Examples

Chapter 10 Class 12 Vector Algebra
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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.