If ABCDEF is a regular hexagon,where two adja | vedclass

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

Question

(A) Given that $ABCDEF$ is a regular hexagon with $\vec{AB} = \vec{a}$ and $\vec{BC} = \vec{b}$.In a regular hexagon,the opposite sides are parallel a

Vedclass · Accepted Answer

$\vec{b}-\vec{a}$

Vedclass · Answer

(A) Given that $ABCDEF$ is a regular hexagon with $\vec{AB} = \vec{a}$ and $\vec{BC} = \vec{b}$.In a regular hexagon,the opposite sides are parallel and equal in magnitude. Thus,$\vec{ED} = \vec{AB} = \vec{a}$ and $\vec{FE} = \vec{BC} = \vec{b}$.Also,the main diagonal $\vec{AD}$ is parallel to $\vec{BC}$ and its magnitude is twice that of $\vec{BC}$. Therefore,$\vec{AD} = 2\vec{BC} = 2\vec{b}$.Using the triangle law of vector addition in $	riangle ABC$,we have:$\vec{AC} = \vec{AB} + \vec{BC} = \vec{a} + \vec{b}$.Now,in $	riangle ACD$,by the triangle law of vector addition:$\vec{AC} + \vec{CD} = \vec{AD}$Substituting the known values:$(\vec{a} + \vec{b}) + \vec{CD} = 2\vec{b}$$\vec{CD} = 2\vec{b} - (\vec{a} + \vec{b})$$\vec{CD} = 2\vec{b} - \vec{a} - \vec{b}$$\vec{CD} = \vec{b} - \vec{a}$

If $ABCDEF$ is a regular hexagon,where two adjacent sides $\vec{AB}$ and $\vec{BC}$ are $\vec{a}$ and $\vec{b}$ respectively. Then $\vec{CD}$ is

Similar Questions

$A$ force is a

If the position vectors of $P$ and $Q$ are $(i + 3j - 7k)$ and $(5i - 2j + 4k)$,then $|\overrightarrow{PQ}|$ is

If $D, E, F$ are the midpoints of the sides $BC, CA$ and $AB$ of the triangle $ABC$,then $\overrightarrow{AD} + \overrightarrow{BE} + \overrightarrow{CF}$ is:

If $a = 2i + j - 8k$ and $b = i + 3j - 4k,$ then the magnitude of $a + b = $

The vectors $3i + j - 5k$ and $ai + bj - 15k$ are collinear if $....$

Vedclass Test Series

Exam Paper Generator

Online Exam Module