<p>If <span class="math-container">$\mathbf{\hat{n}}$</span> points to the north celestial pole, then the eastward tangent vector is</p>

<p><span class="math-container">$$\mathbf{\hat{u} = \frac{\hat{n} \times \hat{w}}{\|\hat{n} \times \hat{w}\|}}$$</span></p>

<p>and the northward tangent vector is</p>

<p><span class="math-container">$$\mathbf{\hat{v} = \frac{\hat{w} \times \hat{u}}{\|\hat{w} \times \hat{u}\|}}$$</span></p>

<p><span class="math-container">$\mathbf{\|\hat{w} \times \hat{u}\|}$</span> analytically should be 1 but can be slightly different numerically.</p>

CC BY-SA 4.0
2019-04-11T21:00:13.380Z
Bingo! renormalize, facepalm, repeat...
uhoh
CC BY-SA 4.0
2019-04-11T16:23:51.583Z
2019-04-11T16:23:51.583Z
Mike G
<p>I am stuck and a little embarrassed. </p>

<p>I'm trying to implement <a href="https://astronomy.stackexchange.com/a/26467/7982">this answer</a>. I have a unit vector <span class="math-container">$w$</span> pointing in the direction of a radio source on the celestial sphere.</p>

<p><span class="math-container">$u$</span> and <span class="math-container">$v$</span> are the two other, mutually orthogonal unit vectors that point <em>east</em> and <em>north</em> from that point, tangent to the sphere.</p>

<p>Without back-converting to R.A. and Dec or using trigonometry, is there a simple way to generate <span class="math-container">$u$</span> and <span class="math-container">$v$</span> vectorially?</p>

<p>Slide 38 <a href="https://www.strw.leidenuniv.nl/radioastronomy/lib/exe/fetch.php?media=radio_astronomy_lec_4_ma_garrett.pdf" rel="nofollow noreferrer">here</a> may or may not be helpful, I think my description is sufficient by itself.</p>

CC BY-SA 4.0
2019-04-11T11:42:59.357Z
How did you determine **w**?
Mike G
CC BY-SA 4.0
2019-04-11T11:49:28.300Z
@MikeG In my question, $w$ *is a given*. It can come from any source. If it comes from orbital mechanics, it's the normalized relative position vector $\mathbf{r}/|r|$, or it can come from an initial RA/Dec value that's been spun around the Earth's axis for 24 hours [like I did here](https://astronomy.stackexchange.com/q/30311/7982) when I built `SgrA_star` (which is what would be $w$), or it can be from somewhere else entirely. From any given $w$ I'm asking how to get $u$ and $v$ *without going back to trigonometry*. Can it be done using just vectors somehow?
uhoh
<mathematics>
Generate East and North unit vectors tangent to a point on the celestial sphere, using vectors?
uhoh
# Generate East and North unit vectors tangent to a point on the celestial sphere, using vectors?

(Asked by: username_0 on 2019-04-11)

I am stuck and a little embarrassed.

I'm trying to implement [this answer](https://astronomy.stackexchange.com/a/26467/7982). I have a unit vector $w$ pointing in the direction of a radio source on the celestial sphere.

$u$ and $v$ are the two other, mutually orthogonal unit vectors that point *east* and *north* from that point, tangent to the sphere.

Without back-converting to R.A. and Dec or using trigonometry, is there a simple way to generate $u$ and $v$ vectorially?

Slide 38 [here](https://www.strw.leidenuniv.nl/radioastronomy/lib/exe/fetch.php?media=radio_astronomy_lec_4_ma_garrett.pdf) may or may not be helpful, I think my description is sufficient by itself.

**Tags:** mathematics
> **Comments:**
> **username_1**: How did you determine **w**?  
> **username_0**: @MikeG In my question, $w$ *is a given*. It can come from any source. If it comes from orbital mechanics, it's the normalized relative position vector $\mathbf{r}/|r|$, or it can come from an initial RA/Dec value that's been spun around the Earth's axis for 24 hours [like I did here](https://astronomy.stackexchange.com/q/30311/7982) when I built `SgrA_star` (which is what would be $w$), or it can be from somewhere else entirely. From any given $w$ I'm asking how to get $u$ and $v$ *without going back to trigonometry*. Can it be done using just vectors somehow?  

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## 1 answers

### Answer by username_1 on 2019-04-11. Score: 1 | Accepted answer

If $\mathbf{\hat{n}}$ points to the north celestial pole, then the eastward tangent vector is

$$\mathbf{\hat{u} = \frac{\hat{n} \times \hat{w}}{\|\hat{n} \times \hat{w}\|}}$$

and the northward tangent vector is

$$\mathbf{\hat{v} = \frac{\hat{w} \times \hat{u}}{\|\hat{w} \times \hat{u}\|}}$$

$\mathbf{\|\hat{w} \times \hat{u}\|}$ analytically should be 1 but can be slightly different numerically.

> **Comments:**
> **username_0**: Bingo! renormalize, facepalm, repeat...  

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