Board 2025 Exams

Example 27 (Supplementary NCERT) Deleted for CBSE Board 2025 Exams

Example 28 (Supplementary NCERT) Important Deleted for CBSE Board 2025 Exams

Example 29 (Supplementary NCERT) Important Deleted for CBSE Board 2025 Exams

Example 30 (Supplementary NCERT) Important Deleted for CBSE Board 2025 Exams

Example 31 (Supplementary NCERT) Important Deleted for CBSE Board 2025 Exams

Last updated at April 16, 2024 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.