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  1. Chapter 11 Class 12 Three Dimensional Geometry
  2. Serial order wise

Transcript

Misc 7 Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane π‘Ÿ βƒ—.(𝑖 Μ‚ + 2𝑗 Μ‚ βˆ’ 5π‘˜ Μ‚) + 9 = 0 . The vector equation of a line passing through a point with position vector π‘Ž βƒ— and parallel to vector 𝑏 βƒ— is 𝒓 βƒ— = 𝒂 βƒ— + πœ†π’ƒ βƒ— Given, the line passes (1, 2, 3) So, 𝒂 βƒ— = 1π’Š Μ‚ + 2𝒋 Μ‚ + 3π’Œ Μ‚ Finding normal of plane π‘Ÿ βƒ—.(𝑖 Μ‚ + 2𝑗 Μ‚ βˆ’ 5π‘˜ Μ‚) + 9 = 0 π‘Ÿ βƒ—. (𝑖 Μ‚ + 2𝑗 Μ‚ βˆ’ 5π‘˜ Μ‚) = βˆ’ 9 βˆ’π‘Ÿ βƒ—. (𝑖 Μ‚ + 2𝑗 Μ‚ βˆ’ 5π‘˜ Μ‚) = 9 π‘Ÿ βƒ—. (βˆ’1𝑖 Μ‚ βˆ’ 2𝑗 Μ‚ + 5π‘˜ Μ‚) = 9 Comparing with π‘Ÿ βƒ—. 𝑛 βƒ— = d, 𝒏 βƒ— = βˆ’π‘– Μ‚ βˆ’ 2𝑗 Μ‚ + 5π‘˜ Μ‚ Since line is perpendicular to plane, the line will be parallel to the normal of the plane ∴ 𝒃 βƒ— = 𝒏 βƒ— = βˆ’1𝑖 Μ‚ βˆ’ 2𝑖 Μ‚ + 5π‘˜ Μ‚ Hence, π‘Ÿ βƒ— = (1𝑖 Μ‚ + 2𝑖 Μ‚ + 3π‘˜ Μ‚) + πœ†(βˆ’1𝑖 Μ‚ βˆ’ 2𝑖 Μ‚ + 5π‘˜ Μ‚) π‘Ÿ βƒ— = (1𝑖 Μ‚ + 2𝑖 Μ‚ + 3π‘˜ Μ‚) βˆ’ πœ†(1𝑖 Μ‚ + 2𝑖 Μ‚ βˆ’ 5π‘˜ Μ‚) 𝒓 βƒ— = (π’Š Μ‚ + 2𝒋 Μ‚ + 3π’Œ Μ‚) + k (π’Š Μ‚ + 2𝒋 Μ‚ βˆ’ 5π’Œ Μ‚)

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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.