# The i-j system

Consider the following figure, which shows a point *P* with coordinates \(\left( {3,4} \right)\) on a rectangular coordinate system with origin *O*:

How can we completely specify the vector \(\overrightarrow {OP} \) in a mathematical sense? Note that to reach *P* from *O*, we can first move along the vector \(\overrightarrow {OQ} \), and then along the vector \(\overrightarrow {QP} \). Using the triangle law, we can write the following relation:

\[\overrightarrow {OP} = \overrightarrow {OQ} + \overrightarrow {QP} \]

Now, \(\overrightarrow {OQ} \) is a vector 3 units long, and points in the positive *x*-direction. Let us denote a unit vector in the positive *x*-direction by \(\hat i\). Then, we can write \(\overrightarrow {OQ} \) as \(3\hat i\).

Similarly, we observe that \(\overrightarrow {QP} \) is a vector 4 units long, and points in the positive *y*-direction. If we denote a unit vector in the positive *y*-direction by \(\hat j\), we can write \(\overrightarrow {QP} \) as \(4\hat j\).

These observations are included in the following figure:

Thus, we can write \(\overrightarrow {OP} \) in terms of \(\hat i\) and \(\hat j\) as follows:

\[\overrightarrow {OP} = \overrightarrow {OQ} + \overrightarrow {QP} = 3\hat i + 4\hat j\]

Now, \(\overrightarrow {OP} \) is completely specified in a mathematical sense, since this representation tells us exactly how to reach *P* from *O*: first move 3 units along the positive *x*-direction ( \(3\hat i\)), and then move 4 units parallel to the positive *y*-direction ( \(4\hat j\)).

The following figure shows a point *A* with coordinates \(\left( { - 4,1} \right)\):

In terms of \(\hat i\) and \(\hat j\), we can write the vector \(\overrightarrow {OA} \) as follows:

\[\overrightarrow {OA} = \overrightarrow {OB} + \overrightarrow {BA} = - 4\hat i + \hat j\]

Consider the following figure, which shows a right-angled triangle *ABC* with the specified coordinates:

Note that vector is 5 units long, and points in the positive *x*-direction. Thus, we can write \(\overrightarrow {AB} = 5\hat i\). Vector \(\overrightarrow {BC} \) points in the negative *y*-direction and is 2 units long, so we can write \(\overrightarrow {BC} = - 2\hat j\). Using the triangle law, we have:

\[\overrightarrow {AC} = \overrightarrow {AB} + \overrightarrow {BC} = 4\hat i - 2\hat j\]

**Example 1:** Observe the following figure, which shows three vectors \(\overrightarrow a \), \(\overrightarrow b \) and \(\overrightarrow c \):

Specify these vectors in terms of \(\hat i\) and \(\hat j\).

**Solution:** To move from the starting point of vector \(\overrightarrow a \) to its tip, we need to move 4 units in the negative *x*-direction, and 2 units in the negative *y*-direction:

Thus,

\[\overrightarrow a = - 4\hat i - 2\hat j\]

Following a similar approach, we have:

\[\begin{array}{l}\overrightarrow b = 2\hat i - 5\hat j\\\overrightarrow c = - 3\hat i + 2\hat j\end{array}\]