Scalars And Vectors

Collinear Vectors

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.

Collinear Vectors: Decomposition of a Vector in a Plane

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.