Subtracting Two Vectors

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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):

Co-initial vectors

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:

Parallelogram law of vector addition

(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 \(\vec c\):

Triangle law of vector addition

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:

Vector subtraction example 1

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 1200 with the east direction:

Vector with magnitude and 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 \(\vec a\):

Vector subtraction example 2

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 1200 with the east direction, measured in a clockwise manner:

Vector subtraction example 3

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:

Vector subtraction example 4

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:

Vector subtraction example 5

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:

Vector subtraction example 6

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 \(\vec a\):

Vector subtraction example 7

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.

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Vectors
grade 10 | Questions Set 1
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Vectors
grade 10 | Questions Set 1
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grade 10 | Answers Set 2
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