Magnitude ##11,61m##, angle ##137,8^@## anticlockwise from unit vector ##hat i##
In rectangular form, we may simply add the corresponding components to get :
##vecA+vecB= (5,4-14) hat i + (1,4+6) hat j = -8,6 hat i + 7,8 hat j m ##
Hence the magnitude may be found from Pythagoras as
##|vecA + vecB|=sqrt((-8,6)^2+7,8^2) = 11,61 m##
The direction may be given by
##theta=tan^(-1)((7,8)/(-8,6))=-42,2^@##
Since tan is negative in both the 2nd and 4th quadrants, we see from the components of the resultant vector that we here in the 2nd quadrant. Hence a reference angle of 42,2 implies that the actual angle is 42,2 degrees above the negative x axis and hence in standard form it is ##180-42,2=137,8^@##
So in full vector form, ##vecA + vecB=11,61angle 137,8^@ m##
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