How can I calculate the equation of a curve if it's tangent was given!

Question

I need the equation for curve (not straight line!!) that passes through $(1,-1,0)$.

The tangent to the curve at that point is vector $-3\vec{i} +\vec{j} +2\vec{k}$.

The answer is parametric ie $x(t) =$ , $y(t)=$ , $z(t) =$ . Thanks

Answer 1

As I said in my comment, there is an infinity of such curves.

Here is a general solution. Begin with the line :

$$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix} \ \ 1\\-1\\ \ \ 0\end{pmatrix}+t\begin{pmatrix}-3\\ \ \ 1\\ \ \ 2\end{pmatrix}$$

then "bend" it by adding functions:

$$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1-3t+\color{red}{f(t)}\\-1+t+\color{red}{g(t)}\\2t+\color{red}{h(t)}\end{pmatrix}$$

such that $f(0)=g(0)=h(0)=0$ and $f'(0)=g'(0)=h'(0)=0$ in order that neither the point nor its tangent are impacted.

For example, take $\color{red}{f(t)=t^2, g(t)=t^3, h(t)=0},$ which corresponds to a Bezier curve.