Vector product - Area

Chapter 10 Class 12 Vector Algebra
Concept wise

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Misc 10 The two adjacent sides of a parallelogram are 2π Μ β 4π Μ + 5π Μ and π Μ β 2π Μ β 3π Μ Find the unit vector parallel to its diagonal. Also, find its area. Let π β and π β are adjacent side of a parallelogram, where π β = 2π Μ β 4π Μ + 5π Μ π β = π Μ β 2π Μ β 3π Μ Let diagonal be π β Hence, π β = π β + π β = (2π Μ β 4π Μ + 5π Μ) + (1π Μ β 2π Μ β 3π Μ) = (2 + 1) π Μ β (4 + 2) π Μ + (5 β 3)π Μ = 3π Μ β 6π Μ + 2π Μ Magnitude of βπ β β = β((3)^2+(β6)^2+(2)^2 ) = β(9+36+4) = β49 = 7 Unit vector in direction of π β = 1/(ππππππ‘π’ππ ππ π β ) Γ π β π Μ = π/π ("3" π Μ" β 6" π Μ" + 2" π Μ ) Finding Area of parallelogram Area of parallelogram = |π β Γ π β | Now, π β Γ π β = |β 8(π Μ&π Μ&π Μ@2&β4&5@1&β2&β3)| = π Μ (12 β (β10) β π Μ (β6 β5) + π Μ (β4 β (β4)) = 22π Μ + 11π Μ + 0π Μ = 22π Μ + 11π Μ So π β Γ π β = 22π Μ + 11π Μ Now, |π β Γ π β | = β(22^2+11^2 ) = β(2^2 (γ11)γ^2+11^2 ) = β( 11^2 (2^2+1)) = 11β5 Hence, Area of parallelogram = |π β Γ π β |= ππβπ