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Vector help

Can someone please help me on EF
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Original post by z_o_e
Can someone please help me on EF


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Vectors OA and EF are both the same length and direction due to symmetry of the regular hexagon
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Original post by RDKGames
Vectors OA and EF are both the same length and direction due to symmetry of the regular hexagon


Thank you is the same rule applied to question b
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Original post by z_o_e
Thank you is the same rule applied to question b
ImageUploadedByStudent Room1489325340.708540.jpg


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Unparseable latex formula:

\overrightarrove{EX} = \overrightarrow{EB}+ \overrightarrow{BX}

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Original post by AishaGirl
Unparseable latex formula:

\overrightarrove{EX} = \overrightarrow{EB}+ \overrightarrow{BX}



I got 6b for EB
How about BX?


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EX = EB + 1/2 BC

EB = a multiple of OB

BC = same length as EF but opposite direction....
Original post by z_o_e
I got 6b for EB
How about BX?


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6b is not correct, remember it's a regular hexagon with equal lengths. OB→=6b\overrightarrow{OB}=6b so what will EB→\overrightarrow{EB} be?
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Original post by AishaGirl
6b is not correct, remember it's a regular hexagon with equal lengths. OB→=6b\overrightarrow{OB}=6b so what will EB→\overrightarrow{EB} be?


I got 6a :/


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Original post by z_o_e


EB→\overrightarrow{EB} is twice the length as OB→\overrightarrow{OB} So you see that if OB→=6b\overrightarrow{OB}=6b then can you see what EB→\overrightarrow{EB} would be?

Likewise BX→\overrightarrow{BX} is half the length of BC→\overrightarrow{BC}
(edited 7 years ago)
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Original post by AishaGirl
EB→\overrightarrow{EB} is twice the length as OB→\overrightarrow{OB} So you see that if OB→=6b\overrightarrow{OB}=6b then can you see what EB→\overrightarrow{EB} would be?

Likewise BX→\overrightarrow{BX} is half the length of BC→\overrightarrow{BC}


Omg 12b right?


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Original post by z_o_e
Omg 12b right?


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Yup :smile:

Now BX→\overrightarrow{BX} ?
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Original post by AishaGirl
Yup :smile:

Now BX→\overrightarrow{BX} ?


Is it 6a?


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Original post by z_o_e


No... If BC→=6a\overrightarrow{BC} = 6a what will BX→\overrightarrow{BX} be?

Or as the bear wrote it, what will 12BC→\frac{1}{2}\overrightarrow{BC} be?
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Original post by AishaGirl
No... If BC→=6a\overrightarrow{BC} = 6a what will BX→\overrightarrow{BX} be?

Or as the bear wrote it, what will 12BC→\frac{1}{2}\overrightarrow{BC} be?


Omg I understand now! :smile:
It's 3a :smile:


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Original post by AishaGirl
No... If BC→=6a\overrightarrow{BC} = 6a what will BX→\overrightarrow{BX} be?

Or as the bear wrote it, what will 12BC→\frac{1}{2}\overrightarrow{BC} be?


Can you check this please
ImageUploadedByStudent Room1489330329.310337.jpg


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Original post by z_o_e
Omg I understand now! :smile:
It's 3a :smile:


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Your final answer should be 12b−3a12b-3a
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Original post by AishaGirl
Your final answer should be 12b−3a12b-3a


Why is it minus3a


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Original post by z_o_e
Why is it minus3a


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As the bear said, it's in the opposite direction so the + becomes a -
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Original post by AishaGirl
As the bear said, it's in the opposite direction so the + becomes a -


Oh okay thank you x

Can you check this please
ImageUploadedByStudent Room1489332021.219392.jpg


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Original post by z_o_e
Oh okay thank you x

Can you check this please
ImageUploadedByStudent Room1489332021.219392.jpg


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That's close.

For AC→\overrightarrow{AC} you can try this AB→=b−a\overrightarrow{AB}=b-a and AC→=2AB→\overrightarrow{AC}=2 \overrightarrow{AB}

OD→\underrightarrow{OD} is very wrong. Try to go back to your notes or the text book you're working from and see where you might be going wrong.
(edited 7 years ago)

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