Author 
Topic 

dmmathwiz
Average Member
USA
7 Posts 
Posted  09/06/2007 : 21:37:59

....../\........(periods only used to keep spacing) ...../..\c...... ...d/....\...... .../......\..... ../......./b.... ./_______/...... .....a..........
is where vector 'a' and 'b' meet. is where vector 'b' and 'c' meet.
Find an equation relating the lengths, d,a,b,c, and the angles , and for figure above. Your expression should give d as a function of a, b, c, cos, cos, and cos(+).
HINT: use the vector dot product and vector sum relation.
I need help, i am assuming you set up an addition equation of the vectors to add up to d. Then i guess you need some how get it in the form of the stuff in the question. THANKS 


dmmathwiz
Average Member
USA
7 Posts 
Posted  09/08/2007 : 14:26:01

I came up with
d * cos( + ) = a + bcos + ccos
any ideas? 


sahsjing
Advanced Member
USA
2399 Posts 
Posted  09/08/2007 : 19:06:39

quote: Originally posted by dmmathwiz
....../\........(periods only used to keep spacing) ...../..\c...... ...d/....\...... .../......\..... ../......./b.... ./_______/...... .....a..........
is where vector 'a' and 'b' meet. is where vector 'b' and 'c' meet.
Find an equation relating the lengths, d,a,b,c, and the angles , and for figure above. Your expression should give d as a function of a, b, c, cos, cos, and cos(+).
HINT: use the vector dot product and vector sum relation.
I need help, i am assuming you set up an addition equation of the vectors to add up to d. Then i guess you need some how get it in the form of the stuff in the question. THANKS
Hint: Assume and are interior angles. a, b, c, d are all vectors. d = a + b + c x direction:d_{x} = a  bcos()  ccos(+) y direction:d_{y} = bsin + csin(+) d = (d_{x} + d_{y}) 
Edited by  sahsjing on 09/08/2007 20:28:44 


dmmathwiz
Average Member
USA
7 Posts 
Posted  09/10/2007 : 22:53:35

well thank you for all you help it was very helpful.
my teacher said to take the approach d.d = (a+b+c) . (a+b+c)
all letters are vectors, all "."'s are dot product.
i ended up getting
d = a+b+c+2a.b+2a.c+2b.c
which i then substituted in the definition of dot products. That is just the approach he wanted us to take, not sure how i would have come up with that, but i guess it works. Thanks Again. 



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