For any two vectors `veca` and `vecb`, we always have `|veca+vecb|le |veca|+|vecb|` [Triangle inequality].

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

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

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

If theta is the angle between any two vectors veca and vecb , then |veca.vecb|=|veca xx vecb| when theta is equal to

Two vectors vecA and vecB are such that vecA+vecB=vecA-vecB . Then

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)|

If vectors vecA and vecB are such that |vecA+vecB|= |vecA|= |vecB| then |vecA-vecB| may be equated to

The angle between the two vectors veca + vecb and veca-vecb is

For two vectors veca and vecb,veca,vecb=|veca||vecb| then (A) veca||vecb (B) veca_|_vecb (C) veca=vecb (D) none of these

Find |veca-vecb| , if two vector veca and vecb are such that |veca|=4,|vecb|=5 and veca.vecb=3