For any two vectors ` vec a\ a n d\ vec b` prove that `| vec axx vec b|^2=| (veca. veca , veca. vecb),(vecb.veca ,vecb.vecb)|`

For any two vectors veca and vecb prove that |veca+vecb|lt+|veca|+|vecb|

For any two vectors veca and vecb prove that |veca-vec|ge|veca|-|vecb|

For any two vectors veca and vecb prove that |veca.vecb|<=|veca||vecb|

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):}|

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)

If two vectors vec aa n d vec b are such that |veca|=3, |vecb|=2 and veca.vecb=6, . , Find | vec a+ vec b|a n d| vec a- vec b|dot

If | vec a|=2,\ | vec b|=5\ a n d\ | vec axx vec b|=8,\ find veca.vecb

Prove that : veca*(vecb+vec c)xx(veca+2vecb+3vec c)=[veca vecb vec c]

If vecl,vecm,vecn are three non coplanar vectors prove that [vec vecm vecn](vecaxxvecb) =|(vec1.veca, vec1.vecb, vec1),(vecm.veca, vecm.vecb, vecm),(vecn.veca, vecn.vecb, vecn)|