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 (i) Important

Ex 11.2, 8 (ii)

Ex 11.2, 9 (i) Important

Ex 11.2, 9 (ii)

Ex 11.2, 10 Important

Ex 11.2, 11

Ex 11.2, 12 Important

Ex 11.2, 13 Important

Ex 11.2, 14

Ex 11.2, 15 Important

Question 1 Important Deleted for CBSE Board 2025 Exams

Question 2 Deleted for CBSE Board 2025 Exams

Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

Last updated at April 16, 2024 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)