1. Chapter 11 Class 12 Three Dimensional Geometry
  2. Serial order wise
  3. Plane
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Question 2 - Plane - Chapter 11 Class 12 Three Dimensional Geometry

Last updated at Dec. 16, 2024 by Teachoo

Transcript

Question 2 Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector 3 + 5 6 . Vector equation of a place at a distance d from the origin and normal to the vector is . = d Unit vector of = = 1 ( ) Distance form origin = d = 7 = 3 + 5 6 Magnitude of = 32+52+ 6 2 = 9+25+36 = 70 Now, = 1 ( ) = 1 70 (3 + 5 6 ) = 3 + 5 6 70 Vector equation is . = d . + = 7 Therefore, the vector equation of the plane is . 3 + 5 6 70 = 7

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

    3. Plane

    Direction cosines and ratios →
Direction cosines and ratios →
Chapter 11 Class 12 Three Dimensional Geometry
Serial order wise

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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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo