How do I find the magnitude and direction angle of the vector v =6i-6j?
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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##
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