Examples

Chapter 10 Class 12 Vector Algebra
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Example 29 (Method 1) Three vectors π β, π β and π β satisfy the condition π β + π β + π β = 0 β . Evaluate the quantity ΞΌ = π β βπ β + π β β π β + π β β π β, if |π β|=1, |π β|= 4 and |c β|= 2.Given |π β|=1, |π β|= 4 and |c β|= 2 Also, 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) = |π β |π + |π β |π + |π β |π + 2 (π β. π β + π β. π β + π β . π β) = 12 + 42 + 22 + 2(π β. π β + π β. π β + π β. π β) = 1 + 16 + 4 + 2(π β. π β + π β. π β + π β. π β) = 21 + 2 (π β. π β + π β. π β + π β. π β) So, |π β+π β+π β |2 = 21 + 2 (π β. π β + π β. π β + π β. π β) Now, given that |π β" + " π β" + " π β | = 0 |π β" + " π β" + " π β |2 = 0 21 + 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21 (π β. π β + π β. π β + π β. π β) = (β21)/2 Therefore, π = π β. π β + π β. π β + π β . π β = (βππ)/π Example 29 (Method 2) Three vectors π β, π β and π β satisfy the condition π β + π β + π β = 0 β . Evaluate the quantity ΞΌ = π β βπ β + π β β π β + π β β π β, if |π β|=1, |π β|= 4 and |c β|= 2.Given |π β| = 1, |π β|= 4 and |c β|= 2 Also, ( π β + π β + π β ) = 0 β Now, π β . (π β + π β + π β) = π β . π β + π β. π β + π β . π β π β . 0 β = π β. π β + π β. π β + π β. π β 0 = π β. π β + π β. π β + π β. π β 0 β = |π β |π + π β. π β + π β. π β (Using prop : π β . π β = |π β |2 ) 0 β = |π β |2 + π β. π β + π β. π β 0 = 12 + π β. π β + π β. π β 0 = 1 + π β. π β + π β. π β π β. π β + π β. π β = β1 Also, π β . (π β + π β + π β) = π β . π β + π β. π β + π β . π β π β . 0 β = π β. π β + π β. π β + π β. π β 0 = π β. π β + π β. π β + π β. π β 0 = π β. π β + π β. π β + π β. π β 0 = π β. π β + |π β |2 + π β. π β 0 = π β. π β + 42 + π β . π β 0 = π β. π β + 16 + π β . π β π β. π β + π β. π β = β16 Also π β . (π β+ π β + π β) = π β . π β + π β . π β + π β . π β π β. 0 β = π β. π β + π β. π β + π β. π β 0 = π β. π β + π β. π β + π β. π β 0 = π β. π β + π β. π β + π β. π β 0 = π β. π β + π β. π β + |π β |2 0 = π β. π β + π β . π β + 22 0 = π β. π β + π β . π β + 4 π β. π β + π β. π β = β4 Adding (1), (2) and (3), (π β. π β + π β. π β) + (π β. π β + π β. π β) + (π β. π β + π β. π β) = β1 + (β16) + (β4) 2π β. π β + 2π β. π β + 2π β. π β = β21 2(π β. π β + π. π β + π β. π β) = β21 π β. π β + π β. π β + π β. π β = (β21)/2 Therefore, π = π β. π β + π β. π β + π β . π β = (βππ)/π