Find the magnitude of two vectors \(\vec{a}\) and \(\vec{b}\), having the same magnitude and such that the angle between them is 60º and their scalar product is 1/2.

|\(\vec{a}\)| =|\(\vec{b}\)|= 1

Let θ be the angle between the vectors \(\vec{a}\) and \(\vec{b}\) ,

It is given that |\(\vec{a}\)| =  |\(\vec{b}\)|and (\(\vec{a}\) .\(\vec{b}\) ) = 1/2,  also θ = 60º

we know that  (\(\vec{a}\) .\(\vec{b}\) ) = (|\(\vec{a}\)| .|\(\vec{b}\)|)/cos(θ )

so,                        1/2 = (|\(\vec{a}\)| .|\(\vec{a}\)|)/cos(60º )         (since |\(\vec{a}\)|  =  |\(\vec{b}\)|)

                                 ⇒|\(\vec{a}\)|² = 1

                                 ⇒|\(\vec{a}\)| =|\(\vec{b}\)|= 1