Prove that ` |vecavecbvecc|=abc `   if and only if ` veca ` , ` vecb ` , ` vecc ` are mutually perpendicular.

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

Vectors 3veca-5vecb and 2veca + vecb are mutually perpendicular. If veca + 4 vecb and vecb - veca are also mutually perpendicular, then the cosine of the angle between veca nad vecb is

Prove that [veca-vecb,vecb-vecc,vecc-veca]=0

[veca, veca+vecb, veca+vecb+vecc] is :

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

Prove that [veca+vecb vecb+vecc vecc+veca]=2[veca vecb vecc]

If veca + 2 vecb + 3 vecc = vec0 " then " veca xx vecb + vecb xx vecc + vecc xx veca=