Finding magnitude (length) and direction (angle) of a vector

[coooool222]

The [imath]x[/imath]- and [imath]y[/imath]-components of a vector [imath]\overrightarrow{P}[/imath] are given as [imath]P_x = -1\textrm{ in.}[/imath] and [imath]P_y = -2\textrm{ in.}[/imath] Find the magnitude and direction, and write the vector [imath]\overrightarrow{P}[/imath] is its rectangular and polar forms.





So i found the magnitue which is
(-1)^2 + (-2)^2 = P^2 =
Sqrt(5)

Then I used the inverse tan function to find the angle (direction)
theta = arctan (-2/-1) = 63.8 degrees

Im confused with my 63.8 degrees since the angle in the graph looks greater than 63.4 degrees

I subtracted 180 by 63.8 and got 116.6

Since it's going clock wise it's -116.6

Am i right?

 

[blamocur]

I get the same answer. In some programming language (C, Python) libraries there is arctan2(y,x) function which uses y/x but also takes into account the signs of 'x' and 'y'.

 

[Steven G]

(-1)^2 + (-2)^2 = P^2 = Sqrt(5) That is NOT true and you know that! Equal signs must be valid!

 

[Steven G]

The [imath]x[/imath]- and [imath]y[/imath]-components of a vector [imath]\overrightarrow{P}[/imath] are given as [imath]P_x = -1\textrm{ in.}[/imath] and [imath]P_y = -2\textrm{ in.}[/imath] Find the magnitude and direction, and write the vector [imath]\overrightarrow{P}[/imath] is its rectangular and polar forms.


View attachment 36564
Then I used the inverse tan function to find the angle (direction)
theta = arctan (-2/-1) = 63.8 degrees

You concluded that an angle is 63.8 degrees. Now you need to figure out which angle is 63.8. Subtracting 63.8 from 180 may or may not be right. It seems that you are just guessing. Please tell us which angle is 63.8. Just ignore that arc labeled theta

 

[pka]

The [imath]x[/imath]- and [imath]y[/imath]-components of a vector [imath]\overrightarrow{P}[/imath] are given as [imath]P_x = -1\textrm{ in.}[/imath] and [imath]P_y = -2\textrm{ in.}[/imath] Find the magnitude and direction, and write the vector [imath]\overrightarrow{P}[/imath] is its rectangular and polar forms.


View attachment 36564

If [imath](x,y)\ne(0,0)[/imath] is the 'tip' of a vector,
then if [imath]x\,\cdot\,y\,\ne 0~\&~\tau=\arctan\left|\dfrac{y}{x}\right|[/imath] then the argument of the vector is
[imath]\theta = \left\{ \begin{gathered} \tau ,\quad (x,y) \in I \\ \pi - \tau ,\quad (x,y) \in II \\ -\pi + \tau ,\quad (x,y) \in III \\ - \tau ,\quad (x,y) \in IV \\ \end{gathered} \right.[/imath]
The length is [imath]\ell=\sqrt{x^2+y^2}[/imath][imath][/imath]