If [math]a^{\rightarrow}[/math] and [math]b^{\rightarrow}[/math] be any two non-zero and non- collinear plane vectors then any vector [math]r^{\rightarrow}[/math] in their plane can be uniquely expressed as the sum of the vectors parallel to the vectors [math]a^{\rightarrow}[/math] and [math]b^{\rightarrow}[/math]. [br]Proof: Let [math]OA^{\rightarrow}=a^{\rightarrow}[/math],[math]OB^{\rightarrow}=b^{\rightarrow}[/math] and [math]OP^{\rightarrow}=r^{\rightarrow}[/math] be the vectors acting at the point O as shown in the figure. Draw PH and PG parallel to OB and OA respectively. By doing so, we obtain a parallelogram OHPG. Now by the parallelogram law of addition of vectors, [math]OP^{\rightarrow}=[/math] [math]OH^{\rightarrow}+OG^{\rightarrow}[/math] or [math]r^{\rightarrow}=OH^{\rightarrow}+OG^{\rightarrow}[/math]......................(i). But [math]OH^{\rightarrow}[/math] and [math]OG^{\rightarrow}[/math] are respectively parallel to the vectors [math]a^{\rightarrow}[/math] and [math]b^{\rightarrow}[/math]. So we can write [math]OH^{\rightarrow}=xa^{\rightarrow}[/math] and [math]OG^{\rightarrow}=yb^{\rightarrow}[/math]. Putting these values in the expression (i), we get, [math]r^{\rightarrow}=xa^{\rightarrow}+yb^{\rightarrow}[/math] which is the required result to prove the statement of the above theorem. For uniqueness, if possible, let [math]r^{\rightarrow}=[/math][math]x'a^{\rightarrow}+y'b^{\rightarrow}[/math] ..............(ii). From (i) and (ii), we get, [math]xa^{\rightarrow}+yb^{\rightarrow}=x'a^{\rightarrow}+y'b^{\rightarrow}[/math] or ([math]x-x'[/math])[math]a^{\rightarrow}+[/math]([math]y-y'[/math])[math]b^{\rightarrow}[/math] =0 which is possible only when [math]x-x'=y-y'=0[/math] or [math]x=x'[/math] and [math]y=y'[/math]. Hence the expression (i) is unique.

Any finite number of vectors are said to be coplanar if a plane can be drawn parallel to all of them. If this is not the case, then they are called non-coplanar vectors. As it is always possible to draw a parallel plane to any number of two-dimensional vectors, so all the two dimensional vectors are always coplanar.[br][b]Three-dimensional vectors or space vectors: [/b] All vectors other than the two dimensional vectors are called the space vectors. The space vectors do not lie on the same plane.[br][b]Theorem on Three dimensional vectors: [/b]If [math]a^{\rightarrow},b^{\rightarrow},c^{\rightarrow}[/math] be any three non-zero and non-collinear vectors and x,y,z be any three scalars then [math]xa^{\rightarrow}+yb^{\rightarrow}^{ }+zc^{\rightarrow}=0[/math] implies x = y = z = 0.

Any finite number of vectors are said to be coplanar if a plane can be drawn parallel to all of them. If this is not the case, then they are called non-coplanar vectors. As it is always possible to draw a parallel plane to any number of two-dimensional vectors, so all the two dimensional vectors are always coplanar.[br][b]Three-dimensional vectors or space vectors: [/b] All vectors other than the two dimensional vectors are called the space vectors. The space vectors do not lie on the same plane.[br][b]Theorem on Three dimensional vectors: [/b]If [math]a^{\rightarrow},b^{\rightarrow},c^{\rightarrow}[/math] be any three non-zero and non-collinear vectors and x,y,z be any three scalars then [math]xa^{\rightarrow}+yb^{\rightarrow}^{ }+zc^{\rightarrow}=0[/math] implies x = y = z = 0.