Β

Β

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 You are here

Ex 11.3, 5 (b)

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

Last updated at Aug. 24, 2021 by Teachoo

Β

Β

Ex 11.3, 5 (Introduction) Find the vector and cartesian equations of the planes (a) that passes through the point (1, 0, β2) and the normal to the plane is π Μ + π Μ β π Μ.Vector equation of a plane passing through a point (x1, y1, z1) and perpendicular to a line with direction ratios A, B, C is [π β β(π₯1π Μ+π¦1π Μ+π§1π Μ)]. (Aπ Μ + Bπ Μ + Cπ Μ) = 0 or (π β β π β).π β = 0 ("A" π) β is perpendicular to "n" β So, ("A" P) β . "n" β = 0 ("r" β β "a" β)."n" β = 0 Ex 11.3, 5 Find the vector and Cartesian equations of the planes (a) that passes through the point (1, 0, β 2) and the normal to the plane is π Μ + π Μ β 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, 0, β2) So, π β = 1π Μ + 0π Μ β 2π Μ Normal to plane = π Μ + π Μ β π Μ π β = π Μ + π Μ β π Μ Vector equation of plane is (π β β π β) . π β = 0 [π ββ("1" π Μ+"0" π Μβ"2" π Μ)].(π Μ + π Μ β π Μ) = 0 [π ββ(π Μβ"2" π Μ)] . (π Μ+π Μβπ Μ) = 0 Now, Finding Cartesian form using two methods Cartesian form (Method 1) Vector equation is [π ββ(π Μβ"2" π Μ)] . (π Μ+π Μβπ Μ) = 0 Put π β = xπ Μ + yπ Μ + zπ Μ [("x" π Μ+"y" π Μ+"z" π Μ )β("1" π Μ+"0" π Μβ"2" π Μ)].(1π Μ + 1π Μ β 1π Μ) = 0 [("x" β1) π Μ+("y" β0) π½ Μ+(π§β(β2))π Μ ].(1π Μ + 1π Μ β 1π Μ) = 0 1(x β 1) + 1(y β 0) β 1 (z + 2) = 0 x β 1 + y β z β 2 = 0 x + y β z = 3 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, 0, β2) x1 = 1, y1 = 0, z1 = 2 And normal is π Μ + π Μ β π Μ So, Direction ratios of line perpendicular to plane = 1, 1, β1 β΄ A = 1, B = 1, C = β1 Therefore, equation of line in Cartesian form is 1(x β 1) + 1(y β 0) β 1(z β (β2)) = 0 x + y β z = 3