`(vecaxxvecb)^2+(veca.vecb)^2=|veca|^2 |vecb|^2`

If veca and vecb are any two vectors , then prove that |vecaxxvecb|^(2)=|veca|^(2)|vecb|^(2)-(veca.vecb)^(2)=|{:(veca.veca,veca.vecb),(veca.vecb,vecb.vecb):}| or |vecaxxvecb|^(2)+(veca.vecb)^(2)=|veca|^(2)|vecb|^(2) (This is also known as Lagrange identily)

|veca pm vecb|^2 = |veca|^2 + |vecb|^2 pm 2|veca||vecb|cos theta and (veca + vecb).(veca - vecb) = |veca|^2 - |vecb|^2

Prove that (vecaxxvecb)^2=veca^2b^2-(veca.vecb)^2 .

Show that (veca xx vecb)^(2) = |veca| ^(2) |vecb|^(2) - (veca.vecb)^(2) = |(veca.veca)/(veca. vecb)(veca.vecb)/(vecb.vecb)|

If veca vecb be any two mutually perpendiculr vectors and vecalpha be any vector then |vecaxxvecb|^2 ((veca.vecalpha)veca)/(veca|^2)+|vecaxvecb|^2 ((vecb.vecalpha)vecb)/(|vecb|^2)-|vecaxxvecb|^2vecalpha= (A) |(veca.vecb)vecalpha|(vecaxxvecb) (B) [veca vecb vecalpha](vecbxxveca) (C) [veca vecb vecalpha](vecaxxvecb) (D) none of these

[(veca,vecb,axxvecb)]+(veca.vecb)^(2)=

If |vecaxxvecb|^(2)+(veca.vecb)^(2)=144and|veca|=4,"then "|vecb| is equal to ……..

The vector (veca-vecb)xx(veca+vecb) is equal to (A) 1/2 (vecaxxvecb) (B) vecaxxvecb (C) 2(veca+vecb) (D) 2(vecaxxvecb)

Prove that: (2veca-vecb)xx (veca+2vecb)=5vecaxxvecb .

If |vecaxxvecb|=2,|veca.vecb|=2 , then |veca|^(2)|vecb|^(2) is equal to