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

[Drain Brain]

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!

 

[Fantini]

The vector itself is responsible for the direction. All you need to do is divide the vector by its magnitude:

$$\vec{u} = \frac{\vec{A}}{\| \vec{A}\|} = \frac{2 \vec{a}_{\rho} - z \vec{a}_z}{\sqrt{4+z^2}}.$$

 

[Drain Brain]

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?

 

[Fantini]

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. :)

 

[Drain Brain]

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?

 

[Prove It]

does this also holds for spherical form?

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