draw points J and K in the coordinate axis, and join them so that the top of the vector is point K.
the magnitude of the vector is just the length of the segment JK, so we can either write it as |JK| or |KJ|, it does not make any change.
Let O be the point in the x axis such that KO is perpendicular to OJ, as shown in the picture.
From the Pythagorean theorem:
[tex]|KJ|= \sqrt{ |OK|^{2} + |OJ|^{2} } = \sqrt{3^{2} + 3^{2} } = \sqrt{2*3^{2}}=3 \sqrt{2} [/tex] units
since |OK|=|OJ|, the right triangle KOJ is isosceles, so the measure of KJO is 45°, which means that the angle of vector JK to the positive x axis is 180°-45°=135°
Answer: magnitude= [tex]3 \sqrt{2} [/tex] units, direction : 135°
