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aggiesrul1 · Feb 13, 2007, 09:04 PM

:confused: does anyone know how to find the magnitude of a vector because I can add and subtract them and all that stuff but I can not find the magnitude and the heading... so if someone could help me that would be nice!:confused:

Capuchin · Feb 13, 2007, 11:19 PM

As Pythagoras' theorem shows: the magnitude of a vector V is
$V^2 = V_x^2 + V_y^2$

For the heading you need to use basic trigonometry, form a triangle and use the lengths you know to work out the angle you need.

asterisk_man · Feb 15, 2007, 06:28 PM

Capuchin's answer is correct but I thought you might like to have it spelled out a bit more.
Lets assume you have a vector V which starts at $\left(x_1 , y_1\right)$ and ends at $\left(x_2 , y_2 \right)$
$\left|V\right|$ means "magnitude of V"
$\left|V\right|=\sqrt {{\left(x_2 - x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

The direction can be found like capuchin said, with trig.

Here's the vector V starting at (x1,y1) and ending at (x2,y2).

Code:

           (x2,y2)
           /|
          / |
        V/  |dy
        /   |
       /----|
(x1,y1)  dx

$dx=x_2 - x_1 \\<br /> dy=y_2 - y_1<br />$

Now you want the angle of the vector V. I found this image that will explain what to do. Replace the x and y with the dx and dy you found above. $\theta$ is the angle of V.

As an aside, Capuchin, |V| is the magnitude of V, do you recall what symbol is used for the angle of V? I feel like there is something but I can only recall how to denote the unit vector.

Capuchin · Feb 16, 2007, 12:30 AM

I don't believe there is one, you just use I, j and k, the direction information is there :/