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

Let veca,vecb and vecc are vectors such that |veca|=3,|vecb|=4and |vecc|=5, and (veca+vecb) is perpendicular to vecc,(vecb+vecc) is perpendiculatr to veca and (vecc+veca) is perpendicular to vecb . Then find the value of |veca+vecb+vecc| .

If veca , vecb are unit vectors such that the vector veca + 3vecb is peependicular to 7 veca - vecb and veca -4vecb is prependicular to 7 veca -2vecb then the angle between veca and vecb is

If veca and vecb are two unit vectors such that veca+2vecb and 5veca-4vecb are perpendicular to each other, then the angle between veca and vecb is

If veca and vecb be two non collinear vectors such that veca=vecc+vecd , where vecc is parallel to vecb and vecd is perpendicular to vecb obtain expression for vecc and vecd in terms of veca and vecb as: vecd= veca- ((veca.vecb)vecb)/b^2,vecc= ((veca.vecb)vecb)/b^2

The formula (veca + vecb)^(2)= a^(-2) + vecb^(2) + 2veca xx vecb valid for non zero vectors veca and vecb .

Show that the vectors 2veca-vecb+3vecc, veca+vecb-2vecc and veca+vecb-3vecc are non-coplanar vectors (where veca, vecb, vecc are non-coplanar vectors).

The vector vecA and vecB are such that |vecA+vecB|=|vecA-vecB| . The angle between vectors vecA and vecB is -

Vector veca is prepedicular to vecb , componets of veca- vecb along veca + vecb will be :

STATEMENT-1 Let veca,vecb bo two vectors such that veca.vecb=0 , then veca and vecb are perpendicular. And STATEMENT-2 Two non-zero vectors are perpendicular if and only if their dot product is zero.