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Consider the following
If $\vec{a}$ and $\vec{b}$ are the vectors forming consecutive sides of a regular hexagon ABCDEF, then
$1.$ $\overrightarrow{\mathrm{CE}}=\overrightarrow{\mathrm{b}}-2 \overrightarrow{\mathrm{a}}$
$2.$ $\overrightarrow{\mathrm{AE}}=2 \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}$
$3.$ $\quad \overrightarrow{\mathrm{FA}}=\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}$
Which of the above are correct?
Options:
If $\vec{a}$ and $\vec{b}$ are the vectors forming consecutive sides of a regular hexagon ABCDEF, then
$1.$ $\overrightarrow{\mathrm{CE}}=\overrightarrow{\mathrm{b}}-2 \overrightarrow{\mathrm{a}}$
$2.$ $\overrightarrow{\mathrm{AE}}=2 \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}$
$3.$ $\quad \overrightarrow{\mathrm{FA}}=\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}$
Which of the above are correct?
Solution:
2954 Upvotes
Verified Answer
The correct answer is:
1,2 and 3
Let ABCDEF be the regular hexagen as shown in the figure.
Let $\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{b}}$
Join $\mathrm{AD}, \mathrm{FC}$ and $\mathrm{EB}$. They meet at a common point $\mathrm{O}$, which is the centre of hexagon.
$\mathrm{AO} \| \mathrm{BC}$ so, $\overrightarrow{\mathrm{AO}}=\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{b}}$
$\mathrm{OC} \| \mathrm{AB} \mathrm{so}, \overrightarrow{\mathrm{OC}}=\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{a}}$
OAB forms a triangle, $\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BO}}=\overrightarrow{\mathrm{AO}}$
$\Rightarrow \overrightarrow{\mathrm{BO}}=\overrightarrow{\mathrm{AO}}-\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}$
$\mathrm{BO}=\mathrm{OE}$ and they are on the same line,
So, $\overrightarrow{\mathrm{BO}}=\overrightarrow{\mathrm{OE}}=\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}$
$\operatorname{In} \Delta \mathrm{OCE}, \overrightarrow{\mathrm{CO}}+\overrightarrow{\mathrm{OE}}=\overrightarrow{\mathrm{CE}}$
$\Rightarrow \overrightarrow{\mathrm{CE}}=-\overrightarrow{\mathrm{OC}}+\mathrm{OE}=-\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{b}}-\overrightarrow{2 \mathrm{a}}$
So, (1) is correct.
$\overrightarrow{\mathrm{BE}}=2 \overrightarrow{\mathrm{OB}}$ In $\Delta \mathrm{AEB}, \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BE}}=\overrightarrow{\mathrm{AE}}$
$\Rightarrow \overrightarrow{\mathrm{AE}}=\overrightarrow{\mathrm{AB}}+2 \overrightarrow{\mathrm{BO}}=\overrightarrow{\mathrm{A}}+2(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}})$
$\Rightarrow \overrightarrow{\mathrm{AE}}=\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}}-2 \overrightarrow{\mathrm{a}}=2 \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}$
So, $(2)$ is also correct.
$\mathrm{FA} \| \mathrm{OB} \Rightarrow \overrightarrow{\mathrm{FA}}=-\overrightarrow{\mathrm{BO}}=-(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}})=\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}$
So, (3) is also correct.
So, $(1),(2) \&(3)$ are correct.
Let $\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{b}}$
Join $\mathrm{AD}, \mathrm{FC}$ and $\mathrm{EB}$. They meet at a common point $\mathrm{O}$, which is the centre of hexagon.
$\mathrm{AO} \| \mathrm{BC}$ so, $\overrightarrow{\mathrm{AO}}=\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{b}}$
$\mathrm{OC} \| \mathrm{AB} \mathrm{so}, \overrightarrow{\mathrm{OC}}=\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{a}}$
OAB forms a triangle, $\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BO}}=\overrightarrow{\mathrm{AO}}$
$\Rightarrow \overrightarrow{\mathrm{BO}}=\overrightarrow{\mathrm{AO}}-\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}$
$\mathrm{BO}=\mathrm{OE}$ and they are on the same line,
So, $\overrightarrow{\mathrm{BO}}=\overrightarrow{\mathrm{OE}}=\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}$
$\operatorname{In} \Delta \mathrm{OCE}, \overrightarrow{\mathrm{CO}}+\overrightarrow{\mathrm{OE}}=\overrightarrow{\mathrm{CE}}$
$\Rightarrow \overrightarrow{\mathrm{CE}}=-\overrightarrow{\mathrm{OC}}+\mathrm{OE}=-\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{b}}-\overrightarrow{2 \mathrm{a}}$
So, (1) is correct.
$\overrightarrow{\mathrm{BE}}=2 \overrightarrow{\mathrm{OB}}$ In $\Delta \mathrm{AEB}, \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BE}}=\overrightarrow{\mathrm{AE}}$
$\Rightarrow \overrightarrow{\mathrm{AE}}=\overrightarrow{\mathrm{AB}}+2 \overrightarrow{\mathrm{BO}}=\overrightarrow{\mathrm{A}}+2(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}})$
$\Rightarrow \overrightarrow{\mathrm{AE}}=\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}}-2 \overrightarrow{\mathrm{a}}=2 \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}$
So, $(2)$ is also correct.
$\mathrm{FA} \| \mathrm{OB} \Rightarrow \overrightarrow{\mathrm{FA}}=-\overrightarrow{\mathrm{BO}}=-(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}})=\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}$
So, (3) is also correct.
So, $(1),(2) \&(3)$ are correct.
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