If |A→ ×B→| = 3A→.B→,  then  the  value  of  |A→+B→|  is

|A→×B→| =3A→.B→ABsinθ = 3    ABcosθ ⇒tanθ = 3    ∴θ = 600Now |R→| = |A→+B→| = A2+B2+2ABcosθ= A2+B2+2AB12 = A2+B2+AB12