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Ex 11.2, 5 - Find equation of line in vector, cartesian form

Ex 11.2, 5 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2
Ex 11.2, 5 - Chapter 11 Class 12 Three Dimensional Geometry - Part 3

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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)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.