The vector representing the position of an object relative to an arbitrary origin is called the **position vector**. In other words, those vectors whose initial and terminal points are fixed are called position vectors. If $O$ is arbitrary origin and $P(x,y,z)$ is the position of a particle, then the position vector of particle $P$ relative to origin $O$ is given by \[\overrightarrow{OP}=(x,y,z)\]

## Vectors in terms of Coordinates

Let $XOX’$ and $YOY’$ be two mutually perpendicular lines representing x-axis and y-axis respectively. Let $P(x,y)$ be a point on the plane. Join $OP$. From $P$, draw $PM\perp OX$ and $PN\perp OY$.

Now, $OM$$=x$$=$ projection of $OP$ on the x-axis and, $ON$$=MP$$=y$$=$ projection of $OP$ on the y-axis. The directed line segment $OP$ is known as $\overrightarrow{OP}$ (read vector $OP$). $\overrightarrow{OP}$ is also defined by an ordered pair $(x,y)$.

Hence, $\overrightarrow{OP}$ is defined as the directed line segment from $O$ to $P$ or it is defined by an ordered pair with x-coordinate and y-coordinate as the projection of $OP$ on the x-axis and y-axis respectively.

There is no difference between the vector $(x,y)$ and the vector $\overrightarrow{OP}$ represented by the directed line segment. \[\therefore \overrightarrow{OP}=(x,y)\] The vector $\overrightarrow{OP}=(x,y)$ is a vector in the Cartesian plane, and hence the name **plane vector**.

We now see how an ordered triple $(x,y,z)$ represents a vector in three dimensional space.

Let $XOX’$, $YOY’$ and $ZOZ’$ represent three rectangular axes. Now, we can represent any vector by a directed line segment starting from the point $O$ and ending at a point $P$ in a direction parallel to the vector such that the length of $OP$ is equal to the magnitude of the vector.

Thus, the vector $\overrightarrow{OP}$ may be identified uniquely with the point $P(x,y,z)$. This vector $\overrightarrow{OP}=(x,y,z)$ is called the **space vector**.

## Position Vectors

The vectors $\overrightarrow{OP}=(x,y)$ and $\overrightarrow{OP}=(x,y,z)$ are also called **position vectors** as their initial and terminal points are fixed at the origin and the point $P$ respectively.

Let $O$ be the origin. Let $P(a_1,a_2)$ and $Q(b_1,b_2)$ be two points in the plane. Join $PQ$. From $P$ and $Q$, draw $PM$ and $QN$ perpendicular to $OX$ and $PM’$ and $QN’$ perpendicular to $OY$.

Then, \[\begin{array}{c}OM=a_1,&MP=a_2\\ON=b_1,&NQ=b_2\end{array}\] Now, projection of $PQ$ on x-axis \[=MN=ON-OM=b_1-a_1\] and, projection of $PQ$ on y-axis \[=M’N’=ON’-OM’=NQ-MP\]\[=b_2-a_2\] \[\therefore\overrightarrow{PQ}=(MN,M’N’)=(b_1-a_1,b_2-a_2)\]

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