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

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)

Prove that |veca xx vecb|^(2)=|{:(veca*veca,veca *vecb),(veca*vecb,vecb*vecb):}| .

Prove that | vecaxxvecb | ^ 2 = det ((veca.veca, veca.vecb), (veca.vecb, vecb.vecb))

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

|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 (veca+3vecb)xx(veca+vecb)+(3veca-5vecb)xx(veca-vecb)=0

If veca and vecb are two vectors , then prove that (vecaxxvecb)^(2)=|{:(veca.veca" ",veca.vecb),(vecb.veca" ",vecb.vecb):}|

If veca and vecb are two vectors , then prove that (vecaxxvecb)^(2)=|{:(veca.veca" ",veca.vecb),(vecb.veca" ",vecb.vecb):}|

The value of the expression |veca xx vecb|^(2) + (veca.vecb)^(2) is-

Show that, (veca-vecb)xx(veca+vecb) = 2(vecaxxvecb) .