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Ex 11.2

Ex 11.2, 1

Ex 11.2, 2

Ex 11.2, 3 Important

Ex 11.2, 4

Ex 11.2, 5 Important You are here

Ex 11.2, 6

Ex 11.2, 7 Important

Ex 11.2, 8

Ex 11.2, 9 Important

Ex 11.2, 10 (i) Important

Ex 11.2, 10 (ii)

Ex 11.2, 11 (i) Important

Ex 11.2, 11 (ii)

Ex 11.2, 12 Important

Ex 11.2, 13

Ex 11.2, 14 Important

Ex 11.2, 15 Important

Ex 11.2, 16

Ex 11.2, 17 Important

Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

Last updated at Feb. 1, 2020 by Teachoo

Ex 11.2, 5 Find the equation of the line in vector and in cartesian form that passes through the point with position vector 2𝑖 ̂ − 𝑗 ̂ + 4𝑘 ̂ and is in the direction 𝑖 ̂ + 2 𝑗 ̂ − 𝑘 ̂ . Equation of a line passing though a point with position vector 𝑎 ⃗ and parallel to vector 𝑏 ⃗ is 𝑟 ⃗ = 𝑎 ⃗ + 𝜆 𝑏 ⃗ Here, 𝑎 ⃗ = 2𝑖 ̂ − 𝑗 ̂ + 4𝑘 ̂ & 𝑏 ⃗ = 𝑖 ̂ + 2𝑗 ̂ − 𝑘 ̂ So, 𝑟 ⃗ = (2𝒊 ̂ − 𝒋 ̂ + 4𝒌 ̂) + 𝜆 (𝒊 ̂ + 2𝒋 ̂ − 𝒌 ̂) ∴ Equation of line in vector form is (2𝑖 ̂ − 𝑗 ̂ + 4𝑘 ̂) + 𝜆 (𝑖 ̂ + 2𝑗 ̂ − 𝑘 ̂) Equation of a line passing though (x1, y1, z1) and parallel to a line having direction ratios a, b, c is (𝑥 − 𝑥1)/𝑎 = (𝑦 − 𝑦1)/𝑏 = (𝑧 − 𝑧1)/𝑐 Since the line passes through a point with position vector 2𝑖 ̂ − 𝑗 ̂ + 4𝑘 ̂, ∴ 𝑥1 = 2, y1 = −1, z1 = 4 Also, line is in the direction of 𝑖 ̂ + 2𝑗 ̂ − 𝑘 ̂, Direction ratios : 𝑎 = 1, b = 2, c = −1 Equation of line in Cartesian form is (𝑥 − 2)/1 = (𝑦 − ( −1))/2 = (𝑧 − 4)/( − 1) (𝒙 − 𝟐)/𝟏 = (𝒚 + 𝟏)/𝟐 = (𝒛 − 𝟒)/( − 𝟏)