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Ex 11.3

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Chapter 11 Class 12 Three Dimensional Geometry
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Ex 11.3, 2 - Find vector equation of a plane which is 7 units - Equation of plane - In Normal Form

Ex 11.3, 2  - Chapter 11 Class 12 Three Dimensional Geometry - Part 2

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Transcript

Ex 11.3, 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

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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.