How do i draw a vector diagram and find magnitude for this question?

A car is driven 215km east and then 85 km northeast. How do i draw a vector diagram and find its magnitude of the resultant displacement?|||Decide on some accceptable scale: perhaps 1cm = 10km


On your cartesian plane take the +X axis as being due east. On the X axis draw a line 21.5cm long.This will represent the 215km east travel


At the end of this line, at the 21.5cm mark, make an angle of 45° which is in the nort east direction. Draw a second line 8.5cm long. This will represent the 85km nort east travel.


Now join the origin to this end mark on the NE line. This line is the displacement of the car from the start in magnitude and direction. Measure this line in cm, multiply by 10 to give you the magnitude of the displacement, and with a protractor, measure the angle between the line and the X-axis, which is the direction of the resultant displacement.|||plot head to tail usin' the xy plane,


then, just use the cosine law to calculate the magnitude.|||First draw a normal x/y plane. Draw a vector of magnitude 215km (vector "a") directed to the right along the x axis. Starting from the tip of that vector, draw another vector of magnitude 85km (vector "b") directed 45 degrees above horizontal.





Splitting vector "b" into components b(x)= bcos(45) and b(y) =bsin(45)


We can now see a right triangle. Base = a + bcos(45) Height = bsin(45)


we will call the resultant displacement vector "x"


using the pythagorean theorem:


x= Square Root [ (a +bcos(45) )^2 + (bsin(45) )^2 ) ]


finding that x = 281.593 km





we can find the angle of this resultant vector by saying:


tan(Theta) = bsin45 / (a+bcos45)


take the arctan of both sides... we find that


Theta = 12.32 degrees





Hope that helps.