Examples

Chapter 10 Class 12 Vector Algebra
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Example 19 For any two vectors π β and π β , we always have |π β βπ β| β€ |π β||π β| (Cauchy - Schwartz inequality). To prove : |π β βπ β| β€ |π β||π β| We check trivially first |π β βπ β| = |"|" π β"||" π β"| cos ΞΈ" | = 0 Γ |π β| cos ΞΈ = 0 "|" π β"||" π β"|" = 0 Γ |π β| = 0 |π β βπ β| = |"|" π β"||" π β"| cos ΞΈ" | = |π β| Γ 0 Γ cos ΞΈ = 0 |π β βπ β| = |"|" π β"||" π β"| cos ΞΈ" | = |π β| Γ 0 Γ cos ΞΈ = 0 "|" π β"||" π β"|" = |π β| Γ 0 = 0 Therefore, the inequality |π β" 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 βπ β = "|" π β"||" π β"| cos ΞΈ" Taking modulus on both sides, |π β βπ β| = |"|" π β"||" π β"| cos ΞΈ" | |π β βπ β| = "|" π β"||" π β"|" |"cos ΞΈ" | |"cos ΞΈ" | = ("|" π β" β" π β"|" )/("|" π β"||" π β"|" ) We know that β1 β€ cos ΞΈ β€ 1 0 β€ "|cos ΞΈ|" β€ 1 β΄ ("|" π β" β" π β"|" )/("|" π β"||" π β"|" ) β€ 1 So, |π ββπ β| β€ "|" π β"||" π β"|" Hence proved.