# A better definition of InfinitePlane [duplicate]

Is there any reason why an InfinitePlane needs a point and two vectors? One would be sufficient. (The one perpendicular to the plane)

• Because in more than three dimensions, a normal and a point do not uniquely determine a plane. Commented May 18, 2020 at 10:54
• As noted in the answer below, you can use Hyperplane if you have a normal vector n and point p. An equivalent form would be InfinitePlane[p, NullSpace[{n}]]. Commented May 18, 2020 at 13:36
• Possible duplicate of How can I create a plane using a point and normal vector? Commented May 18, 2020 at 14:16

• Since the plane is infinite, orientation doesn't matter, they will all be the same plane regardless of how you turn it about the axis, I think. Also, consider Hyperplane which works the way OP suggested. You could say the same about the hyperplane, so why does this work for Hyperplane and not for InfinitePlane? Commented May 18, 2020 at 10:48
• In $n$ dimensions, Hyperplane[] is $n-1$ dimensional, but InfinitePlane[] is always two-dimensional. It is a fortuitous coincidence that these two concepts are equivalent when $n=3$. Commented May 18, 2020 at 10:56
• @J.M. Exactly, they are equivalent for $n = 3$. That is what I was hoping to confirm. Consequently, WRI could have supported the point + normal syntax for InfinitePlane, as they did for Hyperplane, if they wanted to, they just chose not to. Commented May 18, 2020 at 11:31