Get live Maths 1-on-1 Classs - Class 6 to 12
Ex 10.5 (Supplementary NCERT)
Ex 10.5, 2 (Supplementary NCERT) Deleted for CBSE Board 2023 Exams
Ex 10.5, 3 (Supplementary NCERT) Deleted for CBSE Board 2023 Exams
Ex 10.5, 4 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams
Ex 10.5, 5 (Supplementary NCERT) Deleted for CBSE Board 2023 Exams
Ex 10.5, 6 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams You are here
Ex 10.5 , 7 (Supplementary NCERT) Important Deleted for CBSE Board 2023 Exams
Ex 10.5 (Supplementary NCERT)
Last updated at March 23, 2023 by Teachoo
Ex 10.5, 6 (Supplementary NCERT) Find x such that A (3, 2, 1), B = (4, x, 5), C(4, 2, β2) and D (6, 5, β1) are coplanar Four points A, B, C, D are coplanar if the three vectors (π΄π΅) β , (π΄πΆ) β and (π΄π·) β are coplanar. i.e. [(π¨π©) β, (π¨πͺ) β, (π¨π«) β ] = 0 A (3, 2, 1), B (4, x, 5) (π¨π©) β = (4 β 3)π Μ + (x β 2)π Μ + (5 β 1) π Μ = 1π Μ + (x β 2)π Μ + 4π Μ A (3, 2, 1), C (4, 2, β2) (π¨πͺ) β = (4 β 3)π Μ + (2 β 2)π Μ + (β2 β 1) π Μ = 1π Μ + 0π Μ β 3π Μ A (3, 2, 1), D (6, 5, β1) (π¨π«) β = (6 β 3)π Μ + (5 β 2)π Μ + (β1 β 1) π Μ = 3π Μ + 3π Μ β 2π Μ Now, [(π΄π΅) β, (π΄πΆ) β, (π΄π·) β ] = |β 8(1&(π₯β2)&[email protected]&0&β[email protected]&3&β2)| 0 = 1[(0Γβ2)β(3Γβ3) ] β (x β 2) [(1Γβ2)β(3Γβ3)] + 4[(1Γ3)β(3Γ0) ] 0 = 1[0β(β9)]β"(x β 2)" [β2β(β9)]+4[3β0] 0 = 1(9) β "(x β 2)"(β2 + 9) + 4(3) 0 = 9 β (x β 2) (7) + 12 0 = 9 β 7x + 14 + 12 0 = 35 β 7x 7x = 35 x = 35/7 x = 5