Vector product - Area

Chapter 10 Class 12 Vector Algebra
Concept wise

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Ex 10.4, 10 Find the area of the parallelogram whose adjacent sides are determined by the vectors π β = π Μ β π Μ + 3π Μ and b = 2π Μ β 7π Μ + π Μ . π β = π Μ β π Μ + 3π Μ = 1π Μ β 1π Μ + 3k Μ π β = 2π Μ β 7π Μ + π Μ = 2π Μ β 7π Μ + 1k Μ Area of parallelogram ABCD = |π β" Γ " π β | π β Γ π β = |β 8(π Μ&π Μ&π Μ@1&β1&3@2&β7&1)| = π Μ (β1 Γ 1 β (β7) Γ 3) β π Μ (1 Γ 1 β 2 Γ 3) + π Μ (1 Γ β7 β 2 Γ β1) = π Μ (β1β(β21)) β π Μ (1 β 6) + π Μ (β7 β(β2)) = π Μ (β1 + 21) β π Μ (β5) + π Μ (β7 + 2) = 20 π Μ + 5π Μ β 5π Μ Magnitude of π β Γ π β = β(202+52+(β5)2) |π β" Γ " π β | = β(400+25+25) = β450 = β(25Γ9Γ2) = 5 Γ 3 Γ β2 = 15 β2 Therefore, the area of parallelogram is 15βπ .