International Development homework help

International Development homework help. Magnitude = ##|6i-6j|=sqrt(6^2+(-6)^2)=sqrt(72)=6sqrt(2)”##
Angle = ##arctan(-6/6)=arctan(-1)=-pi/4=-45°##
If you draw this vector on the plane, the x-y coordinate axis, as (6,-6) because ##i## is the unit vector in the x-direction and ##j## in the y-direction. Then pictorially it might seem easier.
So if you draw the vector (so join the line between the point (6,-6) and the origin), and draw the perpendicular line between the point and the x-axis. Now, you should see a triangle, more specifically a right-angles triangle.
Now that you have you’re right-angles triangle you can use Pythagoras to find the magnitude of the vector (length of the straight line between the origin and the point (6,-6)) and the use SOHCAHTOA to find the angle between the x-axis and the vector (be careful, here if the vector is the on the left-hand side of the plane i.e. the x-coordinate is negative, then you will have to make sure the you take the angle from the positive side of the x-axis, so do ##pi/2(1-theta)##.
Some quick formulae that may be helpful to do quick calculation:
magnitude of ##ai+bj=|ai+bj|=sqrt(a^2+b^2)##
##theta=arctan(b/a)##
if ##b<0## then angle of ##ai+bj=pi/2(1-theta)## if ##bge0## then angle of ##ai+bj=theta##

International Development homework help