Find a vector of length 3 that is perpendicular to [2,1,-2]

Answer 1

First we need a perpendicular, any one will do as long as the dot product is zero. How about

(2,1,-2)dot(1,4 a) = 0
2 + 4 - 2a = 0
-2a = -6
a = 3

so (1,4,3) is a normal
now we have to stretch to a length of 3

length of my choice = (1/√26)
so a unit vector is (1/√26)[1,4,3)
and one of length 3 is

(3/√26([1,4,3]

of course this answer is not unique, there would be an infinite number of answers.