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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.

Jack Talion
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2 Answers2

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

Tsemo Aristide
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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.

user
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  • 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 '19 at 08:05