Show that `(veca - vecb) `x` (veca + vecb) = 2( veca `x` vecb)`

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

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

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

If veca and vecb are unequal unit vectors such that (veca -vecb) xx [(vecb+veca) xx (2veca+vecb)]=veca+vecb , then angle theta between veca and vecb can be

If veca, vecb and vecc are vectors such that veca. vecb = veca.vecc, veca xx vecb = veca xx vecc, a ne 0. then show that vecb = vecc.

Prove that: |(veca+vecb)xx(veca-vecb)|=2ab if veca_|_vecb

if veca + vecb + vecc=0 , then show that veca xx vecb = vecb xx vecc = vecc xx veca .

If for any two vectors veca and vecb. (veca + vecb )^(2) + (veca - vecb)^(2) =lamda[(veca)^(2)+ (vecb)^(2)] then write the value of lamda.

Let veca , vecb,vecc be three vectors such that veca bot ( vecb + vecc), vecb bot ( vecc + veca) and vecc bot ( veca + vecb) , " if " |veca| =1 , |vecb| =2 , |vecc| =3 , " then " | veca + vecb + vecc| is,

l veca . m vecb = lm(veca . vecb) and veca . vecb = 0 then veca and vecb are perpendicular if veca and vecb are not null vector