- #1
jimbo_durham
- 13
- 0
Hi, i have a known magnitude to give my vector in an xy plane, and i have a desired direction. I need the (vx, vy, vz=0) to describe my vector. I am sure this can be done easily.
an example is,
i have a point at (10,10,0) in cartesian (x,y,z) and will use this as the starting point of my vector. This vector must have a magnitude of 4, and must be in a direction along the line connecting the point (10,10,0) with the origin (0,0,0).
ie my vector has the magnitude 5 and direciton that of a vector (-10,-10,0).
in this example the solution is simply to write
[tex]
z^{2}=x^{2}+y^{2}[/tex], and as x=y,
[tex]
\sqrt{ \frac{z^{2}}{2} }=x=3.5
[/tex] ish.
giving me vector with components (3.5,3.5,0)
this however is in the wrong direction (need (-3.5,-3.5,0))
however if x=/=y, how is this solved? and how is the direction accounted for (+ve or -ve)?
an example is,
i have a point at (10,10,0) in cartesian (x,y,z) and will use this as the starting point of my vector. This vector must have a magnitude of 4, and must be in a direction along the line connecting the point (10,10,0) with the origin (0,0,0).
ie my vector has the magnitude 5 and direciton that of a vector (-10,-10,0).
in this example the solution is simply to write
[tex]
z^{2}=x^{2}+y^{2}[/tex], and as x=y,
[tex]
\sqrt{ \frac{z^{2}}{2} }=x=3.5
[/tex] ish.
giving me vector with components (3.5,3.5,0)
this however is in the wrong direction (need (-3.5,-3.5,0))
however if x=/=y, how is this solved? and how is the direction accounted for (+ve or -ve)?
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