Given two non-collinear vectors, Is it necessary that they define a plane? I read somewhere that any two non-collinear vectors define a plane. But I dont seem to understand that how can two skew vectors define a plane.
 A: I think your problem is with the geometric picture of a vector as an arrow. If your idea is that a vector can be anywhere in space then the two vectors can lie on skew lines. Then your are right: they don't determine a plane. That's often a useful view of vectors in physics.
But ... the convention when you're studying vectors in mathematics is that (in the picture you imagine) they all start out at the origin of the coordinate system. Then you should "see" how two vectors that lie on two different lines through the origin will span a plane. 
Once you have the picture right the other answers (with the algebra) should make more sense.
A: Since the two vectors are non collinear... their cross product is not zero but a vector which is perpendicular to the plane defined by the given two vectors....
A: If you call "plane" a $2-$dimensional vectorial space, then for some vectorial space $V$ if you have two non-colinear vectors $v_1\in V$ and $v_2\in V$, they are a base of $P=\mathrm{Span}(v_1,v_2)$. Indeed, the family $(v_1,v_2)$ is :


*

*free : if $\lambda_1v_1+\lambda_2v_2=0$ with $\lambda_1\neq0$, then $v_1=\frac{\lambda_2}{\lambda_1}v_2$ and your vectors would be colinear, same argument if $\lambda_2\neq 0,$

*generating : by definition.
