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

Rectangular coordinate system

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:

Rectangular coordinate system example

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 \(\left( { - 4,1} \right)\):

Rectangular coordinate system example 2

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:

Specified coordinates right-angled triangle

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 \(\overrightarrow c \):

Specified coordinates three vectors

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:

Specified coordinates vector

Thus,

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

Following a similar approach, we have:

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

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