Ex 11.3, 2 - Chapter 11 Class 12 Three Dimensional Geometry (Term 2)

Last updated at May 29, 2018 by Teachoo

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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Next: Ex 11.3, 3 (a) →

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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