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Answer 1 · 2017-11-10T07:55:13

See the proof below

Let #vecu= < a, b,c ># in the basis #beta=(hati, hatj, hatk)#

and #vecv= < p,q,r > #

It is given that #||vecu|| = ||vecv||#

Then

#vecu+vecv=< a, b,c > + < p,q,r > #

#=< a+p,b+q,c+r >#

#vecu-vecv=< a, b,c > - < p,q,r > #

#=< a-p, b-q, c-r >#

The dot product is

#(vecu+vecv).(vecu-vecv)#

#=< a+p,b+q,c+r > . < a-p, b-q, c-r >#

#= (a+p)(a-p) +(b+q)(b-q) + (c+r)(c-r) #

#=a^2-p^2+b^2-q^2+c^2-r^2#

#=(a^2+b^2+c^2) - (p^2+q^2+r^2)#

#=||vecu||^2- ||vecv||^2=0#

Therefore,

#(vecu+vecv)# and #(vecu-vecv)# are orthogonal since their dots products #=0#

Answer 2 · 2017-11-10T11:30:01

Consider two vectors #vec u# and #vec v#.

Let #vec w = vec u + vec v# and #vec x = vec u - vec v#

Now, #vecw*vecw = u^2 + v^2 + 2uvcos theta#

Similarly,
#vecx*vecx = u^2 + v^2 - 2uvCos theta# where #theta# is the angle between vectors #vec u# and #vec v#.

It is given that #|w| = |x|#
#implies w^2 = x^2#
#implies vecw*vecw = vecx*vecx#
#implies u^2 + v^2 + 2uvcos theta = u^2 + v^2 -2uvcos theta#

Thus, we may obtain #4uvcos theta = 0# but for #u# and #v# non zero, therefore #cos theta = 0# and hence #vec u# and #vec v# are orthogonal.