Question

ABCDEF is regular hexagon with $\underset{AB}{\to }=\overline{a}and\underset{BC}{\to }=\overline{b}.Then\underset{CE}{\to }=$

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

Answer 1

The correct option is B $\overline{b}-2\overline{a}$

Given,

ABCDEF is a regular heragon,

therefore

$\begin{array}{c}AB=BC=CD=DE=EF=FA\\ \text{Given,}\\ AB=a,BC=b\end{array}$

Now,

In $△ABC$

$AB+BC=AC$

$\Rightarrow AC=a+b-\left(i\right)$

$\begin{array}{cc}\therefore AD& \text{is parallel to}BC\text{and}AD\\ & AD=2BC\\ \therefore 0.& AD=2b.\end{array}$

$\text{In}△ACD$

$CD=AD-AC.$

$\begin{array}{cc}& =2b-(a+b)\text{(from (ii) and(i))}\\ =& b-a\end{array}$

Now, $In△CDE$

CE $=CD+DF$

$=b-a-a$

$=b-2a$