Finding a Unit Vector Along $\vec{A}$ - Please Help!

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
5 replies · 2K views
Drain Brain
Messages
143
Reaction score
0
Hello! I need help with this problem

how do you find a unit vector along the direction $\vec{A}=2\hat{a_{\rho}}-z\hat{a_{z}}$(cylindrical)?

do I have to convert it to Cartesian or there is a direct method? please help! Thanks!
 
Physics news on Phys.org
The vector itself is responsible for the direction. All you need to do is divide the vector by its magnitude:

$$\vec{u} = \frac{\vec{A}}{\| \vec{A}\|} = \frac{2 \vec{a}_{\rho} - z \vec{a}_z}{\sqrt{4+z^2}}.$$
 
Fantini said:
The vector itself is responsible for the direction. All you need to do is divide the vector by its magnitude:

$$\vec{u} = \frac{\vec{A}}{\| \vec{A}\|} = \frac{2 \vec{a}_{\rho} - z \vec{a}_z}{\sqrt{4+z^2}}.$$
HI fantini
$\vec{A}=2\hat{a_{\rho}}-z\hat{a_{z}}$ <--- this vector is in cylindrical form. Why did you directly get the unit vector from a cylindrical form? Is that valid?
 
Yes, it is valid. This is because cylindrical coordinates form an ortonormal system. Thus the length of the vector does not change from cartesian. You can compute the length as you would normally. :)
 
Fantini said:
Yes, it is valid. This is because cylindrical coordinates form an ortonormal system. Thus the length of the vector does not change from cartesian. You can compute the length as you would normally. :)

does this also holds for spherical form?
 
Drain Brain said:
does this also holds for spherical form?

Can you prove if spherical co-ordinates form an orthonormal system?