Any number of vectors are said to be * collinear vectors* when all of them are parallel to the same line whatever their magnitudes may be. Otherwise, the vectors are said to be

*.*

**non collinear**Any vector $\overrightarrow{r}$ collinear with a given vector $\overrightarrow{a}$ can be expressed as \[\overrightarrow{r}=k\:\overrightarrow{a}\] where, $k$ is a scalar quantity.

## Two Dimensional Vectors

Consider two non zero and non collinear vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ and two scalars $x$ and $y$ such that \[x\:\overrightarrow{a}+y\:\overrightarrow{b}=0\] then, $x=0$ and $y=0$.

## Decomposition of a Vector in a Plane

If $\overrightarrow{a}$ and $\overrightarrow{b}$ be two given vectors, then every vector $\overrightarrow{r}$ in the plane can uniquely be represented as the sum of the two vectors parallel to $\overrightarrow{a}$ and $\overrightarrow{b}$. \[\text{i.e.}\;\;\overrightarrow{r}=x\:\overrightarrow{a}+y\:\overrightarrow{b}\] where, $x$ and $y$ are scalars.

Let $\overrightarrow{OA}=\overrightarrow{a}$ and $\overrightarrow{OB}=\overrightarrow{b}$ be two given non collinear vectors. Again, let $\overrightarrow{OP}=\overrightarrow{r}$ be a vector in the plane of $\overrightarrow{a}$ and $\overrightarrow{b}$. Through $P$, draw $PM$ and $PN$ parallel to $BO$ and $AO$ respectively.

Since $\overrightarrow{OM}$ and $\overrightarrow{OA}$ are collinear, we have \[\overrightarrow{OM}=x\:\overrightarrow{OA}=x\:\overrightarrow{a}\]

Similarly, \[\overrightarrow{ON}=y\:\overrightarrow{OB}=y\:\overrightarrow{b}\] where $x$ and $y$ are scalars.

Now, by parallelogram law of vector addition, we have, \[\overrightarrow{OP}=\overrightarrow{OM}+\overrightarrow{ON}=x\:\overrightarrow{a}+y\:\overrightarrow{b}\] \[\therefore\overrightarrow{r}=x\:\overrightarrow{a}+y\:\overrightarrow{b}\;\text{__(1)}\]

Now, we shall show that the relation is unique. For this, if possible, let \[\overrightarrow{r}=x’\:\overrightarrow{a}+y’\:\overrightarrow{b}\;\text{__(2)}\]

where $x’$ and $y’$ are scalars different from $x$ and $y$ respectively.

From $\text{(1)}$ and $\text{(1)}$, we have, \[x\:\overrightarrow{a}+y\:\overrightarrow{b}=x’\:\overrightarrow{a}+y’\:\overrightarrow{b}\] \[(x-x’)\:\overrightarrow{a}+(y-y’)\:\overrightarrow{b}=0\]

Since $\overrightarrow{a}$ and $\overrightarrow{b}$ are non zero and non collinear, so \[\begin{array}{c}\therefore x-x’=0&\therefore x=x’\\ \therefore y-y’=0&\therefore y=y’\end{array}\]

Hence, the relation is unique.

**Previous:** Unit Vector

**Next:** Coplanar Vectors