CBSE Class 12 Sample Paper for 2020 Boards

Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

## If a + b + c = 0 and |a| = 3, |b| = 5, |c| = 7, then find the value of a.bΒ  + b.cΒ  + c.a

Note : - This is similar to Example 29 of NCERT β Chapter 10 Class 12

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Question 24 (OR 2nd Question) if π β + π β + π β = 0 and |π β | = 3, |π β | = 5, |π β | = 7, then find the value of π β. π β + π β. π β + π β. π β . Given, π β + π β + π β = 0 β So, |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 + π β . π β + π β . π β + 2π β. π β + 2π β. π β + 2π β. π β = π β . π β + π β . π β + π β . π β + 2(π β. π β + π β. π β + π β. π β) Prop : π β . π β = |π β |2 = |π β |π + |π β |π + |π β |π + 2 (π β. π β + π β. π β + π β . π β) = 32 + 52 + 72 + 2(π β. π β + π β. π β + π β. π β) = 9 + 25 + 49 + 2(π β. π β + π β. π β + π β. π β) = 83 + 2 (π β. π β + π β. π β + π β. π β) β΄ |π β+π β+π β|2 = 83 + 2 (π β. π β + π β. π β + π β. π β) Now, |π β" + " π β" + " π β | = 0 |π β" + " π β" + " π β |^2 = 0 83 + 2 (π β. π β + π β. π β + π β. π β) = 0 2(π β. π β + π β. π β + π β. π β) = β83 (π β. π β + π β. π β + π β. π β) = (βππ)/π