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)

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 You are here

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 Feb. 4, 2020 by Teachoo

Ex 11.3, 11 Find the equation of the plane thro ugh the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0.Equation of a plane passing through the intersection of planes A1x + B1y + C1z = d1 and A2x + B2y + C2z = d2 is (A1x + B1y + C1z − d1) + 𝜆 (A2x + b2y + c2z − d2) = 0 x + y + z = 1 1x + 1y + 1z = 1 Comparing with 𝐴_1 "x" + 𝐵_1 "y" + 𝐶_1 "z" = d1 𝐴_1= 1 , 𝐵_1= 1 , 𝐶_1= 1 , 𝑑_1= 1 2x + 3y + 4z = 5 Comparing with A_2 "x" + B_2 "y" + C_2 "z" = d2 A_2= 2 , B_2= 3 , C_2= 4 , d_2= 5 Equation of plane is (𝐴_1 "x" + 𝐵_1 "y" + 𝐶_1 "z" = d1) + 𝜆 (𝐴_2 "x" + 𝐵_2 "y" + 𝐶_2 "z" = d2) = 0 Putting values (1x + 1y + 1z − 1) + 𝜆 (2x + 3y + 4z − 5) = 0 x + y + z − 1 + 2𝜆 x + 3𝜆y + 4𝜆z − 5𝜆 = 0 (1 + 2𝜆) x + (1 + 3𝜆)y + (1 + 4𝜆) z + (−1 −5𝜆) = 0 Also, the plane is perpendicular to the plane x − y + z = 0 So, the normal vector 𝑁 ⃗ to be the plane is perpendicular to the normal vector of x − y + z = 0. Theory : Two lines with direction ratios 𝑎1, b1, c1 and 𝑎2, b2, c2 are perpendicular if 𝑎1 𝑎2 + b1b2 + c1 c2 = 0 𝑁 ⃗ = (1 + 2𝜆) 𝑖 ̂ + (1 + 3𝜆)𝑗 ̂ + (1 + 4𝜆)𝑘 ̂ Direction ratios = 1 + 2𝜆, 1 + 3𝜆, 1 + 4𝜆 ∴ 𝑎_1 = 1 + 2𝜆 , 𝑏_1 = 1 + 3𝜆 , 𝑐_1 = 1 + 4𝜆 𝑛 ⃗ = 1𝑖 ̂ – 1𝑗 ̂ + 1𝑘 ̂ Direction ratios = 1 , –1 , 1 ∴ 𝑎_2 = 1, 𝑏_2 = –1, 𝑐_2 = 1 Since, 𝑵 ⃗ is perpendicular to 𝒏 ⃗, 𝑎1 𝑎2 + b1b2 + c1 c2 = 0 (1 + 2𝜆) × 1 + (1 + 3𝜆) × − 1 + (1 + 4𝜆) × 1 = 0 1 + 2𝜆 − 1 − 3𝜆 + 1 + 4𝜆 = 0 1 + 3𝜆 = 0 −1 = 3𝜆 ∴ 𝜆 = (−𝟏)/𝟑 Putting value of 𝜆 in (1), (1 + 2𝜆) x + (1 + 3𝜆)y + (1 + 4𝜆) z + (−1 −5𝜆) = 0 (1+2 ×(−1)/3)x + (1+3×(−1)/3) y + (1+4 ×(−1)/3) z +(−1−5×(−1)/3) = 0 (1− 2/3) x + (1−1) y + (1−4/3) z + (− 1+ 5/3) = 0 1/3 x + 0y − 1/3 z + 2/3 = 0 1/3 x − 1/3 z + 2/3 = 0 1/3 (x − z + 2) = 0 x − z + 2 = 0 Therefore, the equation of the plane is x − z + 2 = 0