Question 1206661
<pre>
Draw the vector. Remember that negative angles are measured rotating
CLOCKWISE from the right side of the x-axis. And the length, or magnitude
of the vector is 4. 

{{{drawing(410,1600/7, -4.9,4.9,-2.8,2.8,
red(locate(.2,.75,-390^o)),
graph(410,1600/7, -4.9,4.9,-2.5,2.5),
line(0,0,4,-2),line(3.8,-1.7,4,-2) ,line(3.75,-2,4,-2) ,
red(arc(0,0,2.4,-2,-30,180),arc(-.3,0,1.8,-1.8,180,360)))
}}}

Draw a perpendicular from the tip of the vector to the x-axis:

{{{drawing(400,1600/7, -4.9,4.9,-2.8,2.8,

green(line(3.9,0,3.9,-1.9)),
red(locate(.2,.75,-390^o)),
graph(415,1600/7, -4.9,4.9,-2.5,2.5),
line(0,0,4,-2),line(3.8,-1.7,4,-2) ,line(3.75,-2,4,-2) ,
red(arc(0,0,2.4,-2,-30,180),arc(-.3,0,1.8,-1.8,180,360)))
}}}
{{{matrix(1,3,x=r*cos(theta), ",",y=r*sin(theta))}}}
{{{matrix(1,3,x=4*cos(-390^o), ",", y=4*sin(-390^o))}}}

To get an equivalent positive angle add 360 twice, or 720 to the angle,
getting -390+360+360 = 330 and the reference angle is 30

{{{matrix(1,3,x=4*cos(330^o), ",", y=4*sin(330^o))}}}

{{{matrix(1,3,x=4*(sqrt(3)/2), ",", y=4*(-1/2))}}}

{{{matrix(1,3,x=2sqrt(3), ",", y=-2)}}}

The horizontal-component is the x-value, a multiple of the unit vector i.
 
The vertical-component is the y-value, a multiple of the unit vector j.

Answer: {{{2sqrt(3)*i+-2*j}}}

Edwin</pre>