Is there a method to determine which quadrant a vector is in?

Question

For example, Bob pulls with a force of $250$ Newtons due West and Tim pulls with a force of $500$ Newtons due North. What is the resultant vector?

I use the Pythagorean theorem on $250$ and $500$. Then I find $~\tan^{-1}\left(\frac{500}{250}\right)$, which equals $27^{\circ}$. But then I have to add $27^{\circ}$ to $90^{\circ}$. The answer is $117^{\circ}$. But how do I know when to add or subtract to/from $90^{\circ}$. What about $180^{\circ}$, $270^{\circ}$, or $360^{\circ}$?

Answer 1

You can use the signal for the quadrants of the sum of two vectors in the following way:

• N and E will result in a vector in the first quadrant (both coordinates positive);

• N and W will result in a vector in the second quadrant (the horizontal coordinate is negative);

• S and W will result in a vector in the third quadrant (both coordinates negative);

• S and E will result in a vector in the fourth quadrant (the vertical coordinate is negative).

And you may also use the magnitude for the octants: If the absolute value of the vertical coordinate is greater than the absolute value of the horizontal coordinate, then you are in the second (NNE), third (NNW), sixth (SSW) or seventh (SSE) octant.

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Edit:

If you compute the angle using the absolute value of the horizontal coordinate $x$ and the absolute value of the vertical coordinate $y$:

$$ \theta = \arctan{\frac{|y|}{|x|}} $$

then the "true" angle is:

• $\theta$ if you are in the first quadrant;

• $90^{\circ} - \theta$ if you are in the second quadrant (if $x$ is positive but very close to zero, that angle will be the vertical symmetric of the angle if the horizontal coordinate was $-x$);

• $180^{\circ} + \theta$ if you are in the first quadrant (if you switch the signal of both coordinates, the angle "points" in the opposite direction, creating a straight line);

• $270^{\circ} - \theta$ (for a similar reason to an angle in the second quadrant).

Answer 2

Draw a force diagram. You have a vector of length $500$ going north and one of length $250$ going east.

Complete the parallelogram and draw the resultant vector. It will naturally be in the correct quadrant already.

Now you just need to measure the angle from the correct starting direction. When I see north and east components I'm usually in an actual geographic system and want to measure the compass direction, which is measured clockwise from due north. But you seem to be working in a system where you have to measure counterclockwise from due east.

In more advanced physics classes you will rarely see north, south, east, or west. You will just have $x$ and $y$ components (plus a $z$ component for a three-dimensional problem) and you will almost never need to ask what quadrant you're in.