# Vector Geometry

#### Tarmac27

##### New member
Hello.

I came across this vector geometry question and I haven't made much progess on it.

Question:
The figure shows a regular hexagon ABCDEF. Let a = →AF and b =→AB.

Express →CD in terms of a and b.

My Progress:

→FB = -b+a

All angles are 120 degrees.

Im assuming this answer is simple like →CD = →AF but how do we know this?

Anything helps.

Thanks

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#### skeeter

##### Elite Member
in fact, $$\displaystyle \vec{CD} = \vec{AF}$$ ... because, a translated vector does not change its magnitude or direction

#### Tarmac27

##### New member
in fact, $$\displaystyle \vec{CD} = \vec{AF}$$ ... because, a translated vector does not change its magnitude or direction

Thanks,

But how do we know that CD is the same length as AF?

#### skeeter

##### Elite Member
Thanks,

But how do we know that CD is the same length as AF?
what’s the definition of a regular hexagon?

#### Tarmac27

##### New member
what’s the definition of a regular hexagon?
Ah yes thank you.

Yes, a regular hexagon has all sides equal so →CD = →AF=a. But this does not mean →AB = →AF correct? because of the direction?

Are these the vector equalities?:

→AF = →CD
→AB = →ED
→FE = →BC

#### Tarmac27

##### New member
Also there is a part two I am now having trouble with.

Part Two:

Find →BE

My Progress:

Let →FE = c
therefore, →BC = c

→FE is parallel to →BC

→BE = -b+a+c

This is where I'm up to, not much but I haven't been able to add to this for an hour so I'm not really sure. Any hints would be appreciated

Thanks

#### skeeter

##### Elite Member
Ah yes thank you.

Yes, a regular hexagon has all sides equal so →CD = →AF=a. But this does not mean →AB = →AF correct? because of the direction?
yes, $$\displaystyle \vec{AB} \ne \vec{AF}$$ because they are not in the same direction

Are these the vector equalities?:

→AF = →CD
→AB = →ED
→FE = →BC
yes, note there are three more pairs that are equal (reverse the direction for each vector in the above equations)

#### skeeter

##### Elite Member
Also there is a part two I am now having trouble with.

Part Two:

Find →BE

My Progress:

Let →FE = c
therefore, →BC = c

→FE is parallel to →BC

→BE = -b+a+c

This is where I'm up to, not much but I haven't been able to add to this for an hour so I'm not really sure. Any hints would be appreciated

Thanks
$$\displaystyle \vec{BE} = 2 \cdot \vec{AF}$$

#### Tarmac27

##### New member
Thanks,

Thats obviously simple now but how did you find that equality? Is it another hexagon geometric property because I have tried almost every vector addition possible in the question and none of them relate →AF to →BE.

#### skeeter

##### Elite Member
what kind of triangles are formed by connecting the opposite vertices of the regular hexagon?

recommend you review your geometry before tackling these vector questions

#### Tarmac27

##### New member
what kind of triangles are formed by connecting the opposite vertices of the regular hexagon?

View attachment 25237

recommend you review your geometry before tackling these vector questions

Ah right thanks,

That clears up all my confusion. Do you really need to know these properties off the top of your head? I realise these hexagon ones aren't so complex properties but other shapes may have more complex properties and remembering all of them must surely be difficult?

#### skeeter

##### Elite Member
not that difficult ...