`bbv = 3(cos(60^@) bbi + sin(60^@) bbj)` Find the magnitude and direction angle of the vector `bbv`.

The magnitude of a vector `v=v_x*i + v_y*j` is given by the following formula, such that:

`|v| = sqrt(v_x^2+v_y^2)`

`|v| = sqrt(9cos^2 60^o + 9sin^2 60^o)`

`|v| = sqrt(9(cos^2 60^o + sin^2 60^o))`

`|v| = sqrt(9*1)`

`|v| = 3`

You may evaluate the direction angle of the vector v, such that:

`tan alpha = (v_y)/(v_x)`

`tan alpha = (3 sin 60^o)/(3 cos 60^o)`

`tan alpha = tan 60^o => alpha = 60^o`

Hence, evaluating the magnitude and the direction angle of the vector v, yields `|v| = 3` and  `alpha = 60^o.`