ABCDEF is a regular hexagon. Show that vecotr AB+AD+FA+EA=3AB
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ABCDEF is a regular hexagon. Show that vecotr AB+AD+FA+EA=3AB
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To show that vector AB + AD + FA + EA = 3AB in a regular hexagon ABCDEF, we can use vector addition and properties of a regular hexagon.
A regular hexagon has all sides of equal length and all internal angles of equal measure (each angle measuring 120 degrees). Let's break down the given expression step by step:
1. Start by drawing a regular hexagon with labeled vertices: ABCDEF.
```
A _________ B
/\ / \
/ \ / \
/ \ / \
/ \ / \
F _______ G _______ C
\ / \ /
\ / \ /
\ / \ /
\/_______\/
E D
```
2. We are given vector AB + AD + FA + EA.
Expressing each of these vectors in terms of the common vector AB:
AB + AD = AB + AG = AB + (BC × AB)
In vector notation, BC × AB represents the cross product of vectors BC and AB.
FA = FE + EB = -AB + AB = (0)AB
EA = EG + GC = AB + AB = (2)AB
3. Substitute these values back into the original expression:
AB + AD + FA + EA = AB + (BC × AB) + (0)AB + (2)AB
= AB + BC × AB + 0AB + 2AB
= (1 + BC + 0 + 2)AB
= (3)AB
= 3AB
Therefore, we have shown that AB + AD + FA + EA = 3AB in a regular hexagon ABCDEF.