Expected Distance of Unit Vectors
Two unit vectors in two-dimensional space
a
^
and
b
^
are added together. The expected magnitude of the resulting vector
a+b
is equal to
E
.
What is
⌊
1
0
0
E
⌋
?
Also try Daniel Liu's Expected Distance on a Circle .
Image credit: Wikipedia P.Fraundorf
The answer is 127.
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3 solutions
Discussions for this problem are now closed
I am not convinced with the answer.
Image
Fix one of the vectors as i ^ (orange). The other vector (blue) which is at an angle θ with the first one can be picked from a "cone" of vectors (the vectors are in 3-D space) with the probability 4 π 2 π sin θ d θ = 2 sin θ d θ . The magnitude is given by: 2 + 2 cos θ . Hence, the expected value is:
∫ 0 π 2 + 2 cos θ 2 sin θ d θ
@Avineil Jain and @Trevor B. , can you please clarify?
I think I needed to specify that the vectors were in 2-D space. Can you try that again and tell me if you get π 4 ? Sorry for the inconvenience.
Thanks for the quick response!
I got π 4 when I assumed the vectors to be in 2-D space. The problem is OK now, I guess. My argument could be wrong too, though.
@Pranav Arora – I don't know anything about 3-D vectors, so can someone confirm that Pranav's integral would be correct if the vectors were in 3-dimensional space and that the answer would be 3 4 ?
i think ur correct
Well I thought that 2 vectors are always co planar, so this problem becomes limited for 2-D.
Yes, two vectors are always coplanar but you cannot say that there is only one vector which is at some angle to the other, you have infinite choices.
@Pranav Arora – But I still think the "cone" logic is wrong. I am still not convinced what is the harm in approaching this problem by 2-D.
I am not sure if this is correct. Sqrt(x^2) = mod(x) and not x.
It is easy to see angle between two vectors changes from 0 to 2pi, but it is the same for two ranges: [0, pi] and [pi, 2pi]. So can use the expression to calculate: 2 ( 2 π 1 ) ∫ 0 π 2 c o s ( 2 x ) = π 4 ⇒ r e s u l t = 1 2 7
Experimental Mathematics!!
>>> s = 0
>>> import math
>>> addvector = lambda x: math.sqrt(1+1+2*1*1*math.cos(x))
>>> for i in xrange(0,628):
... x = float(i)/100
... s += addvectors(x)
...
>>> int(100*s/628)
127
The net resultant of vectors is given by the cosine law.
Let | a + b | = S
S = 1 + 1 + 2 c o s θ
S = 2 c o s 2 θ
Therefore, average or expected value is given by-
S a v g = ∫ 0 π d θ ∫ 0 π 2 c o s 2 θ d θ
The limits are taken from 0 to π since θ can vary from 0 to π .
S a v g = 1 . 2 7 3
[ 1 0 0 S a v g ] = 1 2 7