Examples On Addition And Subtraction Of Vectors

Example – 1

From any two vectors \(\vec a\,\,{\rm{and}}\,\,\vec b\) , prove that

(i) \(\left| {\,\vec a + \vec b\,} \right| \le \left| {\,\vec a\,} \right| + \left| {\,\vec b\,} \right|\)

(ii) \(\left| {\,\vec a - \vec b\,} \right| \le \left| {\,\vec a\,} \right| + \left| {\,\vec b\,} \right|\)

(iii) \(\left| {\,\vec a + \vec b\,} \right| \ge \left| {\left| {\,\vec a\,} \right| - \left| {\,\vec b\,} \right|} \right|\)

When does the equality hold in these cases?

Solution: Consider this figure:

The first two relations follow from the fact that in any triangle, the sum of two sides is greater than the third side:

In \(\Delta \) ABC:                   \(AC \le AB + BC\)(we’ll soon talk about how and when the equality comes)

                  \( \Rightarrow \qquad  \quad \left| {\vec a + \vec b} \right| \le \left| {\vec a} \right| + \left| {\vec b} \right|\)

In \(\Delta \) ABC ':                \(AC' \le AB + BC' = AB + BC\)

                 \( \Rightarrow \qquad \quad \left| {\vec a - \vec b} \right| \le \left| {\vec a} \right| + \left| {\vec b} \right|\)

In the first relation, the equality can hold only if the two vectors have the same direction; this should be intuitively obvious:

The equality in the second relation holds if the two vectors are exactly opposite:

To prove the third relation, we use in \(\Delta \)ABC in Fig - 11, the geometrical fact that the difference of any two sides of a triangle is less than its third side:

\[\qquad\left| {AB - BC} \right| \le AC\]

\[ \Rightarrow  \quad \left| {\left| {\vec a} \right| - \left| {\vec b} \right|} \right| \le \left| {\vec a + \vec b} \right|\]

The equality holds when \(\vec a\,\,{\rm{and}}\,\,\vec b\) are precisely in the opposite direction

The main point to understand from this example is how easily vector relations follows from corresponding geometrical facts.

Example – 2

Suppose that the vectors \(\vec a\,\,{\rm{and}}\,\,\vec b\) represent two adjacent sides of a regular hexagon. Find the vectors representing the other sides.

Solution: Let the hexagon be A₁A₂A₃A₄A₅A₆, as shown:

First of all, we note an important geometrical property of a regular hexagon:

\[\begin{align}&\quad{Diagonal{\rm{ }}\qquad =\qquad {\rm{ }}2{\rm{ }} \times {\rm{ }}side}\\&{ \Rightarrow \quad\;\;  {A_1}{A_4}{\rm{ }}\qquad =\qquad {\rm{ }}2{\rm{ }} \times {A_2}{A_3}}\end{align}\]

Also, since  \({A_1}{A_{4}}||{A_2}{A_3}\), we have

\[\begin{align}\quad\overrightarrow {{A_1}{A_4}}\qquad&\rm{ =\qquad   2} \times \overrightarrow {{A_2}{A_3}} \\&=\qquad 2\,\vec b\end{align}\]

Now we use the triangle law to determine the various sides:

\[\begin{align}\;\;\overrightarrow {{A_3}{A_4}}\quad & =\quad \overrightarrow {{A_1}{A_4}}  - \overrightarrow {{A_1}{A_3}} \\\;\;\;\quad\qquad&=\quad 2\vec b - \left( {\vec a + \vec b} \right)\\\qquad\quad\;\;\;&=\quad\vec b - \vec a\\\;\;\;
\overrightarrow {{A_4}{A_5}}\quad &=\;\;\;\;  - \vec a\left( {only{\rm{ \;}}the{\rm{\; }}sense{\rm{\; }}differs;{\rm{\;}}support{\rm{ \;}}is{\rm{\; }}parallel{\rm{ \;}}to{\rm{\; }}the{\rm{\;}}support{\rm{ \;}}of{\rm{\; }}\vec a{\rm{ }}} \right)\\\;\;\;\overrightarrow {{A_5}{A_6}}\quad  &=\quad  - \,\overrightarrow {{A_2}{A_3}} \\\qquad\quad &= \quad - \vec b\\~~\overrightarrow {{A_6}{A_1}}\quad & = \quad - \,\overrightarrow {{A_3}{A_4}} \\ \qquad~~~&= \quad\vec a - \vec b\end{align}\]

Thus, all sides are expressible in terms of \(\vec a\,\,{\rm{and}}\,\,\vec b\).

Example – 3

What can be interpreted about  \(\vec a\,\,{\rm{and}}\,\,\vec b\) if they satisfy the relation:

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

Solution: Make  \(\vec a\,\,{\rm{and}}\,\,\vec b\)  co-initial so that they form the adjacent sides of a parallelogram:

We have,

\[\left| {\vec a + \vec b} \right| = \left| {\overrightarrow {OC} } \right| = OC\]

\[and\;\left| {\vec a - \vec b} \right| = \left| {\overrightarrow {BA} } \right| = BA\]

Thus, the stated relation implies that the two diagonals of the parallelogram OACB are equal, which can only happen if OACB is a rectangle.

This implies that \(\vec a\,\,{\rm{and}}\,\,\vec b\) form the adjacent sides of a rectangle. In other words, \(\vec a\,\,{\rm{and}}\,\,\vec b\) are perpendicular to each other.