Vector product - Area

Chapter 10 Class 12 Vector Algebra
Concept wise

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Area of parallelogram whose diagonals are given Let us consider a parallelogram ABCD Here, π β + π β = (π_1 ) β and π β + (βπ β) = (π_2 ) β π β β π β = (π_2 ) β Letβs find (π_1 ) β Γ (π_2 ) β (π_1 ) β Γ (π_2 ) β = (π β + π β) Γ (π β β π β) = π β Γ (π β β π β) + π β Γ (π β β π β) = π β Γ π β β π β Γ π β + π β Γ π β β π β Γ π β Since π β Γ π β = 0 = π β Γ π β β 0 + 0 β π β Γ π β = π β Γ π β β π β Γ π β Since π β Γ π β = β (π β Γ π β) = π β Γ π β β (β (π β Γ π β)) = (π β Γ π β) + (π β Γ π β) = 2 (π β Γ π β) Therefore, (π_1 ) β Γ (π_2 ) β = 2 (π β Γ π β) Now, we know that Area of parallelogram = |π β" Γ " π β | Writing in terms of diagonals Area of parallelogram = π/π |(π_π ) βΓ(π_π ) β |