Find the vector equation of the plane that passes through the point (1, 0, 0) and contains the line r = ⋏i Μ‚ j Μ‚

 

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  1. Class 12
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Transcript

Equation of plane passing through point A whose position vector is 𝒂 βƒ— & perpendicular to 𝒏 βƒ— is (𝒓 βƒ— βˆ’ 𝒂 βƒ—) . 𝒏 βƒ— = 0 Given Plane passes through (1, 0, 0) So 𝒂 βƒ— = 1𝑖 Μ‚ + 0𝑗 Μ‚ βˆ’ 0π‘˜ Μ‚ 𝒂 βƒ— = 𝑖 Μ‚ Thus, equation of plane is (π‘Ÿ βƒ— βˆ’ π‘Ž βƒ—) . 𝑛 βƒ— = 0 (𝒓 βƒ— βˆ’ π’Š Μ‚) . 𝒏 βƒ— = 0 Now, The plane contains the line π‘Ÿ βƒ—= πœ†π‘— Μ‚ So, Line is perpendicular to normal of plane 𝒋 Μ‚ . 𝒏 βƒ— = 0 And, Lines point (0, 0, 0) lies on plane as well Putting π‘Ÿ βƒ— = 0 βƒ— in equation of plane (0 βƒ— βˆ’ 𝑖 Μ‚) . 𝑛 βƒ— = 0 βˆ’ 𝑖 Μ‚ . 𝑛 βƒ— = 0 π’Š Μ‚ . 𝒏 βƒ— = 0 Since 𝒏 βƒ— perpendicular to both π’Š Μ‚ and 𝒋 Μ‚ Thus, 𝒏 βƒ— = π’Œ Μ‚ So, our equation of plane becomes (π‘Ÿ βƒ— βˆ’ 𝑖 Μ‚) . 𝑛 βƒ— = 0 (𝒓 βƒ— βˆ’ π’Š Μ‚) . π’Œ Μ‚ = 0 π‘Ÿ βƒ— . π‘˜ Μ‚ βˆ’ 𝑖 Μ‚ . π‘˜ Μ‚ = 0 𝒓 βƒ— . π’Œ Μ‚ = 0

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.