Vector

ABCDE is a regular pentagon . If $\displaystyle \overrightarrow{{AB}}\ =\overrightarrow{a}$ and $\displaystyle \overrightarrow{{BC}}\ =\overrightarrow{b}$, find $\displaystyle \overrightarrow{{CD}}\ ,\overrightarrow{{DE}}\ \text{and }\overrightarrow{{EA}}$ in terms of $\displaystyle \overrightarrow{a}\ \ \text{and }\overrightarrow{b}$.

Solution

$\displaystyle \overrightarrow{{AB}}=\overrightarrow{a}\ \ ,\ \ \overrightarrow{{BC}}=\overrightarrow{b}$

$\displaystyle \overrightarrow{{FC}}=-\frac{{1+\sqrt{5}}}{2}\overrightarrow{a}+\overrightarrow{b}$

$\displaystyle \overrightarrow{{CD}}=\frac{2}{{1+\sqrt{5}}}\overrightarrow{{FC}}$

$\displaystyle \ \ \ \ \ =-\overrightarrow{a}+\frac{2}{{1+\sqrt{5}}}\overrightarrow{b}$

$\displaystyle \overrightarrow{{GD}}=-\frac{{1+\sqrt{5}}}{2}\overrightarrow{b}+\frac{2}{{1+\sqrt{5}}}\overrightarrow{b}-\overrightarrow{a}$

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

$\displaystyle \overrightarrow{{DE}}=\frac{2}{{1+\sqrt{5}}}\overrightarrow{{GD}}$

$\displaystyle \ \ \ \ \ =-\frac{2}{{1+\sqrt{5}}}(\overrightarrow{a}+\overrightarrow{b})$

$\displaystyle \overrightarrow{{HE}}=\overrightarrow{{HD}}+\overrightarrow{{DE}}$

$\displaystyle \ \ \ \ \ \ =-\frac{{1+\sqrt{5}}}{2}\overrightarrow{{CD}}-\frac{2}{{1+\sqrt{5}}}(\overrightarrow{a}+\overrightarrow{b})$

$\displaystyle \ \ \ \ \ =\frac{{1+\sqrt{5}}}{2}\overrightarrow{a}-\overrightarrow{b}-\frac{2}{{1+\sqrt{5}}}\overrightarrow{a}-\frac{2}{{1+\sqrt{5}}}\overrightarrow{b}$

$\displaystyle \ \ \ \ \ =\overrightarrow{a}-\frac{{3+\sqrt{5}}}{{1+\sqrt{5}}}\overrightarrow{b}$

$\displaystyle \overrightarrow{{EA}}=\frac{2}{{1+\sqrt{5}}}\overrightarrow{{HE}}$

$\displaystyle \ \ \ \ \ =\frac{2}{{1+\sqrt{5}}}\overrightarrow{a}-\overrightarrow{b}$