If `abs(veca+vecb)=abs(veca-vecb)` prove that `veca` and `vecb` are perpendicular.

Prove that (veca+vecb)*(veca+vecb)=|veca|^(2)+|vecb|^(2) , if and only if veca,vecb are perpendicular , given vecanevec0,vecbnevec0 .

If |veca+vecb|^(2)-|veca-vecb|^(2)=4 .Show that veca and vecb are coincident, if veca and vecb are unit vectors.

If (veca xx vecb)xx vecc= veca xx (vecb xx vecc) and vecb perpendicular to vecc then which is true ?

If |veca| = |vecb|=|veca+vecb|=1 then prove that |veca-vecb|=sqrt(3) .

If veca, vecb and vecc are vectors such that |veca|=3,|vecb|=4 and |vecc|=5 and (veca+vecb) is perpendicular to vecc, (vecb+vecc) is perpendicular to veca and (vecc+veca) is perpendicular to vecb then |veca+vecb+vecc|= (A) 4sqrt(3) (B) 5sqrt(2) (C) 2 (D) 12

If veca and vecb are unit vectors such that (veca +vecb). (2veca + 3vecb)xx(3veca - 2vecb)=vec0 then angle between veca and vecb is

If |veca+ vecb| lt | veca- vecb| , then the angle between veca and vecb can lie in the interval

If |veca|=13, |vecb|=5" and "veca*vecb=60^(@) then |veca xx vecb|=