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 of an angle 2 theta such that |veca -vecb| lt 1 and 0 le theta le pi then theta in the interval

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 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

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

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

If veca and vecb are unit vectors inclined at an angle theta ,then prove that sin(theta/2)=1/2|veca-vecb|

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