Can a plane be defined by a vector on the plane and a point on the plane? I can find many references that provide for a definition of a plane given a normal vector and a point, for example, this is explained in this question.  However, can a plane be defined by a vector on the plane, perpendicular to its normal vector, and a point on the plane (that is not coincident with the vector)?
I would think so since from the vector, one can deduce two points on the plane, which when combined with the given point, gives three points, which defines a plane.
 A: In a vector space if a vector stay on a plane than the plane passes thorough the origin.
The other point define another vector and, if the two vectors are linearly independent,  the plane is the span of the two vectors.
If we are in an affine space than a vector on a plane is defined as the ''difference'' between two points of this plane, so we know two point ( the tail and the tip of the vector) and if we know also another (not aligned) point,  we have three points and the plane is well defined.
A: Answer is NO. Vector is family of line segments with same length and direction. So with one vector and a point you define a family of planes all of which contain the line which is passing through the point and has the direction of the vector.
But if you mean by vector - particular line segment, then see Emilio's answer
A: No. A vector (direction) and a point define a line in space. A line can have a family of planes where the line is tangent, or a family of planes where the line is perpendicular.
To fully specify the plane you need another piece of information, such as the minimum distance of the plane to the origin, or another point where the plane goes through.
Mathematically to define a plane you need two items


*

*The plane normal (vector $\mathbf{n}$)

*The minimum distance to the origin (scalar $d$)


Then a point $\mathbf{r} = \pmatrix{x\\y\\z}$ belongs to the plane if $$ \mathbf{n} \cdot \mathbf{r} = d$$


*

*To define a plane using the normal and another point $\mathbf{p}$, use $d = \mathbf{n} \cdot \mathbf{p}$.

*To define a plane using a point $\mathbf{p}$, a tangent vector $\mathbf{e}$ and another point $\mathbf{q}$, use the vector normal $\mathbf{n} = (\mathbf{p}-\mathbf{q}) \times \mathbf{e}$ and the distance $d = \mathbf{n} \cdot \mathbf{p}$.
The second point $\mathbf{q}$ is required to define the plane normal.
