Suppose that \(\vec a\) and \(\vec b\) are two vectors. How can we interpret the subtraction of these vectors? That is, what meaning do we attach to \(\vec a - \vec b\)?

To start with, we note that \(\vec a - \vec b\)will be a vector which when added to \(\vec b\) should give back \(\vec a\):

\[\left( {\vec a - \vec b} \right) + \vec b = \vec a\]

But how do we determine the vector \(\vec a - \vec b\), given the vectors and \(\vec b\)? The following figure shows vectors \(\vec a\) and \(\vec b\) (we have drawn them to be co-initial):

We need a way to determine the vector \(\vec a - \vec b\).

(i) Using the parallelogram law of vector addition, we can determine the vector as follows. We interpret \(\vec a - \vec b\) as \(\vec a + \left( { - \vec b} \right)\), that is, the vector sum of \(\vec a\) and \( - \vec b\). Now, we reverse vector \(\vec b\), and then add \(\vec a\) and \( - \vec b\) using the parallelogram law:

(ii) We can also use the triangle law of vector addition. Denote the vector drawn from the end-point of \(\vec b\) to the end-point of \(\vec a\) by \(\vec c\):

Note that \(\vec b + \vec c = \vec a\,\). Thus, \(\vec c = \vec a\, - \vec b\). In other words, the vector \(\vec a - \vec b\) is the vector drawn from the tip of \(\vec b\) to the tip of \(\vec a\) (if \(\vec a\) and \(\vec b\) are co-initial).

Note that both ways described above give us the same vector for \(\vec a - \vec b\). This becomes clearer from the figure below:

The vector \(\overrightarrow {PT} \) is obtained by adding \(\vec a\) and \( - \vec b\) using the parallelogram law. The vector \(\overrightarrow {RQ} \) is obtained by drawing the vector from the tip of \(\vec b\) to the tip of \(\vec a\). Clearly, both vectors are the same (they are translated versions of each other).

**Example 1:** Vector \(\vec a\) has a magnitude of 2 units and points towards the west. Vector \(\vec b\) has a magnitude of 2 units and makes an angle of 120^{0} with the east direction:

Find \(\vec a\, - \vec b\).

**Solution:** Make \(\vec a\) and \(\vec b\) co-initial, and draw the vector from the tip of \(\vec b\) to the tip of \(\vec a\):

Clearly, the triangle formed by these three vectors is equilateral. Thus, \(\vec a\, - \vec b\) is a vector of magnitude 2 units, and makes an angle of 120^{0} with the east direction, measured in a clockwise manner:

**Example 2:** A **unit vector** is a vector with unit magnitude. A unit vector is generally denoted by a cap on top of a letter. For example, whenever you encounter symbols like \(\widehat a\), \(\widehat b\), \(\widehat c\) etc., you should interpret these as unit vectors. \(\widehat a\) and \(\widehat b\) are two unit vectors inclined at an angle of \(\theta \) to each other:

Find \(\left| {\widehat a - \widehat b} \right|\).

**Solution:** We need to find the magnitude of \(\widehat a - \widehat b\). Let us make \(\widehat a\) and \(\widehat b\) co-initial, and draw the vector \(\vec c\) from the tip of \(\widehat b\) to the tip of \(\widehat a\), as shown below:

To find the magnitude of \(\vec c\), we use the cosine law:

\[\begin{align}&\left| {\overrightarrow c } \right| = \sqrt {{{\left| {\widehat a} \right|}^2} + {{\left| {\widehat b} \right|}^2} - 2\left| {\widehat a} \right|\left| {\widehat b} \right|\cos \theta } \\\,\,\, &\;\;\;\;\;= \sqrt {1 + 1 - 2\left( 1 \right)\left( 1 \right)\cos \theta } \\\,\,\, &\;\;\;\;\;= \sqrt {2 - 2\cos \theta } = \sqrt {2\left( {1 - \cos \theta } \right)} \\\,\,\, &\;\;\;\;\;= \sqrt {4{{\sin }^2}\frac{\theta }{2}} = 2\sin \frac{\theta }{2}\end{align}\]

**Example 3:** What can you say about non-zero vectors \(\vec a\) and \(\vec b\), if \(\left| {\vec a\, + \vec b} \right| = \left| {\vec a\, - \vec b} \right|\)?

**Solution:** The relation \(\left| {\vec a\, + \vec b} \right| = \left| {\vec a\, - \vec b} \right|\) says that the magnitude of the sum of vectors \(\vec a\) and \(\vec b\) is equal to the magnitude of their difference. Consider the following figure:

Note that in this particular figure, vectors \(\vec a\, + \vec b\) and \(\vec a\, - \vec b\) have unequal lengths. Is there any case possible when the two have equal lengths? A little thinking will show that this is possible *only* when \(\vec a\) and \(\vec b\) are perpendicular. In such a scenario, \(\vec b\) and \( - \vec b\) have a symmetry about \(\vec a\):

Clearly, \(\vec a\, + \vec b\) and \(\vec a\, - \vec b\) have equal lengths in this case. Thus, for two non-zero vectors \(\vec a\) and \(\vec b\), \(\left| {\vec a\, + \vec b} \right| = \left| {\vec a\, - \vec b} \right|\) only if \(\vec a\) and \(\vec b\) are perpendicular.