In a regular hexagon $A B C D E F, \overrightarrow{A B}=\vec{a}$ and $\overrightarrow{B C}=\vec{b}$, then…

In a regular hexagon $A B C D E F, \overrightarrow{A B}=\vec{a}$ and $\overrightarrow{B C}=\vec{b}$, then $\overrightarrow{F A}=$
  1. $\vec{a}-\vec{b}$
  2. $\vec{a}+\vec{b}$
  3. $\vec{b} \quad \vec{a}$
  4. $2 \vec{b}-\vec{a}$

Solution

Since, $\overrightarrow{\mathrm{AC}}=\vec{a}+\vec{b}$
Now, $\overrightarrow{\mathrm{FA}}+\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{FC}}$ $\overrightarrow{\mathrm{FA}}+\vec{a}+\vec{b}=2 \vec{a} \Rightarrow \overrightarrow{\mathrm{FA}}=\vec{a}-\vec{b}$

Asked in: AP EAMCET 2024 (18 May Shift 1)

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