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, 2 You are here
Ex 11.3, 3
Ex 11.3, 4 Important
Ex 11.3, 5 Important
Ex 11.3, 6 Important
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Ex 11.3, 11 Important
Ex 11.3, 12 Important Deleted for CBSE Board 2021 Exams only
Ex 11.3, 13 (a) Important
Ex 11.3, 13 (b)
Ex 11.3, 13 (c)
Ex 11.3, 13 (d)
Ex 11.3, 13 (e)
Ex 11.3, 14 (a) Important
Ex 11.3, 14 (b)
Ex 11.3, 14 (c)
Ex 11.3, 14 (d)
Ex 11.3
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