Ex 10.5 (Supplementary NCERT)

Chapter 10 Class 12 Vector Algebra
Serial order wise

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Ex 10.5, 5 (Supplementary NCERT) Show that the four points with position vectors 4π Μ + 8π Μ + 12π Μ, 2π Μ + 4π Μ + 6π Μ, 3π Μ + 5π Μ + 4π Μ & 5π Μ + 8π Μ + 5π Μ are coplanarLet points be A = 4π Μ + 8π Μ + 12π Μ B = 2π Μ + 4π Μ + 6π Μ C = 3π Μ + 5π Μ + 4π Μ D = 5π Μ + 8π Μ + 5π Μ Four points A, B, C, D are coplanar if the three vectors (π΄π΅) β , (π΄πΆ) β and (π΄π·) β are coplanar. i.e. [(π¨π©) β, (π¨πͺ) β, (π¨π«) β ] = 0 A (4π Μ + 8π Μ + 12π Μ) B (2π Μ + 4π Μ + 6π Μ) (π¨π©) β = (2π Μ + 4π Μ + 6π Μ) β (4π Μ + 8π Μ + 12π Μ) = (2 β 4) π Μ + (4 β 8) π Μ + (6 β 12)π Μ = β2π Μ β 4π Μ β 6π Μ A (4π Μ + 8π Μ + 12π Μ) C (3π Μ + 5π Μ + 4π Μ) (π¨πͺ) β = (3π Μ + 5π Μ + 4π Μ) β (4π Μ + 8π Μ + 12π Μ) = (3 β 4) π Μ + (5 β 8) π Μ + (4 β 12) π Μ = βπ Μ β 3π Μ β 8π Μ A (4π Μ + 8π Μ + 12π Μ) D (5π Μ + 8π Μ + 5π Μ) (π¨π«) β = (5π Μ + 8π Μ + 5π Μ) β (4π Μ + 8π Μ + 12π Μ) = (5 β 4) π Μ + (8 β 8) π Μ + (5 β 12) π Μ = π Μ + 0π Μ β 7π Μ Now, [(π΄π΅) β, (π΄πΆ) β, (π΄π·) β ] = |β 8(β2&β4&β[email protected]β1&β3&β[email protected]&0&β7)| = β2[(β3Γβ7)β(0Γβ8) ] β (β4) [(β1Γβ7)β(1Γβ8)] + (β6)[(β1Γ0)β(1Γβ3) ] = β2[21β0]+4[7+8]β6[0+3] = β2[21]+4[15]β6[3] = β 42 + 60 β 18 = 60 β 60 = 0 β΄ [(π¨π©) β, (π¨πͺ) β, (π¨π«) β ] = 0 Therefore, points A, B, C and D are coplanar.