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vectors
hi im stuck on the second part of this question would be great if someone could help
a) write a vector equation in terms of a, b, c and d to express the fact that
AB=DC
the answer to this is
b-a=c-d
b) use this equation to prove that if e is the point such that OAEC is a parallelogram then OBED is a parallelogram
thanks
a) write a vector equation in terms of a, b, c and d to express the fact that
AB=DC
the answer to this is
b-a=c-d
b) use this equation to prove that if e is the point such that OAEC is a parallelogram then OBED is a parallelogram
thanks
OBED is a parallelogram if OB = DE, or BE = OD.
To prove OB = DE
OB = OA + AB
AB = DC (given)
OAEC is a parallelogram (given)
=> OA = CE
therefore,
OB = OA + AB = CE + DC = DC + CE = DE
OB = DE QED
======
Edit: Changed the line OBED is a parallelogram if OB = DE, or BE = OS. to OBED is a parallelogram if OB = DE, or BE = OD.
To prove OB = DE
OB = OA + AB
AB = DC (given)
OAEC is a parallelogram (given)
=> OA = CE
therefore,
OB = OA + AB = CE + DC = DC + CE = DE
OB = DE QED
======
Edit: Changed the line OBED is a parallelogram if OB = DE, or BE = OS. to OBED is a parallelogram if OB = DE, or BE = OD.
Reply 2
Fermat
OBED is a parallelogram if OB = DE, or BE = OS.
To prove OB = DE
OB = OA + AB
AB = DC (given)
OAEC is a parallelogram (given)
=> OA = CE
therefore,
OB = OA + AB = CE + DC = DC + CE = DE
OB = DE QED
======
To prove OB = DE
OB = OA + AB
AB = DC (given)
OAEC is a parallelogram (given)
=> OA = CE
therefore,
OB = OA + AB = CE + DC = DC + CE = DE
OB = DE QED
======
thanks alot
Reply 3
another vectors question im stuck on
ABC is a triangle. D is the midpoint of BC, E is the mid point of AC, F is the mid point of AB and G is the mid point of EF
express the displacement vector AG in terms of a,b and c
thanks alot
ABC is a triangle. D is the midpoint of BC, E is the mid point of AC, F is the mid point of AB and G is the mid point of EF
express the displacement vector AG in terms of a,b and c
thanks alot
AG = AF + FG
AG = ½AB + ½FE = ½AB + ¼BC
AG = ½(b - a) + ¼(c - b)
AG = ¼(2b - 2a + c - b)
AG = ¼(b - 2a + c)
===============
AG = ½AB + ½FE = ½AB + ¼BC
AG = ½(b - a) + ¼(c - b)
AG = ¼(2b - 2a + c - b)
AG = ¼(b - 2a + c)
===============
Reply 5
Fermat
AG = AF + FG
AG = ½AB + ½FE = ½AB + ¼BC
AG = ½(b - a) + ¼(c - b)
AG = ¼(2b - 2a + c - b)
AG = ¼(b - 2a + c)
===============
AG = ½AB + ½FE = ½AB + ¼BC
AG = ½(b - a) + ¼(c - b)
AG = ¼(2b - 2a + c - b)
AG = ¼(b - 2a + c)
===============
how did you know that 1/2 FE was equal to 1/4 BC?
thanks
mmm.. I think I just knew it! It's a sort of standard thing. If F and E are the mid-pts of AB and AC, then FE = ½BC. You can prove by using the cosine rule, for example.
let a = BC. then
a² = b² + c² - 2bc.cosA
let a' = FE, then
FE² = AF² + AE² - 2.AF.AE.cosA
a'² = (½b)² + (½c)² - 2(½b)(½c).cosA
a'² = ¼{b² + c² - 2bc.cosA}
a'² = ¼a²
a' = ½a
FE = ½BC
=======
let a = BC. then
a² = b² + c² - 2bc.cosA
let a' = FE, then
FE² = AF² + AE² - 2.AF.AE.cosA
a'² = (½b)² + (½c)² - 2(½b)(½c).cosA
a'² = ¼{b² + c² - 2bc.cosA}
a'² = ¼a²
a' = ½a
FE = ½BC
=======
Reply 7
Fermat
mmm.. I think I just knew it! It's a sort of standard thing. If F and E are the mid-pts of AB and AC, then FE = ½BC. You can prove by using the cosine rule, for example.
let a = BC. then
a² = b² + c² - 2bc.cosA
let a' = FE, then
FE² = AF² + AE² - 2.AF.AE.cosA
a'² = (½b)² + (½c)² - 2(½b)(½c).cosA
a'² = ¼{b² + c² - 2bc.cosA}
a'² = ¼a²
a' = ½a
FE = ½BC
=======
let a = BC. then
a² = b² + c² - 2bc.cosA
let a' = FE, then
FE² = AF² + AE² - 2.AF.AE.cosA
a'² = (½b)² + (½c)² - 2(½b)(½c).cosA
a'² = ¼{b² + c² - 2bc.cosA}
a'² = ¼a²
a' = ½a
FE = ½BC
=======
thankyou!
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