Last updated at May 29, 2018 by Teachoo

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

Ex 11.3

Ex 11.3, 1
Important

Ex 11.3, 2 You are here

Ex 11.3, 3

Ex 11.3, 4 Important

Ex 11.3, 5 Important

Ex 11.3, 6 Important

Ex 11.3, 7

Ex 11.3, 8

Ex 11.3, 9

Ex 11.3, 10 Important

Ex 11.3, 11 Important

Ex 11.3, 12 Important Deleted for CBSE Board 2022 Exams

Ex 11.3, 13 (a) Important Deleted for CBSE Board 2022 Exams

Ex 11.3, 13 (b)

Ex 11.3, 13 (c)

Ex 11.3, 13 (d)

Ex 11.3, 13 (e) Deleted for CBSE Board 2022 Exams

Ex 11.3, 14 (a) Important

Ex 11.3, 14 (b)

Ex 11.3, 14 (c)

Ex 11.3, 14 (d)

Chapter 11 Class 12 Three Dimensional Geometry (Term 2)

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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 10 years. He provides courses for Maths and Science at Teachoo.