Question

Let $ABCDEF$ be a regular hexagon and let $\overrightarrow{AB}=\overrightarrow{a},\overrightarrow{BC}=\overrightarrow{b}$, then $\overrightarrow{CD}$ is equal to

• A. $\overrightarrow{a}-\overrightarrow{b}$
• B. $\overrightarrow{a}+\overrightarrow{b}$
• C. $\overrightarrow{b}-\overrightarrow{a}$
• D. $2\overrightarrow{a}$

Answer 1

Answer: (C)

$\overrightarrow{b}-\overrightarrow{a}$

Consider figure I,

The arrows show the directions of the vectors of the sides of the triangle.

It has been taken in the clockwise sense.

Now consider the second figure.

Since the direction of $\overrightarrow{a}$ is in clockwise sense, the direction of $\overrightarrow{-a}$ is in anticlockwise sense.

The doted represents the $\overrightarrow{b}$ for triangular law of vector addition, since we need to join the vectors is head to tail manner.

Hence from the figure we get

$\overrightarrow{c}=\overrightarrow{b}-\overrightarrow{a}$.

$\overrightarrow{c}||\overrightarrow{CD}$

Since it is a regular hexagon $AB=BC$

Hence

$a=b$

Thus

$\overrightarrow{CD}=\overrightarrow{c}$.