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Last updated at Aug. 24, 2021 by

Β

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Ex 11.3, 5 Find the vector and cartesian equations of the planes (b) That passes through the point (1, 4, 6) and the normal vector to the plane is π Μ β 2π Μ + π Μ Vector equation Equation of plane passing through point A whose position vector is π β & perpendicular to π β is (π β β π β) . π β = 0 Given, Plane passes through (1, 4, 6) So, π β = 1π Μ + 4π Μ + 6π Μ Normal to plane = π Μ β 2π Μ + π Μ π β = π Μ β 2π Μ + π Μ Vector equation of plane is (π β β π β).π β = 0 [π ββ(π Μ+ππ Μ+ππ Μ)] . (π Μ β 2π Μ + π Μ) = 0 Now, Finding Cartesian form using two methods Cartesian form (Method 1) : Vector equation is [π ββ(π Μ+4π Μ+6π Μ)] . (π Μ β 2π Μ + π Μ) = 0 Put π β = xπ Μ + yπ Μ + zπ Μ [(π₯π Μ+π¦π Μ+π§π Μ )β(1π Μ+4π Μ+6π Μ)]. (1π Μ β 2π Μ + 1π Μ) = 0 [(π₯β1) π Μ +(π¦β4) π Μ+(π§β6)π Μ ]. (1π Μ β 2π Μ + 1π Μ) = 0 1(x β 1) + (β2)(y β 4) + 1 (z β 6) = 0 x β 1 β 2(y β 4) + z β 6 = 0 x β 2y + z + 1 = 0 β΄ Equation of plane in Cartesian form is x β 2y + z + 1 = 0. Cartesian form (Method 2) Equation of plane passing through (x1, y1, z1) and perpendicular to a line with direction ratios A, B, C is A(x β x1) + B(y β y1) + C (z β z1) = 0 Since the plane passes through (1, 4, 6) x1 = 1, y1 = 4, z1 = 6 And normal is π Μ β 2π Μ + π Μ So, Direction ratios of line perpendicular to plane = 1, β2, 1 β΄ A = 1, B = β2, C = 1 Therefore, equation of line in Cartesian form is 1(x β 1) β 2 (y β 4) + 1(x β 6) = 0 x β 2y + z + 1 = 0

Ex 11.3

Ex 11.3, 1 (a)

Ex 11.3, 1 (b)

Ex 11.3, 1 (c) Important

Ex 11.3, 1 (d) Important

Ex 11.3, 2

Ex 11.3, 3 (a)

Ex 11.3, 3 (b)

Ex 11.3, 3 (c) Important

Ex 11.3, 4 (a) Important

Ex 11.3, 4 (b)

Ex 11.3, 4 (c)

Ex 11.3, 4 (d) Important

Ex 11.3, 5 (a) Important

Ex 11.3, 5 (b) You are here

Ex 11.3, 6 (a) Important

Ex 11.3, 6 (b)

Ex 11.3, 7

Ex 11.3, 8

Ex 11.3, 9

Ex 11.3, 10 Important

Ex 11.3, 11 Important

Ex 11.3, 12 Important Deleted for CBSE Board 2022 Exams

Ex 11.3, 13 (a) Important Deleted for CBSE Board 2022 Exams

Ex 11.3, 13 (b) Important

Ex 11.3, 13 (c)

Ex 11.3, 13 (d)

Ex 11.3, 13 (e) Deleted for CBSE Board 2022 Exams

Ex 11.3, 14 (a) Important

Ex 11.3, 14 (b)

Ex 11.3, 14 (c)

Ex 11.3, 14 (d) Important

Chapter 11 Class 12 Three Dimensional Geometry (Term 2)

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.