assume that two vector given with relationship below: $$\vec{n}\cdot \vec{u} = 0 $$ Then $\vec{n}$ and $\vec{u}$ are orthogonal vectors. Assume that we now have the direction of $\vec{n}$
Then there are infinitely many possible direction for $\vec{u}$
How do we choose the vector $\vec{u}$?
When we consider vector valued functions, we choose Normal vector to be directed inner side of the curve.
When we consider grad $\nabla$ we know that it is perpendicular to any level curve. Then we choose $\nabla$ to be directed towards outward of the curve.
How do we decide direction? What is the algorithm behind that? What am I missing? Or is it application dependent?
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$\begingroup$It is very much application dependent. Note that in $\Bbb{R}^3$ there are infinitely many vectors orthogonal to a given vector, not just $4$. Indeed, if $\vec{n} = (a,b,c)$ and $\vec{u} = (x,y,z)$, then $$ 0 = \vec{n}\cdot\vec{u} = ax + by + cz $$ is the equation of a plane through the origin.
In the case of the normal vector $\vec{n}$ to a curve $\gamma$ at some point $p$, you can describe the choice in the following (impractical) way:
- Consider the tangent vector $\vec{t}$ at $\gamma$ in $p$.
- The unit vectors orthogonal to $\vec{t}$ parametrise all the planes containing $\vec{t}$. Consider the osculating plane $\pi$ of $\gamma$ at $p$ (which is uniquely determined).
- Unless $\vec{n} = 0$ (in which case $\gamma$ has a flex point at $p$), the line through $p$ with direction $\vec{t}$ divides $\pi$ in two half-planes. The projection of $\gamma$ on $\pi$ is contained in one of them, which we call $H$.
- The vector $\vec{n}/\|\vec{n}\|$ is uniquely determined as the unit vector orthogonal to $\vec{t}$ which is contained in $H$.
There is a special case for twice-differentiable curves acting in $R^3$ For any point of such a curve you can explicitly define an orthonormal basis in $R^3$ that includes the normalized direction and two orthonormal vectors.
The trick is to realize that if $f''$ exists and is and not zero
$(f'(t) \cdot f'(t))' = 0 = 2f''(t)\cdot f'(t) = 0$
and it follows that
$(\frac{f'(t)}{||f'(t)||}, \frac{f''(t)}{||f''(t)||}, \frac{f'(t)}{||f'(t)||} \times \frac{f''(t)}{||f''(t)||})$
is an orthonormal basis in $R^3$.
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