# How to know if a curve is plane without calculating its torsion

Given $$\alpha(t)=\left(t,\frac{1+t}{t},\frac{1-t^2}{t}\right)$$ I want to know if there is way of knowing if this curve is plane or not without calculating its torsion.

I considered the option of trying to know if its contained in a plane. But I don't know how to proceed. Any ideas? Thanks in advance.

$${{1+t}\over t}-{{1-t^2}\over t}-t=1$$ so the curve is in the plane $$-x+y-z=1$$

If we can’t guess the equation by inspection, we can proceed as follows by three points on the line

• $$\alpha(1)=(1,2,0)$$
• $$\alpha(-1)=(-1,0,0)$$
• $$\alpha(2)=(2,3/2,-3/2)$$

and the $$2$$ vectors

• $$v_1=\alpha(1)-\alpha(-1)=(2,2,0)$$
• $$v_2=\alpha(2)-\alpha(-1)=(3,3/2,-3/2)$$

then since

$$v_1\times v_2=(-3,3,-3)$$

to determine if the curve is contained in a plane, we need to check if $$x-y+z$$ is a constant.

• Equivalently, since the three points are not colinear, check that the matrix $\small{\begin{bmatrix}\alpha(t)&1\\\alpha(1)&1\\\alpha(-1)&1\\\alpha(2)&1\end{bmatrix}}$ is singular. – amd Oct 18 at 8:05