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Example 26 (Supplementary NCERT) Deleted for CBSE Board 2023 Exams

Example 27 (Supplementary NCERT) Deleted for CBSE Board 2023 Exams

Example 28 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams

Example 29 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams

Example 30 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams

Example 31 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams

Last updated at March 22, 2023 by Teachoo

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