Yes it does work for 3 numbers.
So you have
where and
Give an example of a vector perpendicular to [6,2,3] that has the same length.
If the answer was [a,b,c]
a squared + b squared + c squared = 49
this is all i can get so far...
I know for [a,b] and [m,n] to be perpendicular am + bn =0
b/a X n/m = -1
bn = - am
am+bn = 0
Then, would this work for three numbers too?
[a,b,c,] and [ 6,2,3] 6a + 2b + 3c =0
First, you must understand that the is not one single correct answer. There is a whole plane perpendicular to a given vector and so an infinite number of vectors in that plane that are all perpendicular to given vector. Even restricting the length, you have a whole "circle" of such vectors.
Yes, for [a, b, c] to be perpendicular to [6, 2, 3], we must have 6a+ 2b+ 3c= 0. Since that is a single equation in three unknowns, just choose two of them to be whatever number you want and solve for the third. Whatever vector you get there,
divide by its length to get a unit vector perpendicular to [6, 2, 3].
Yes, . So what is the length of [6, 2, 3]? Whatever unit vector you got perpendicular to [6, 2, 3], multiply by that length.