Vector product - Area

Chapter 10 Class 12 Vector Algebra
Concept wise

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### Transcript

Example 25 Find the area of a parallelogram whose adjacent sides are given by the vectors (π ) β = 3π Μ + π Μ + 4π Μ and π β = π Μ β π Μ + π Μ Given (π ) β = 3π Μ + 1π Μ + 4π Μ π β = 1π Μ β 1π Μ + 1k Μ Area of parallelogram ABCD = |π β Γ π β | Now, (π ) βΓ π β = |β 8(π Μ&π Μ&π Μ@3&1&[email protected]&β1&1)| = π Μ (1 Γ 1 β (β1) Γ 4) β π Μ (3 Γ 1 β 1 Γ 4) + π Μ (3 Γ β1 β 1 Γ 1) = π Μ(1 β (-4)) β j Μ (3 β 4) + π Μ (β3 β1) = π Μ(1 + 4) β j Μ (β1) + π Μ (β4) = 5π Μ + π Μ β 4π Μ Magnitude of π β Γ π β = β(52+1^2+(β4)2) |π β Γ π β | = β(25+1+16) = βππ Area of parallelogram ABCD = |π β Γ π β | = β42 Therefore, the required area is βππ .