Show that the components of `vecb` parallel to `veca` and perpendicular to it are `((veca.vecb)veca)/veca^2 and ((vecaxxvecb)veca))/a^2` respectively.

Prove that: (2veca-vecb)xx (veca+2vecb)=5vecaxxvecb .

The resolved part of the vector veca along the vector vecb is veclamda and that perpendicular to vecb is vecmu . Then (A) veclamda=((veca.vecb).veca)/veca^2 (B) veclamda=((veca.vecb).vecb)/vecb^2 (C) vecmu=((vecb.vecb)veca-(veca.vecb)vecb)/vecb^2 (D) vecmu=(vecbxx(vecaxxvecb))/vecb^2

If vectors veca and vecb are two adjacent sides of parallelograsm then the vector representing the altitude of the parallelogram which is perpendicular to veca is (A) vecb+(vecbxxveca)/(|veca|^2) (B) (veca.vecb)/(vecb|^2) (C) vecb-(vecb.veca)/(|veca|)^2) (D) (vecaxx(vecbxxveca))/(vecb|^20

Find veca.vecb if |veca|=2, |vecb|=5,and |vecaxxvecb|=8

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

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

The resultant of vecA and vecB is perpendicular to vecA . What is the angle between vecA and vecB ?

The resultant of vecA and vecB is perpendicular to vecA . What is the angle between vecA and vecB ?

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

Show that |veca|vecb+|vecb|veca is perpendicular to |veca|vecb-|vecb|veca for any two non zero vectors veca and vecb.