Examples

Chapter 10 Class 12 Vector Algebra
Serial order wise

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Example 28 Let π β, π β and π β be three vectors such that |π β|= 3, |π β|= 4, |π β|= 5 and each one of them being perpendicular to the sum of the other two, find |π β" + " π β" + " π β |.Given, |π β| = 3 , |π β| = 4 , |π β|= 5 Also, Each one of them being perpendicular to the sum of the other two π β is perpendicular to π β + π β π β. (π β + π β) = 0 π β. π β + π β. π β = 0 π β is perpendicular to π β + π β π β. (π β + π β) = 0 π β. π β + π β. π β = 0 π β is perpendicular to π β + π β π β. (π β + π β) = 0 π β. π β + π β. π β = 0 Now, |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 + 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 + (0) + π β . π β + (0) + π β . π β + (0) = π β. π β + π β. π β + π β. π β = "|" π β"|"2 + "|" π β"|"2 + "|" π β"|"2 = 32 + 42 + 52 = 9 + 16 + 25 = 50 So, |π β+π β+π β |2 = 50 Taking square root both sides, "|" π β "+ " π β" + " π β"|" = β50 "|" π β "+ " π β "+ " π β"|" = β25 Γ β2 "|" π β "+ " π β "+ " π β"|" = πβπ = π β. π β + (0) + π β . π β + (0) + π β . π β + (0) = π β. π β + π β. π β + π β. π β = "|" π β"|"2 + "|" π β"|"2 + "|" π β"|"2 = 32 + 42 + 52 = 9 + 16 + 25 = 50 So, |π β+π β+π β |2 = 50 Taking square root both sides, "|" π β "+ " π β" + " π β"|" = β50 "|" π β "+ " π β "+ " π β"|" = β25 Γ β2 "|" π β "+ " π β "+ " π β"|" = πβπ