Question

Help here, i know how to do it when they're parallele, just not perpendicular.

Answer 1

has to do with the angle right?

Answer 2

a-mb and a+mb are parallel when -m=+m

a-mb and a_mb are per if -m*+m = -1

Answer 3

but yes; when the vectors a and b are dotted; a.b = 0 is perp and arccos(a.b/15) = 0 or pi (0, 180) they are parallel

Answer 4

a.b/15 = 1 or -1 would seem to make it

Answer 5

\[\theta = cos^{-1}\left(\frac{\vec a.\vec b}{|a||b|}\right) \]
\[\frac{\vec a.\vec b}{|a||b|}=0;\ perped\]
\[\frac{\vec a.\vec b}{|a||b|}=1,-1;\ parallel\]

Answer 6

thank you :))