Angle between the vectors `vecA and vecB` is `theta`. By resolving them into mutually perpendicular components show that the magnitude of the resultant vector is `(A^2+B^2+2AB cos theta )^(1//2)`.

Find the angle between the vectors vecA=2hati+3hatj and vecB=-3hati+2hatj .

if veca, vecb and vecc are there mutually perpendicular unit vectors and veca ia a unit vector then find the value of |2veca+ vecb + vecc |^2

If veca and vecb are mutually perpendicular unit vectors, then the value of (3veca-4vecb)*(2veca+5vecb) is -

The maximum magnitude of the resultant of two vectors, vec(P) and vec(Q) ("where" Pgt Q) is x times the minimum magnitude ot the resultant. When the angle between vec(P) and vec(Q)" ""is" " "theta , the magnitude of the resultant is equal to half the sum of the magnitudes of the two vectors. Prove that, cos theta =(x^2+2)/(2(1-x^2))*

Show that, the resultant and the difference of two mutually perpendicular vectors of equal magnitude are equal in magnitude and perpendicular to each other.

vecR is the resultant of the vectors vecA and vecB . If R =B/(sqrt(2)) then the angle theta is

If veca,vecb,vecc are mutually perpendicular vectors of equal magnitude show that veca+vecb+vecc is equally inclined to veca, vecb and vecc

Let veca ,vecb and vecc be pairwise mutually perpendicular vectors, such that |veca|=2, |vecb|=3, |vecc| = 6 , the find the length of veca +vecb + vecc .

If veca,vecb and vecc are three mutually perpendicular vectors of equal magnitude, show that vec a + vecb + vecc is equally inclined to veca, vecb and vecc . Also find the angle.