For line to be tangent to a given curve, they should pass through a common point and both should have same slope at that point. But how do we compare slope in 3D?

E.g.=> For a circle in 2-Dimensions, we know which lines will be it's tangents. But what if I ask, which lines will be tangent to it in 3-Dimensions? Will the answer be same for both cases, or will many others lines, lying in some other planes, also be included?

All I could think of is that, we need to find tangents in planes, and in every plane the curve will be set of points, except the one in which the curve is a circle. And because slope of a point is not defined, there will be no new tangents added when we go from 2D to 3D.

  • $\begingroup$ I truly do not understand how a circle becomes a "ring (without thickness)" when we look at it in 3-D. It's still a curve. You need to explain what you're actually doing here. $\endgroup$ Aug 23, 2019 at 22:48
  • $\begingroup$ @Sciencisco If you deem a given answer satisfactory, please mark it as the accepted answer by clicking the tick mark to the left of the answer body (it'll turn green once you click it). This way the question will not show up in the "unanswered queue" in the future. $\endgroup$ Aug 24, 2019 at 9:30
  • $\begingroup$ @G.Chiusole I know it very well, Relax. I'm just looking if someone else has a better, more satisfying answer. I'll surely mark yours accepted after some hours. $\endgroup$ Aug 24, 2019 at 10:06
  • $\begingroup$ @Sciencisco No worries. Just making sure, since questions often unnecessary end up in the unanswered queue. $\endgroup$ Aug 24, 2019 at 10:08

2 Answers 2


Formally, a curve in $\mathbb{R}^3$ is a continuous map $\gamma: [0,1] \rightarrow \mathbb{R}^3$ defined by $t \mapsto (\gamma_1(t), \gamma_2(t), \gamma_3(t))$. Then one can define the tangent vector to this curve at a point $t \in [0,1]$ to be

$$ \gamma'(t) = \lim_{h \rightarrow 0, h \neq 0} \frac{\gamma(t+h) - \gamma(t)}{h} = (\gamma_1'(t), \gamma_2'(t), \gamma_3'(t))$$

For a curve in $\mathbb{R}^2$ this amounts to the usual idea of a derivative (slope to a curve): enter image description here

For curves in $\mathbb{R}^3$ however, we get the following: (here the red vector is the tangent)enter image description here

Also, if you were to consider not curves but surfaces in $\mathbb{R}^3$ a simple "line" would not be sufficient as a tangent. You will need a plane. For a much more general treatise of these things have a look here.

  • $\begingroup$ Can you please clarify on the example of circle I asked? Your answer suggest, there will be no new tangent lines. $\endgroup$ Aug 23, 2019 at 18:22
  • $\begingroup$ I think one good way of viewing 'the' tangent line is that it's the intersection of all the planes which are tangent to the curve at that point. $\endgroup$ Aug 23, 2019 at 18:25
  • $\begingroup$ @Sciencisco That is the case, yes. What is defined as “slope” at a point $x$ for a graph is the tangent vector for a more general curve. With this definition, there is only one such line. $\endgroup$ Aug 23, 2019 at 18:26
  • $\begingroup$ Maybe think of it this way: the tangent line should be the path a particle takes when no forces act upon it anymore. For a circle that’s clear. For a curve in 3D that’s exactly what the tangent vector gives $\endgroup$ Aug 23, 2019 at 18:27

You can define slope at a point of a curve just as you did in calculus 1 - take the limit of approximating secant lines. The only difference this time is that the limiting object will be a line pointing in, well, any direction. So, at this point, it is best to use vectors.

Two curves in space will be tangent to eachother at a point if they both pass through this point and if they both have the same tangent vectors at this point.

Now, with a little bit of imagination, you can see why you will have new curves which are tangent to the circle beyond the curves which are coplanar with the circle.


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