
Ex 11.3
Ex 11.3, 1 (b)
Ex 11.3, 1 (c) Important
Ex 11.3, 1 (d) Important
Ex 11.3, 2 You are here
Ex 11.3, 3 (a)
Ex 11.3, 3 (b)
Ex 11.3, 3 (c) Important
Ex 11.3, 4 (a) Important
Ex 11.3, 4 (b)
Ex 11.3, 4 (c)
Ex 11.3, 4 (d) Important
Ex 11.3, 5 (a) Important
Ex 11.3, 5 (b)
Ex 11.3, 6 (a) Important
Ex 11.3, 6 (b)
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) Important
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) Important
Last updated at May 29, 2018 by Teachoo
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