If the unit vectors ` veca and vecb ` are inclined of an angle ` 2 theta` such that ` |veca -vecb| lt 1 and 0 le theta le pi` then ` theta` in the interval

If unit vectors veca and vecb are inclined at an angle 2 theta such that |veca - vecb| lt 1 and 0 le theta le pi , then theta lies in the interval

If the unit vectors veca and vecb are inclined at an angle 2theta such that |veca-vecb|lt1 and 0lethetalepi then theta lies in the intervasl. (A) [0,pi/6] (B) (5pi/6,pi] (C) [pi/2,5pi/6] (D) [pi/6,pi/2]

If veca and vecb are unit vectors inclined at an angle theta , then the value of |veca-vecb| is

Let veca and vecb are unit vectors inclined at an angle alpha to each other , if |veca+vecb| lt 1 then

If the unit vectors vec(e_(1)) and vec(e_(2)) are inclined at an angle 2theta and |vec(e_(1)) - vec(e_(2))| lt 2 , then for theta in [0,pi], theta may lie in the interval

If two vectors vecA and vecB are at an angle theta ne 0^(@) , then

Vectors veca and vec b are inclined at an angle theta = 120^@ . If |veca| = |vecb| = 2 , then [(veca + 3vecb) xx (3veca + vecb)]^2 is equal to

If veca and vecb are two vectors and angle between them is theta , then

vectors veca and vecb are inclined at an angle theta = 60^(@). " If " |veca|=1, |vecb| =2 , " then " [ (veca + 3vecb) xx ( 3 veca -vecb)] ^(2) is equal to