Question 1178661
<pre>

{{{drawing(400,2800/9,-6,3,-3,4,

graph(400,2800/9,-6,3,-3,4), 

line(0,0,-2.5sqrt(3),2.5), 

line(-2.5sqrt(3),2.5,-4,2.5), line(-2.5sqrt(3),2.5,-4.2,2.25),

red(line(-2.5sqrt(3),2.5,-2.5sqrt(3),0),

line(-4.2,2.4,-2.5sqrt(3),2.5),line(-4.5,2.3,-2.5sqrt(3),2.5)),

green(line(0,0,-2.5sqrt(3),0),line(-4,-.2,-2.5sqrt(3),0),line(-4,.2,-2.5sqrt(3),0), locate(-3,.9,-expr(5/2)*sqrt(3)) ),

red(locate(-4.6,1.4,5/2)),

locate(1,2.2,150^o),locate(-1.4,.6,30^o),
locate(-2.3,1.7,5),

red(arc(0,0,4,-4,0,150)) )}}}

The black vector <i>v</i> is 5 units long.

The right triangle formed is a 30-60-90 right triangle.
Its hypotenuse is 5, the length of the black vector.
The length of the short leg, the red vector, is 1/2 the hypotenuse or 5/2.
It is +5/2 because it goes up.
The length of the long leg, the green vector, is √3 times the short leg,
which is -(5/2)√3.
It is negative because it goes left.

The black vector is the vector sum of the green vector plus the red vector.

The green vector is {{{-expr(5/2)*sqrt(3)}}}<font size=5><i>i</i></font>.

The red vector is {{{5/2}}}<font size=5><i>j</i></font> 

So

<font size=5><i>v</i></font> = {{{-expr(5/2)*sqrt(3)}}}<font size=5><i>i</i></font> + {{{5/2}}}<font size=5><i>j</i></font>  <--ANSWER

Edwin</pre>