`|( veca xx vecb )|^2=a^2b^2-( veca . vec b )^2`

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

If veca" and "vecb are any two vectors, then (veca xx vecb)^(2)= |veca|^(2)|vecb|^(2)-(veca* vecb)^(2) .

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 vec a\ a n d\ vec b prove that | vec axx vec b|^2=| (veca. veca , veca. vecb),(vecb.veca ,vecb.vecb)|

vecb and vecc are non- collinear if veca xx (vecb xx vecc) + (veca .vecb) vecb = ( 4-2x- sin y) vecb + ( x^(2) -1) vecc andd (vec. vecc) veca =veca then

vecb and vecc are non- collinear if veca xx (vecb xx vecc) + (veca .vecb) vecb = ( 4-2x- sin y) vecb + ( x^(2) -1) vecc andd (vec. vecc) veca =veca then

Two vectors are presented as veca = 3.0 hati + 5.0 hatj and vecb = 2. 0 hati + 4.0 hatj. Find (a) vec a xx vecb , (b) veca. vecb ,(c ) (veca + vecb).vecb ,

Prove that |veca xx vecb|^(2) = abs(veca)^(2)abs(vecb)^(2) - (veca * vecb)^(2)

If (vec a xx vec b)^2 + (veca .vecb)^2 = 676 and |vecb| = 2 , then |veca| is equal to