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Last updated at Feb. 4, 2020 by Teachoo

Transcript

Misc 7 Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane π β.(π Μ + 2π Μ β 5π Μ) + 9 = 0 . The vector equation of a line passing through a point with position vector π β and parallel to vector π β is π β = π β + ππ β Given, the line passes (1, 2, 3) So, π β = 1π Μ + 2π Μ + 3π Μ Finding normal of plane π β.(π Μ + 2π Μ β 5π Μ) + 9 = 0 π β. (π Μ + 2π Μ β 5π Μ) = β 9 βπ β. (π Μ + 2π Μ β 5π Μ) = 9 π β. (β1π Μ β 2π Μ + 5π Μ) = 9 Comparing with π β. π β = d, π β = βπ Μ β 2π Μ + 5π Μ Since line is perpendicular to plane, the line will be parallel to the normal of the plane β΄ π β = π β = β1π Μ β 2π Μ + 5π Μ Hence, π β = (1π Μ + 2π Μ + 3π Μ) + π(β1π Μ β 2π Μ + 5π Μ) π β = (1π Μ + 2π Μ + 3π Μ) β π(1π Μ + 2π Μ β 5π Μ) π β = (π Μ + 2π Μ + 3π Μ) + k (π Μ + 2π Μ β 5π Μ)

Miscellaneous

Misc 1
Important

Misc 2

Misc 3

Misc 4 Important

Misc 5 Important

Misc 6 Important

Misc 7 You are here

Misc 8 Important

Misc 9 Important

Misc 10

Misc 11 Important

Misc 12 Important

Misc 13 Important

Misc 14 Important

Misc 15 Important

Misc 16 Important

Misc 17 Important

Misc 18 Important

Misc 19 Important

Misc 20 Important

Misc 21 Important

Misc 22 Important

Misc 23 Important

Chapter 11 Class 12 Three Dimensional Geometry

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About the Author

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.