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

${{1+t}\over t}-{{1-t^2}\over t}-t=1$ 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.