if a , b,c are three mutually perpendicular vectors of equal magnitude then prove

let θ be the angle between a+b+c and a. then

cos θ = (a+b+c).(a) / |a| |a+b+c|

=a.a/ |a| 3 |a|

=1 / 3

since a ,b c are mutually perpendicular so | a +b+c | will be |a|² + |b|² + | c |² = 3 |a|    ( |a| = |b| = |c| )