ABCDEF is a regular hexagon whose centre is O | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) In a regular hexagon $ABCDEF$ with centre $O$, we have the following vector relations:$\vec{AB} = \vec{ED}$ and $\vec{AF} = \vec{CD}$.Now, conside

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(D) In a regular hexagon $ABCDEF$ with centre $O$, we have the following vector relations:$\vec{AB} = \vec{ED}$ and $\vec{AF} = \vec{CD}$.Now, consider the sum $\vec{S} = \vec{AB} + \vec{AC} + \vec{AD} + \vec{AE} + \vec{AF}$.Substituting the relations, we get:$\vec{S} = \vec{ED} + \vec{AC} + \vec{AD} + \vec{AE} + \vec{CD}$Rearranging the terms:$\vec{S} = (\vec{AE} + \vec{ED}) + (\vec{AC} + \vec{CD}) + \vec{AD}$Using the triangle law of vector addition, $\vec{AE} + \vec{ED} = \vec{AD}$ and $\vec{AC} + \vec{CD} = \vec{AD}$.Therefore, $\vec{S} = \vec{AD} + \vec{AD} + \vec{AD} = 3 \vec{AD}$.Since $O$ is the centre of the regular hexagon, $\vec{AD} = 2 \vec{AO}$.Thus, $\vec{S} = 3(2 \vec{AO}) = 6 \vec{AO}$.

$ABCDEF$ is a regular hexagon whose centre is $O$. Then, $\vec{AB} + \vec{AC} + \vec{AD} + \vec{AE} + \vec{AF}$ is equal to (in $\vec{AO}$)

Similar Questions

If the vector $a = 2 \hat{i} + 3 \hat{j} + 6 \hat{k}$ and $b$ are collinear and $|b| = 21$, then $b$ is equal to:

Let $u$ and $v$ be two vectors in a plane. Then any vector $w$ in the plane can be written as $w = au + bv$ for some scalars $a$ and $b$ if and only if

If $a + b$ bisects the angle between $a$ and $b$,then $a$ and $b$ are

Let $PQRS$ be a quadrilateral. If $M$ and $N$ are midpoints of the sides $PQ$ and $RS$ respectively,then $\overrightarrow{PS} + \overrightarrow{QR} =$

$ABCDEF$ is a regular hexagon. Find the sum of the vectors $\vec{BE} + \vec{BC} + \vec{EF} + \vec{BA} + \vec{CF} + \vec{AF}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module